Answer:
NaN3 = 47.2 g
Explanation:
Given:
2 NaN3 ⇒ 2 Na + 3 N2
Find:
Amount of NaN3
Computation:
N2 moles = Product of N2 / molar mass of N2
N2 moles =30.5/28
N2 moles = 1.0893
2NaN3 makes 3(N2 )
So,
NaN3 moles = (2/3) moles of N2
NaN3 moles = ( 2/3) × 1.0893
NaN3 moles = = 0.7262
NaN3 mass = 0.7262 x 65
NaN3 = 47.2 g
Answer:
NaN3 = 47.2 g
Explanation:
Given:
2 NaN3 ⇒ 2 Na + 3 N2
Find:
Amount of NaN3
Computation:
N2 moles = Product of N2 / molar mass of N2
N2 moles =30.5/28
N2 moles = 1.0893
2NaN3 makes 3(N2 )
So,
NaN3 moles = (2/3) moles of N2
NaN3 moles = ( 2/3) × 1.0893
NaN3 moles = = 0.7262
NaN3 mass = 0.7262 x 65
NaN3 = 47.2 g
Explanation:
Determine whether the following statement about equilibrium is true or false.
(a) When a reaction system reaches a state of equilibrium, the concentration of the products is equal to the concentration of the reactants.
(b) When a system is at equilibrium, Keq = 1.
(c) At equilibrium, the rates of the forward reaction and the reverse reaction are equal.
(d) Adding a catalyst to a reaction system will shift the position of equilibrium to the right so there are more products at equilibrium than if there was no catalyst present.
Answer:
(a) when a reaction system reaches a state of equilibrium, the concentration of the products is equal to the concentration of the reactants
Determining whether the statements about equilibrium is True or False
A) The concentration of the products is equal to the concentration of the reactants at equilibrium : TRUE
B) When a system is at equilibrium, Keq = 1 : TRUE
C) The rates of the forward reaction and the reverse reaction are equal at equilibrium : TRUE
D) Adding a catalyst to a reaction system will shift the position of equilibrium to the right : FALSE
Reaction at equilibriumIn a chemical reaction at equilibrium the value of Keq will be equal to 1 because the concentration of the products is equal to the concentration of the reactants in the chemica reaction. Also at equilibrium the rate of forward reaction is same as the rate of reverse reaction.
A catalyst can only affect the rate of reaction and not the amount of product ( yield of reaction).
Hence we can conclude that the answers to your questions are as listed above.
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A mixture of 50ml of 0.1M HCOOH and 50ml of 0.05M NaOH is equivalent to
Answer:
d) a solution that is 0.025M in HCOOH and 0.025M in HCOONa
Explanation:
The reaction of a weak acid (HOOH) with NaOH is as follows:
HCOOH + NaOH → HCOONa + H₂O
Based on the reaction, 1 mole of the acid reacts with 1 mole of the base (Ratio 1:1).
The initial moles of both species are:
HCOOH: 0.050L × (0.1mol / L) = 0.0050 moles of HCOOH
NaOH: 0.050L × (0.05 mol / L) = 0.0025 moles NaOH
After the reaction, all NaOH reacts with HCOOH producing HCOONa (Because moles of NaOH < moles HCOOH).
Final moles:
HCOOH: 0.0050 moles - 0.0025 moles (After reaction) = 0.0025 moles
HCOONa: Moles HCOONa = Initial Moles NaOH: 0.0025 moles
As volume of the mixture is 100mL (50 from the acid + 50 from NaOH), molarity of both HCOOH and HCOONa is:
0.0025 moles / 0.100L = 0.025M of both HCOOH and HCOONa
Thus, the initial mixture is equivalent to:
d) a solution that is 0.025M in HCOOH and 0.025M in HCOONaDetermine the oxidation state for each of the elements below. The oxidation state of ... silver ... in ... silver oxide Ag2O ... is ... ___ . The oxidation state of sulfur in sulfur dioxide SO2 is ___ . The oxidation state of iron in iron(
Answer:
The oxidation state of silver in [tex]\rm Ag_2O[/tex] is [tex]+1[/tex].
The oxidation state of sulfur in [tex]\rm SO_2[/tex] is [tex]+4[/tex].
Explanation:
The oxidation states of atoms in a compound should add up to zero.
Ag₂OThere are two silver [tex]\rm Ag[/tex] atoms and one oxygen [tex]\rm O[/tex] atom in one formula unit of [tex]\rm Ag_2O[/tex]. Therefore:
[tex]\begin{aligned}&\rm 2 \times \text{Oxidation state of $\rm Ag$}+ \rm 1 \times \text{Oxidation state of $\rm O$} = 0\end{aligned}[/tex].
