Answer: The concentration of fluoride ions after the addition of 9.00 mL of 0.0100 M HCl to 25.0 mL of this solution is 0.0709 M.
Explanation:
Given: Concentration of hydrogen fluoride = 0.126 M
Concentration of fluoride ions = 0.1 M
Volume of HCl = 9.0 mL
Concentration of HCl = 0.01 M
Volume of HCl = 25.0 mL
Moles of [tex]F^{-}[/tex] ions are calculated as follows.
[tex]Moles of F^{-} = molarity \times volume\\= 0.1 M \times 0.025 L\\= 0.0025 mol[/tex]
Moles of HF are as follows.
[tex]Moles of HF = Molarity \times Volume\\= 0.126 M \times 0.025 L\\= 0.00315 mol[/tex]
Moles of HCl are as follows.
[tex]Moles of HCl = Molarity \times volume\\= 0.01 M \times 0.009 L\\= 0.00009 mol[/tex]
Now, reaction equation with initial and final moles will be as follows.
[tex]H^{+} + F^{-} \rightarrow HF[/tex]
Initial: 0.00009 0.0025 0.00315
Equilibrium: (0.0025 - 0.00009) (0.00315 + 0.00009)
= 0.00241 = 0.00324
Total volume = (9.00 mL + 25.0 mL) = 34.0 mL = 0.034 L
Hence, concentration of fluoride ions is calculated as follows.
[tex]Concentration = \frac{moles}{volume}\\= \frac{0.00241 mol}{0.034 L}\\= 0.0709 M[/tex]
Thus, we can conclude that concentration of fluoride ions after the addition of 9.00 mL of 0.0100 M HCl to 25.0 mL of this solution is 0.0709 M.
For the following list of acids, rank the acids in strength from weakest acid to strongest acid.
a. FCH2OH
b. F2CHOH
c. CH3OH
d. F3COH
Answer:
CH3OH < FCH2OH < F2CHOH < F3COH
Explanation:
Let us recall that, for a carboxylic acid, the dissociation of the acid yields;
RCOOH ⇄RCOO^- + H^+
The ease of dissociation and release of the hydrogen ion depends on the nature of the group designated R.
When R is is a highly electronegative element, the -I inductive effect causes the hydrogen to become less tightly held by the C-Cl bond.
As the number of electron withdrawing substituents increaseses, the acid ionizes much more and becomes stronger.CH3OH < FCH2OH < F2CHOH < F3COH
Hence, the order of decreasing acid strength is;
When 52 grams of O2 react with excess C3H8, how many grams of CO2 would be produced?
Answer:35.1g of CO2
Explanation:
C3H8+5O2------=3CO2+4H2O
Classify the processes as endothermic or exothermic.
a. Ice melting
b. Water condensing on surface
c. Baking a cake
d. The chemical reaction inside an instant cold pack.
e. A car using gasoline
endothermic absorbs heat
exothermic gives heat
a. endothermic
b. exothermic
c. endothermic
d. exothermic
a. Ice melting - endothermic
b. Water condensing on the surface - exothermic
c. Baking a cake - endothermic
d. The chemical reaction inside an instant cold pack - endothermic
e. A car using gasoline - exothermic
What is an exothermic and endothermic reaction?An exothermic reaction can be described as a thermodynamic chemical reaction that emits energy from the system to its surroundings usually in the form of light, heat, or sound.
While an endothermic reaction can be described as an opposite of an exothermic reaction where the energy gains in the form of heat. In exothermic chemical reactions, the bond energy is transformed into thermal energy.
In exothermic reactions, the reaction happens the form of the kinetic energy of molecules when the energy is released. The release of energy is due to the electronic transition of electrons from one energy level to another.
The burning of gasoline, and water condensation is also an exothermic reaction in which energy is released while ice melting and baking cake is an endothermic reaction.
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Na2CO3 + CaCl2•2H2O -> CaCO3 + 2NaCl + 2H2O
Calculate how many moles of CaCl2•2H2O are present in 1.50 g of CaCl2•2H2O and then calculate how many moles of pure CaCl2 are present in 1.50 g of CaCl2•2H2O. Then determine how many grams of Na2CO3 are necessary to reach stoichiometric quantities.
For CaCl2 I got 0.0135 mol but I have seen some put 0.0102 mol. Which is it?
For the initial mol of Na2CO3 I got 0.0102 mol but again I’m not sure if I’m right.
For the grams of Na2CO3 I got 1.08 g
Can someone help me figure out if I have this correct?
Answer:
See explanation
Explanation:
Number of moles = reacting mass/molar mass
Number of moles of CaCl2•2H2O = 1.50 g/147.02 = 0.0102 moles
From the equation;
Na2CO3 + CaCl2•2H2O -> CaCO3 + 2NaCl + 2H2O
We can see is 1:1
1 mole of Na2CO3 reacts with 1 mole of CaCl2•2H2O
x moles of Na2CO3 reacts with 0.0102 moles of CaCl2•2H2O
x = 1 × 0.0102 moles/1
x = 0.0102 moles of Na2CO3
Mass of Na2CO3 = 0.0102 moles of Na2CO3 × 106g/mol = 1.08 g of Na2CO3
To determine the concentration of an EDTA solution, 4.11 g of Zn metal was used. The volume of EDTA solution needed to reach the endpoint was 28.26 mL. What was the concentration (in molarity) of the EDTA solution?