The oxidation state of oxygen in most compounds (with the exception of peroxides and fluorides) is [tex]-2[/tex]. Silver oxide [tex]\rm Ag_2O[/tex] isn't an exception. Therefore:
[tex]\begin{aligned}&\rm 2 \times \text{Oxidation state of $\rm Ag$}+ \rm 1 \times \text{Oxidation state of $\rm O$} = 0\\ &\rm 2 \times \text{Oxidation state of $\rm Ag$}+ \rm 1 \times (-2) = 0\end{aligned}[/tex].
Solve this equation for the (average) oxidation state of [tex]\rm Ag[/tex]:
[tex]\text{Oxidation state of $\rm Ag$} = 1[/tex].
SO₂Similarly, because there are one sulfur [tex]\rm S[/tex] atom and two oxygen [tex]\rm O[/tex] atoms in each [tex]\rm SO_2[/tex] molecules:
[tex]\begin{aligned}&\rm 1\times \text{Oxidation state of $\rm S$}+ \rm 2 \times \text{Oxidation state of $\rm O$} = 0\end{aligned}[/tex].
The oxidation state of [tex]\rm O[/tex] in [tex]\rm SO_2[/tex] is also [tex]-2[/tex], not an exception, either.
Therefore:
[tex]\begin{aligned}&\rm 1 \times \text{Oxidation state of $\rm S$}+ \rm 2 \times \text{Oxidation state of $\rm O$} = 0\\ &\rm 1 \times \text{Oxidation state of $\rm S$}+ \rm 2 \times (-2) = 0\end{aligned}[/tex].
Solve this equation for the oxidation state of [tex]\rm S[/tex] here:
[tex]\text{Oxidation state of $\rm S$} = 4[/tex].
A chemist fills a reaction vessel with 0.978 g aluminum hydroxide AlOH3 solid, 0.607 M aluminum Al+3 aqueous solution, and 0.396 M hydroxide OH− aqueous solution at a temperature of 25.0°C.
Under these conditions, calculate the reaction free energy ΔG for the following chemical reaction:
Al(OH)3(s) = A1+ (aq) +30H (aq)
Use the thermodynamic information in the ALEKS Data tab. Round your answer to the nearest kilojoule.
KJ
Answer: [tex]\Delta G^{0}[/tex] = 168.12 kJ
Explanation: Gibbs Free Energy, at any time, is defined as the enthalpy of the system minus product of temperature and entropy of the reaction, i.e.:
[tex]\Delta G^{0} = \Delta H^{0} - T.\Delta S^{0}[/tex]
Enthalpy is defined as internal heat existent in the system. It is calculated as:
[tex]\Delta H^{0} = \Sigma H^{0}_{product} - \Sigma H^{0}_{reagent}[/tex]
Using Enthalpy Formation Table:
[tex]\Delta H^{0} = [3*(-299.9)+(-524.7)] - (-1277)[/tex]
[tex]\Delta H^{0} = 62,6 kJ[/tex]
Entropy is the degree of disorder in the system. It is found by:
[tex]\Delta S^{0} = \Sigma S^{0}_{products} - \Sigma S^{0}_{reagents}[/tex]
Calculating:
[tex]\Delta S^{0} = (-321.7) + 3(-10.8) - 0[/tex]
[tex]\Delta S^{0} = -354.1J[/tex]
And so, Gibbs Free energy will be:
[tex]\Delta G^{0} = \Delta H^{0} - T.\Delta S^{0}[/tex]
[tex]\Delta G^{0} = 62600 - [298.(-354.1)][/tex]
[tex]\Delta G^{0} = 168121.8 J[/tex]
Rounding to the nearest kJ:
[tex]\Delta G^{0}[/tex] = 168.12 kJ
Arrange the compounds in order of decreasing magnitude of lattice energy:
a. LiBr
b. KI
c. CaO.
Rank from largest to smallest.
Answer:
The correct answer is CaO > LiBr > KI.
Explanation:
Lattice energy is directly proportional to the charge and is inversely proportional to the size. The compound LiBr comprises Li+ and Br- ions, KI comprises K+ and I- ions, and CaO comprise Ca²⁺ and O²⁻ ions.
With the increase in the charge, there will be an increase in lattice energy. In the given case, the lattice energy of CaO will be the highest due to the presence of +2 and -2 ions. K⁺ ions are larger than Li⁺ ion, and I⁻ ions are larger than Br⁻ ion.
The distance between Li⁺ and Br⁻ ions in LiBr is less in comparison to the distance between K⁺ and I⁻ ions in KI. As a consequence, the lattice energy of LiBr is greater than KI. Therefore, CaO exhibits the largest lattice energy, while KI the smallest.
Arranging the chemical compounds in order of decreasing magnitude of lattice energy, we have:
c. CaO.
a. LiBr
b. KI
Lattice energy can be defined as a measure of the energy required to dissociate one (1) mole of an ionic compound into its constituent anions and cations, in the gaseous state.
Hence, it is typically used to measure the bond strength of ionic compounds.
Generally, lattice energy is inversely proportional to the size of the ions and directly proportional to their electric charges.
Lithium bromide (LiBr) comprises the following ions:
[tex]Li^+[/tex] and [tex]Br^-[/tex]Potassium iodide (KI) comprises the following ions:
[tex]K^+[/tex] and [tex]I^-[/tex]Calcium oxide (CaO) comprises the following ions:
[tex]Ca^{2+}[/tex] and [tex]O^{2-}[/tex]From the above, we can deduce that there is an increase in the charge possessed by the ionic chemical compounds and as such this would result in an increase in the lattice energy.
In order of decreasing magnitude of lattice energy, the chemical compounds are arranged as:
I. CaO.
II. KI.
III. LiBr.
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One way the U.S. Environmental Protection Agency (EPA) tests for chloride contaminants in water is by titrating a sample of silver nitrate solution. Any chloride anions in solution will combine with the silver cations to produce bright white silver chloride precipitate. Suppose an EPA chemist tests a sample of groundwater known to be contaminated with nickel(II) chloride, which would react with silver nitrate solution like this:
Answer:
6.5 mg/L.
Explanation:
Step one: write out and Balance the chemical reaction in the Question above:
NiCl2 + 2AgNO3 =====> 2AgCl + Ni(NO3)2.
Step two: Calculate or determine the number of moles of AgCl.
So, we are given that the mass of AgCl = 3.6 mg = 3.6 × 10^-3 g. Therefore, the number of moles of AgCl can be calculated as below:
Number of moles AgCl = mass/molar mass = 3.6 × 10^-3 g / 143.32. = 2.5118 × 10^-5 moles.
Step three: Calculate or determine the number of moles of NiCl2.
Thus, the number of moles of NiCl2 = 2.5118 × 10^-5/ 2 = 1.2559 × 10^-5 moles.
Step four: detemine the mass of NiCl2.
Therefore, the mass of NiCl2 = number of moles × molar mass = 1.2559 × 10^-5 moles × 129.6 = 1.6 × 10^-3 g.
Step five: finally, determine the concentration of NiCl2.
1000/ 250 × 1.6 × 10^-3 g. = 6.5 mg/L.
Find the density if the volume is 15 mL and the mass is 8.6 g. (5 pts)
Find the volume if the density is 2.6 g/mL and the mass is 9.7 g.(5 pts)
Find the mass if the density is 1.6 g/cm3 and the volume is 4.1 cm3 (5 pts)
Find the density if the initial volume of water is 12.8 mL, the final volume is 24.6 mL and the mass of the object is 4.3 g. Make a drawing to show the water displacement using a graduated cylinder. (gdoc, gdraw)
Answer:
[tex]\large \boxed{\text{0.57 g/mL; 3.7 mL; 6.6 g; 0.366 g/mL}}[/tex]
Explanation:
1. Density from mass and volume
[tex]\text{Density} = \dfrac{\text{mass}}{\text{volume}}\\\\\rho = \dfrac{m}{V}\\\\\rho = \dfrac{\text{8.6 g}}{\text{15 mL}} = \text{0.57 g/mL}\\\text{The density is $\large \boxed{\textbf{0.57 g/mL}}$}[/tex]
2. Volume from density and mass
[tex]V = \text{9.7 g}\times\dfrac{\text{1 mL}}{\text{2.6 g}} = \text{3.7 mL}\\\\\text{The volume is $\large \boxed{\textbf{3.7 mL}}$}[/tex]
3. Mass from density and volume
[tex]\text{Mass} = \text{4.1 cm}^{3} \times \dfrac{\text{1.6 g}}{\text{1 cm}^{3}} = \textbf{6.6 g}\\\\\text{The mass is $\large \boxed{\textbf{6.6 g}}$}[/tex]
4. Density by displacement
Volume of water + object = 24.6 mL
Volume of water = 12.8 mL
Volume of object = 11.8 mL
[tex]\rho = \dfrac{\text{4.3 g}}{\text{11.8 mL}} = \text{0.36 g/mL}\\\text{The density is $\large \boxed{\textbf{0.36 g/mL}}$}[/tex]
Your drawing showing water displacement using a graduated cylinder should resemble the figure below.
Zeros laced at the end of the significant number are...
Answer:
Zeros located at the end of significant figures are significant.
Explanation:
Hope it will help :)
243
Am
95
1. The atomic symbol of americium-243 is shown. Which of the following is correct?
• A. The atomic mass is 243 amu, and the atomic number is 95.
B. The atomic mass is 338 amu, and the atomic number is 95.
• C. The atomic mass is 95 amu, and the atomic number is 243.