Answer:
2.23M
Explanation:
Molarity of a solution is calculated thus
Molarity = number of moles (n) ÷ volume (V)
According to this question, 4.11g of Zn metal was used in order to reach a volume of EDTA solution of 28.26 mL.
28.26mL = 28.26/1000
= 0.02826L
Using mole = mass/molar mass to calculate no. of moles of Zn
Mole = 4.11/65.4
mole = 0.0628mol
Molarity = 0.0628 ÷ 0.02826
Molarity = 2.23M
The concentration of the EDTA solution used is 2.23M
Elements that have the same number of electron rings are ?
Answer:
are in the same orbital
Explanation:
Answer:
are in the same orbit
Explanation:
The homework question reads:
"A sample of gas in a cylinder of volume 3.42 L at 298 K
and 2.57 atm expands to 7.39 L by two different pathways.
Path A is an isothermal, reversible expansion. Path B has two
steps. In the fi rst step, the gas is cooled at constant volume to
1.19 atm. In the second step, the gas is heated and allowed to
expand against a constant external pressure of 1.19 atm until
the final volume is 7.39 L. Calculate the work for each path.
Answer:
Explanation:
this guy on brainly already did it:
Alleei
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4.8K answers
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Answer : The work done for path A and path B is -685.3 J and -478.1 J respectively.
Explanation :
To calculate the work done for path A :
First we have to calculate the moles of the gas.
where,
= initial pressure of gas = 2.57 atm
= initial volume of gas = 3.42 L
n = moles of gas = ?
R = gas constant = 0.0821 atm.L/mol.K
T = temperature of gas = 298 K
Now put all the given values in the above formula, we get:
According to the question, this is the case of isothermal reversible expansion of gas.
As per first law of thermodynamic,
where,
= internal energy
q = heat
w = work done
As we know that, the term internal energy is the depend on the temperature and the process is isothermal that means at constant temperature.
So, at constant temperature the internal energy is equal to zero.
The expression used for work done will be,
where,
w = work done on the system = ?
n = number of moles of gas = 0.359 mole
R = gas constant = 8.314 J/mole K
T = temperature of gas = 298 K
= initial volume of gas = 3.42 L
= final volume of gas = 7.39 L
Now put all the given values in the above formula, we get :
Thus, the work done of path A is, -685.3 J
To calculate the work done for path B :
The formula used for isothermally irreversible expansion is :
where,
w = work done
= external pressure = 1.19 atm
= initial volume of gas = 3.42 L
= final volume of gas = 7.39 L
Now put all the given values in the above formula, we get :
Thus, the work done of path B is, -478.1 J
please help me with b and c.
Answer:
c.sf4 b.2
Explanation:
A compound, C20H28O, produces a 1H NMR spectrum with 11 distinct signals. The steps made by the integral trace measure 52, 17, 17, 26, 17, 25, 26, 9, 9, 35, and 8 mm. Complete the following table.
Integral # Products
52 mm
17 mm
17 mm
26 mm
17 mm
25 mm
26 mm
35 mm
8 mm
Solution :
The smallest integer value represents the smaller number of protons.
In this case, in the given values, the smallest numbers are 8 mm and 9 mm, so both contains 1H each. Then next highest value is 17 mm, which contains 1H more. Thus 17 mm contains 2H each. Then the next highest is 25 mm and 26 mm which contains 3M each and so on.
Thus the tables is :
Integral Protons
52 mm [tex]6[/tex]
17 mm [tex]2[/tex]
17 mm 2
26 mm 3
17 mm 2
25 mm 3
26 mm 3
9 mm 1
9 mm 1
35 mm 4
8 mm 1
Classify each of the following chemical reaction as a synthesis, decomposition, single-displacement, or double-displacement reaction. Drag the appropriate items to their respective bins.
CH3Br → CH3(g) + Br(g)
Zn(s) + CoCl2(aq) → ZnCl2(aq)
First Reaction is a Decomposition reaction as a single reactant hets decompoesed to form two products.Second reaction is a Synthesis reaction as two Reactant reacts together to form one product.
What is Decomposition Reaction ?Decomposition reactions are processes in which chemical species break up into simpler parts.
Usually, decomposition reactions require energy input.
Hence, First Reaction is a Decomposition reaction as a single reactant hets decompoesed to form two products.Second reaction is a Synthesis reaction as two Reactant reacts together to form one product.
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Which functional group is used in other functional groups?
A. Ester
B. Carbonyl
c. Hydroxyl
D. Amino
q9
Answer:
The answer is B. Carbonyl
Carbonyl is the functional group is used in other functional groups. Therefore, option (B) is correct.
What do you mean by carbonyl functional group?A functional group with the formula C=O that is composed of a carbon atom double-bonded to an oxygen atom and is divalent at the C atom is known as a carbonyl group in organic chemistry.