D. The atomic mass is 243 amu, and the atomic number is 338.
Answer:
A. The atomic mass is 243 amu, and the atomic number is 95.
When the nuclide bismuth-210 undergoes alpha decay:
The name of the product nuclide is_____.
The symbol for the product nuclide is_____
Fill in the nuclide symbol for the missing particle in the following nuclear equation.
_____ rightarrow 4He+ 234Th
2 90
Write a balanced nuclear equation for the following:
The nuclide radium-226 undergoes alpha emission.
Explanation:
An atom undergoes alpha decay by losing a helium atom.
So when bismuth undergoes alpha decay, we have;
²¹⁰₈₃Bi --> ⁴₂He + X
Mass number;
210 = 4 + x
x = 206
Atomic number;
83 = 2 + x
x = 81
The element is Thallium. The symbol is Ti.
For the second part;
X --> ⁴₂He + ²³⁴₉₀Th
Mass number;
x = 4 + 234 = 238
Atomic Number;
x = 2 + 90 = 92
The balanced nuclear equation is;
²³⁸₉₂U --> ⁴₂He + ²³⁴₉₀Th
2. Which one is the odd one
out and why?
o Water
• Hydrogen
Chlorine
o Aluminum
Answer:
Reaction of Chlorine with Hydrogen Chlorine and Hydrogen mixed together explodes when exposed to sunlight, which produces Hydrogen Chloride. In the dark away from sunlight, no reaction occurs, so light energy is required for a reaction. Cl2 + H2 = 2 HCl Reaction of Chlorine with Non-Metals Chlorine directly combines with most non-metals.
Explanation:
I hope this helps bro
Consider the reaction: C(s) + O2(g)CO2(g) Write the equilibrium constant for this reaction in terms of the equilibrium constants, Ka and Kb, for reactions a and b below: a.) C(s) + 1/2 O2(g) CO(g) Ka b.) CO(g) + 1/2 O2(g) CO2(g) Kb
Answer:
A. Ka = [CO2] / [C] [O2]^1/2
B. Kb = [CO2] / [CO] [O2]^1/2
Explanation:
Equilibrium constant is simply defined as the ratio of the concentration of the products raised to their coefficient to the concentration of the reactants raised to their coefficient.
Now, we shall obtain the expression for the equilibrium constant for the reaction as follow:
A. Determination of the expression for equilibrium constant Ka.
This is illustrated below:
C(s) + 1/2 O2(g) <==> CO(g)
Ka = [CO2] / [C] [O2]^1/2
B. Determination of the expression for equilibrium constant Kb.
This is illustrated below:
CO(g) + 1/2 O2(g) <==> CO2(g)
Kb = [CO2] / [CO] [O2]^1/2
Calculate the concentration of H3O+ in a solution that contains 5.5 × 10-5 M OH- at 25°C. Identify the solution as acidic, basic, or neutral.
Explanation:
To calculate [H3O+] in the solution we must first find the pH from the [ OH-]
That's
pH + pOH = 14
pH = 14 - pOH
To calculate the pOH we use the formula
pOH = - log [OH-]
And [OH-] = 5.5 × 10^-5 M
So we have
pOH = - log 5.5 × 10^ - 5
pOH = 4.26
Since we've found the pOH we can now find the pH
That's
pH = 14 - 4.26
pH = 9.74
Now we can find the concentration of H3O+ in the solution using the formula
pH = - log H3O+
9.74 = - log H3O+
Find the antilog of both sides
H3O+ = 1.8 × 10^ - 10 MThe solution is basic since it's pH lies in the basic region.
Hope this helps you
While balancing a chemical equation, we change the _____ to balance the number of atoms on each side of the equation.
Answer:
While balancing a chemical equation, we change the coefficient to balance the number of atoms on each side of the equation
Explanation:
While balancing a chemical equation, we change the coefficient to balance the number of atoms on each side of the equation.
What is chemical equation?To summarize in chemistry terms, a chemical equation depicts the initial chemicals, or reactants, on the left-hand side and the final compounds, or products, just on right-hand side, divided by an arrow. In the chemical equation, the number of atoms in each element as well as the total charge are the same on opposite of the equation's sides.
Chemical equations are used in chemistry to depict chemical processes by writing the reactants and products in terms of their corresponding chemical formulas. While balancing a chemical equation, we change the coefficient to balance the number of atoms on each side of the equation.
Therefore, while balancing a chemical equation, we change the coefficient to balance the number of atoms on each side of the equation.
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What type of bond would form between two atoms of phosphorus? A. Single covalent bond B. Single ionic bond C. Triple covalent bond D. Double covalent bond
Answer:
A double bond is formed when two pairs of electrons are shared between the two participating atoms. It is represented by two dashes (=). It is represented by two dashes (=). Double covalent bonds are much stronger than a single bond, but they are less stable
Explanation:
g Does a reaction occur when aqueous solutions of barium hydroxide and aluminum sulfate are combined
Answer:
3BaO + Al₂(SO₄)₃ → Al₂O₃+ 3BaSO₄
Explanation:
Yes! A reactiin occurs between barium hydroxide and auminium sulphate.
barium sulfate (BaSO4) and aluminum hydroxide (Al(OH)3) are the products obtained in this reaction.
The reaction is given by the equation below;
3BaO + Al₂(SO₄)₃ → Al₂O₃+ 3BaSO₄
The insoluble salts below are put into 0.10 M hydrochloric acid solution. Do you expect their solubility to be more, less, or about the same as in a pure water solution?
1. Zinc sulfide
2. Silver chloride
3. Lead iodide
4. Silver hydroxide
Answer:
1. Zinc sulfide : about the same solubility, no common ion is found.
2. Silver chloride : less solubility due to the presence of chloride ions provided by the 0.10 M hydrochloric acid.
3. Lead iodide : about the same solubility, no common ion is found.
4. Silver hydroxide : about the same solubility, no common ion is found.
Explanation:
Hello,
In this case, we first must remember that adding a common ion (which is related with the dissolving solid) decreases the solubility of the insoluble solid due to the fact Le Chatelier's principle states the reaction will shift leftwards (reactants) to reestablish equilibrium, therefore, we have:
1. Zinc sulfide : about the same solubility, no common ion is found.
2. Silver chloride : less solubility due to the presence of chloride ions provided by the 0.10 M hydrochloric acid.
3. Lead iodide : about the same solubility, no common ion is found.
4. Silver hydroxide : about the same solubility, no common ion is found.
Best regards.
whts the ph of po4 9.78
Answer:
4.22
Explanation:
We know from the question, that the pOH of the solution is 9.78. Now the pOH is defined as -log [OH^-].
If the pOH of a solution is given, one may obtain the pH of such solution from the formula;
pH + pOH =14
Hence we can write;
pH = 14-pOH
pH = 14 - 9.78 = 4.22
Hence the pH of the solution is 4.22.
The substance formed on addition of water to an aldehyde or ketone is called a hydrate or a/an:_______
A) vicinal diol
B) geminal diol
C) acetal
D) ketal
Answer:
B) geminal diol
Explanation:
Hello,
In this case, considering the attached picture, you can see that the substance resulting from the hydrolysis of an aldehyde or a ketone is a geminal diol since the two hydroxyl groups are in the same carbon. Such hydrolysis could be carried out in either acidic or basic conditions depending upon the equilibrium constant.
Regards.
How does the spontaneity of the process below depend on temperature? PCl5(g)+H2O(g)→POCl3(g)+2HCl(g) ΔH=−126 kJ mol−1, ΔS=146 J K−1mol−
The given question is incomplete, the complete question is:
How does the spontaneity of the process below depend on temperature? PCI5(9)+H2O(g)POCI3(g) +2HCI(g) -126 kJ mol1, AS = 146 J K-'mol1 ΔΗ Select the correct answer below: nonspontaneous at all temperatures spontaneous at all temperatures spontaneous at high temperatures and nonspontaneous at low temperatures spontaneous at low temperatures and nonspontaneous at high temperatures
Answer:
The correct answer is spontaneous at all the temperatures.
Explanation:
Gibbs Free energy is an essential relation that determines the spontaneity of any reaction, that is, ΔG = ΔH - TΔS
When ΔG is less than zero, that is, negative, the reaction is considered to be in spontaneous state. Based on the given information, ΔH = -126 kJ/mol
= -126000 J/mol, it is negative
ΔS = 146 J/K/mol, it is positive
Now, ΔG = ΔH-TΔS
= (-ve) - T (+ve), Thus, when ΔH, is -ve, ΔS is +ve, -TΔS is -ve, the ΔG will be -ve. Therefore, reaction will be spontaneous at all the temperatures.
Using the standard reduction potentials Ni2+(aq) + 2 e‑Ni(s) ‑0.25 volt Fe3+(aq) + e‑Fe2+(aq) +0.77 volt Calculate the value of E°cell for the cell with the following reaction. Ni2+(aq) + 2 Fe2+(aq) →Ni(s) + 2 Fe3+(aq)
Answer:
The correct answer is - 1.02 V
Explanation:
From the reduction-oxidation reaction:
Ni²⁺(aq) + 2 Fe²⁺(aq) → Ni(s) + 2 Fe³⁺(aq)
Ni²⁺ is reduced to Ni(s) while Fe²⁺ is oxidized to Fe³⁺. Thus, the half reactions are:
Reduction (cathode) : Ni²⁺(aq) + 2 e‑ → Ni(s) Eº= ‑0.25 V
Oxidation (anode) : 2 x (Fe²⁺ → Fe³⁺ + e-)(aq) Eº= -0.77 V
-------------------------------------
Ni²⁺(aq) + 2 Fe²⁺(aq) → Ni(s) + 2 Fe³⁺(aq)
In order to calculate the Eºcell, we have to add the reduction potential of the reaction in cathode (reduction) to the oxidation potential of the anode (oxidation):
Eºcell= Eºr + Eºo= (-0.25 V) + (-0.77 V) = - 1.02 V
A 1.0 L buffer solution is 0.250 M HC2H3O2 and 0.050 M LiC2H3O2. Which of the following actions will destroy the buffer?
A. adding 0.050 moles of NaOH
B. adding 0.050 moles of LiC2H3O2
C. adding 0.050 moles of HC2H3O2
D. adding 0.050 moles of HCl
E. None of the above will destroy the buffer.
Answer:
D
Explanation:
Addition of 0.05 M HCl, will react with all of the C2H3O2- from LiAc which will give 0.05 M more HAc. So there will be no Acetate ion left to make the solution buffer. Hence, the correct option for the this question is d, which is adding 0.050 moles of HCl.
The action that destroys the buffer is option c. adding 0.050 moles of HCl.
What is acid buffer?It is a solution of a weak acid and salt.
Here, The buffer will destroy at the time when either HC2H3O2 or NaC2H3O2 should not be present in the solution.
The addition of equal moles of HCl finishly reacts with equal moles of NaC2H3O2. Due to this, there will be only acid in the solution.
Since
moles of HC2H3O2 = 1*0.250 = 0.250
moles of NaC2H3O2 = 1*0.050 = 0.050.
moles of HCl is added = 0.050
Now
The reaction between HCl and NaC2H3O2
[tex]HCl + NaC_2H_3O_2 \rightarrow HC_2H_3O_2 + NaCl[/tex]
Now
BCA table is
NaC2H3O2 HCl HC2H3O2
Before 0.050 0.050 0.250
Change -0.050 -0.050 +0.050
After 0 0 0.300
Now, the solution contains the acid (HC2H3O2 ) only.
Therefore addition of 0.050 moles of HCl will destroy the buffer.
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A piece of solid Fe metal is put into an aqueous solution of Cu(NO3)2. Write the net ionic equation for any single-replacement redox reaction that may be predicted. Assume that the oxidation state of in the resulted solution is 2 . (Use the lowest possible coefficients for the reaction. Use the pull-down boxes to specify states such as (aq) or (s). If a box is not needed, leave it blank. If no reaction occurs, leave all boxes blank and click on Submit.)
Answer:
Fe(s) + Cu^2+(aq) ---> Fe^2+(aq) + Cu(s)
Explanation:
The ionic equation shows the actual reaction that took place. It excludes the spectator ions. Spectator ions are ions that do not really participate in the reaction even though they are present in the system.
For the reaction between iron and copper II nitrate, the molecular reaction equation is;
Fe(s) + Cu(NO3)2(aq)----> Fe(NO3)2(aq) +Cu(s)
Ionically;
Fe(s) + Cu^2+(aq) ---> Fe^2+(aq) + Cu(s)
. You have two solutions, both with a concentration of 0.1M. Solution A contains a weak acid with a pKa of 5. ThepH of solution A is 3. Solution B contains a weak acid with a pKa of 9. The pH of solution B is:
Answer:
pH of solution B is 5
Explanation:
A weak acid, HA, is in equilibrium with water as follows:
HA(aq) + H₂O(l) ⇄ A⁻(aq) + H₃O⁺(aq)
Where Ka (10^-pKa = 1x10⁻⁹) is:
Ka = 1x10⁻⁹ = [A⁻] [H₃O⁺] / [HA]
Where concentrations of this species are equilibrium concentrations
As initial concentration of HA is 0.1M, the equilibrium concentrations of the species are:
[HA] = 0.1M - X
[A⁻] = X
[H₃O⁺] = X
Where X is the amount of HA that reacts until reach the equilibrium, X is reaction coordinate.
Replacing in Ka expression:
1x10⁻⁹ = [A⁻] [H₃O⁺] / [HA]
1x10⁻⁹ = [X] [X] / [0.1 - X]
1x10⁻¹⁰ - 1x10⁻⁹X = X²
1x10⁻¹⁰ - 1x10⁻⁹X - X² = 0
Solving for X:
X = -0.00001 → False solution, there is no negative concentrations.
X = 1x10⁻⁵ → Right solution.
As [H₃O⁺] = X
[H₃O⁺] = 1x10⁻⁵M
And pH = -log[H₃O⁺]
pH = 5
pH of solution B is 5
From the following balanced equation, CH4(g)+2O2(g)⟶CO2(g)+2H2O(g) how many grams of H2O can be formed when 1.25g CH4 are combined with 1.25×10^23 molecules O2? Use 6.022×10^23 mol−1 for Avogadro's number.
Answer:
2.81 g of H2O.
Explanation:
We'll begin by calculating mass of O2 that contains 1.25×10²³ molecules O2.
This can be obtained as follow:
From Avogadro's hypothesis, we understood that 1 mole of any substance contains 6.022×10²³ molecules. This implies that 1 mole of O2 also contains 6.022×10²³ molecules.
1 mole of O2 = 16x2 = 32 g.
Thus 6.022×10²³ molecules is present in 32 g of O2,
Therefore, 1.25×10²³ molecules will be present in =
(1.25×10²³ × 32) / 6.022×10²³ = 6.64 g of O2.
Therefore, 1.25×10²³ molecules present in 6.64 g of O2.
Next, the balanced equation for the reaction. This is given below:
CH4(g) + 2O2(g) —> CO2(g) + 2H2O(g)
Next, we shall determine the masses of CH4 and O2 that reacted and the mass of H2O produced from the balanced equation.
This can be obtained as follow:
Molar mass of CH4 = 12 + (4x1) = 16 g/mol.
Mass of CH4 from the balanced equation = 1 x 16 = 16 g
Molar mass of O2 = 16x2 = 32 g/mol.
Mass of O2 from the balanced equation = 2 x 32 = 64 g
Molar mass of H2O = (2x1) + 16 = 18 g/mol.
Mass of H2O from the balanced equation = 2 x 18 = 36 g
From the balanced equation above,
16 g of CH4 reacted with 64 g of O2 to produce 36 g if H2O.
Next, we shall determine the limiting reactant.
This can be obtained as follow:
From the balanced equation above,
16 g of CH4 reacted with 64 g of O2.
Therefore, 1.25 g of CH4 will react with = (1.25 x 64)/16 = 5 g of O2.
From the above calculations, we can see that only 5 g out of 6.64 g of O2 is needed to react completely with 1.25 g of CH4.
Therefore, CH4 is the limiting reactant.
Finally, we shall determine the mass of H2O produced from the reaction.
In this case, the limiting reactant will be used because it will give the maximum yield of H2O.
The limiting reactant is CH4 and the mass of H2O produced from the reaction can be obtained as follow:
From the balanced equation above,
16 g of CH4 reacted to produce produce 36 g if H2O.
Therefore, 1.25 g of CH4 will react to produce = (1.25 x 36)/16 = 2.81 g of H2O.
Therefore, 2.81 g of H2O were obtained from the reaction.
The mass in grams of H₂O which can be formed when 1.25g CH₄ are combined with 1.25×10²³ molecules O₂ is 2.8 grams.
What is stoichiometry?Stoichiometry of any reaction tells about the amount of species present before and after the completion of the reaction.
Given chemical reaction is:
CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)
Moles of CH₄ will b calculate as:
n = W/M, where
W = given mass = 1.25g
M = molar mass = 16g/mol
n = 1.25/16 = 0.078 moles
Molecues of CH₄ in 0.078 moles = 0.078×6.022×10²³ = 0.46×10²³
Given molecules of O₂ = 1.25×10²³
Required molecules of CH₄ is less as compared to the molecules of O₂, so here CH₄ is the limiting reagent and formation of water is depends on it only.
From the stoichiometry of the reaction it is clear that:
1 mole of CH₄ = will produce 2 moles of H₂O
0.078 moles of CH₄ = will produce 2×0.078=0.156 moles of H₂O
Mass of H₂O will be calculated by using its moles as:
W = (0.156)(18) = 2.8g
Hence required mass of H₂O is 2.8g.
To know more about limiting reagent, visit the below link:
https://brainly.com/question/1163339
Of the following two gases, which would you predict to diffuse more rapidly? PLZZ HELPP PLZ PLZ PLZ
Answer:
CO2 will diffuse more rapidly.
Explanation:
From Graham's law of diffusion, we understood that the rate of diffusion of a gas is inversely proportional to the square root of its density as shown below:
Rate (R) & 1/√Density (d)
R & 1/√d
But, the density of a gas is directly proportional to the relative molecular mass (M) of the gas.
Thus, we can say that the rate of diffusion of a gas is inversely proportional to the square root of the molar mass of the gas. This can be represented mathematically as:
Rate (R) & 1/√Molar mass (M)
R & 1/√M
From the above illustration, we can say that the lighter the gas, the faster the rate of diffusion and the heavier the gas, the slower the rate of diffusion.
Now, to answer the question given above,let us determine the molar mass of Cl2 and CO2.
This is illustrated below:
Molar mass of Cl2 = 2 x 35.5 = 71 g/mol
Molar mass of CO2 = 12 + (2x16) = 12 + 32 = 44 g/mol
Summary
Gas >>>>>> Molar mass
Cl2 >>>>>> 71 g/mol
CO2 >>>>> 44 g/mol
From the illustration above, we can see that CO2 is lighter than Cl2.
Therefore, CO2 will diffuse more rapidly.
Answer: CO2
Explanation:
To calculate changes in concentration for a system not at equilibrium, the first step is to determine the direction the reaction will proceed. To do so, we calculate Q and compare it to the equilibrium concentration, K. We can then determine that a reaction will shift to the right if:__________
Answer:
We can then determine that a reaction will shift to the right if Q<K
Explanation:
Comparing Q with K allows to find out the status and evolution of the system:
If the reaction quotient is equal to the equilibrium constant, Qc = Kc, the system has reached chemical equilibrium. If the reaction quotient is greater than the equilibrium constant, Qc> Kc, the system is not in equilibrium and will evolve spontaneously, decreasing the value of Qc until it equals the equilibrium constant. In this way, the concentrations of the products will decrease and the concentrations of the reagents will increase. In other words, the reverse reaction is favored to achieve equilibrium. Then the system will evolve to the left (ie products will be consumed and more reagents will be formed).If the reaction quotient is less than the equilibrium constant, Qc <Kc, the system is not in equilibrium and will evolve spontaneously increasing the value of Qc until it equals the equilibrium constant. This implies that the concentrations of the products will increase and those of the reagents will decrease. In other words, to achieve balance, direct reaction is favored. Then the reaction will shift to the right, that is, reagents will be consumed and more products will be formed.In this case, we can then determine that a reaction will shift to the right if Q<K
1500 L has how many significants figures
Answer:
It has 2
Explanation:
The significant figures are 1 and 5!
Hope this helps:)
Provide the name(s) for the tertiary alcohol(s) with the chemical formula C6H14O that have a 4-carbon chain. Although stereochemistry may be implied in the question, DO NOT consider stereochemistry in your name. Alcohol #1______ Alcohol #2: ______Alcohol #3______
Answer:
Explanation:
A tertiary alcohol is a compound (an alcohol) in which the carbon atom that has the hydroxyl group (-OH) is also bonded (saturated) to three different carbon atoms.
Based on the question, the only tertiary alcohol that can result from C₆H₁₄O that have a 4-carbon chain is
2-hydroxy-2,3-dimethylbutane
H OH H H
| | | |
H - C - C - C - C - H
| | | |
H CH₃ CH₃ H
From the above, we can see that the carbon atom having the hydroxyl group is also bonded to three other carbon atoms. And since we aren't considering stereochemistry, this is the only tertiary alcohol we can have with a 4-carbon chain
Explain your reasoning. Match each explanation to the appropriate blanks in the sentences on the right.
1. the atomic radius decreases
2. the number of gas molecules decreases
3. molar mass and structure complexity decreases
4. structure complexity decreases
5. molar mass decreases
6. each phase (gas, liquid, solid) becomes more ordered
A (I_2(g), Br_2 (g), Cl_2 (g), F_2 (B): The ranking can best be explained by the trend entropy decreases as______.
B (H_2O_2 (g), H_2S(g), H_2O(g): The ranking can best be explained by the decreases a trend entropy decreases as_______.
C. (C(s, amorphous), C(s, graphite), C(s, diamond): The ranking can best be explained by the trend entropy decreases as_______.
Answer:
A (I_2(g), Br_2 (g), Cl_2 (g), F_2 (B): The ranking can best be explained by the trend entropy decreases as 5. molar mass decreases.
B (H_2O_2 (g), H_2S(g), H_2O(g): The ranking can best be explained by the decreases a trend entropy decreases as 3. molar mass and structure complexity decreases.
C. (C(s, amorphous), C(s, graphite), C(s, diamond): The ranking can best be explained by the trend entropy decreases as 4. structure complexity decreases.
Explanation:
Hello.
In this case, we can understand a higher entropy when more disorder is present and a lower entropy when less disorder is present, thus:
A (I_2(g), Br_2 (g), Cl_2 (g), F_2 (B): The ranking can best be explained by the trend entropy decreases as 5. molar mass decreases since iodine has the greatest molar mass (254 g/mol) and fluorine the least molar mass (38 g/mol).
B (H_2O_2 (g), H_2S(g), H_2O(g): The ranking can best be explained by the decreases a trend entropy decreases as 3. molar mass and structure complexity decreases since hydrogen peroxide weights 34 g/mol as well as hydrogen sulfide but the peroxide has more bonds (more complex, higher entropy).
C. (C(s, amorphous), C(s, graphite), C(s, diamond): The ranking can best be explained by the trend entropy decreases as 4. structure complexity decreases since diamond has a well-ordered structure and amorphous carbon has a very disordered one.
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