A carbonyl gathering is a synthetically natural utilitarian gathering made out of a carbon iota twofold clung to an oxygen molecule - - > [C=O] The most straightforward carbonyl gatherings are aldehydes and ketones typically connected to another carbon compound.
A functional group with a carbon double bonded to an oxygen is called a carbonyl group. They have unsurprising properties, like extremity, instability, and reactivity.
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If 52.5 mL of lead(II) nitrate solution reacts completely with excess sodium iodide solution to yield 0.248 g of precipitate, what is the molarity of lead(II) ion in the original solution
Explanation:
The volume of given lead nitrate solution is:
52.5 mL.
The amount of lead iodide formed is ---0.248 g.
To get the molarity of lead (II) ion follow the below-shown procedure:
The number of moles of lead iodide formed is:
[tex]number of moles of lead iodide(n)=mass of lead iodide/its molecular mass\\n=0.248 g/461.01g/mol\\n=0.000537mol[/tex]
0.000537 mole of lead iodide contains --- 0.000537 moles of lead (II) ion.
Thus, the number of moles are there, volume is there, and to get the molarity of lead (II) ion use the formula:
[tex]Molarity=\frac{number of moles}{volume in L.} \\M=0.000537 mol / 0.0525 L.\\M=0.0102mol/L[/tex]
Molarity of lead iodide is --- 0.0102 M.
Compute the equilibrium constant for the spontaneous reaction between Cd2 (aq) and Zn(s).
Answer:
Kc = [Zn²⁺] / [Cd²⁺]
Explanation:
Let's consider the spontaneous redox reaction between Cd²⁺ and Zn.
Cd²⁺(aq) + Zn(s) ⇄ Cd(s) + Zn²⁺(aq)
The equilibrium constant, Kc, is the ratio of the equilibrium concentrations of products over the equilibrium concentrations of reactants each raised to the power of their stoichiometric coefficients. It only includes gases and aqueous species.
Kc = [Zn²⁺] / [Cd²⁺]
Apakah ciri-ciri ikan
Answer:
fishes are cold blooded
A sample of pure tin metal is dissolved in nitric acid to produce 15.00 mL of solution containing Sn2+. When this tin solution is titrated, a total of 42.1 mL of 0.145 mol/L KMnO4 is required to reach the equivalence point. a. What is the concentration of the Sn2+ solution?b. Find the concentration of the Sn2+(aq) in mol/L: (give your answer to 3 decimal places)
Answer:
1.00 M
Explanation:
Sn^2+ reacts with KMNO4 as follows;
5Sn^2+(aq) + 2MnO4^-(aq) + 16H^+(aq) ----> 5Sn^4+(aq) + 2Mn^+(aq) + 8H2O(l)
The number of moles of MnO4^- reacted = 42.1/1000 L × 0.145 mol/L
= 0.0061 moles
If 5 moles of Sn^2+ reacts with 2 moles of MnO4^-
x moles of Sn^2+ reacts with 0.0061 moles of MnO4^-
x= 5 × 0.0061/2
x= 0.015 moles
Since the volume of the Sn^2+ solution is 15.00mL or 0.015 L
number of moles = concentration × volume
Concentration = number of moles/volume
Concentration= 0.015 moles/0.015 L
Concentration = 1 M
Question 16 of 25
What is the energy of a photon emitted with a wavelength of 654 nm?
A. 1.01 * 10-27 J
B. 4.33 x 10-22
C. 1.30 x 10-22 j
D. 3.04 10-19 J
SUBMIT
Answer:
D. 3.04 × 10^-19 J
Explanation:
c = lambda * nunu = c/lambda
E = h* nu
E = h * c/lambda = 6.626x10^-34 J-sec * 3x10^8 m/sec / 6.54x10^-7 m
E = 3.04x10^-19 J
Write the structure of methanamine
Answer: CH3NH2
Explanation:
Each set of quantum numbers to the correct sub shell description
42 Organic compound may have names ending in -ane, -ene, -ol or -oic acid. How many of these endings indicate the compounds contain double bonds in their molecules? * (1 Point)
Answer: Organic compounds ending with the name (-ene) indicate that the compounds contain double bonds in their molecules.
Explanation:
Organic compounds are those molecules that contains carbon atoms (as their main element), hydrogen and oxygen which are usually present. The presence of numerous organic compounds is due to the following properties of carbon:
--> the exceptional ability of carbon atoms to catenate, that is, to combine with one another to form straight chains, branched chains or ring compounds containing many carbon atoms.
--> The ease with which carbon combines with hydrogen, oxygen, Nitrogen and halogens
--> The ability of carbon atoms to form single, DOUBLE or triple bonds.
The organic compound that has the name ending with -ene are known as the alkenes. The members of the alkene series are formed from the alkanes by the removal of two hydrogen atoms and the introduction of a DOUBLE BOND in the carbon chain. They are named after the corresponding alkanes by changing the -ane ending to -ene.
Note: the systematic name of a compound is formed from the root hydrocarbon by adding a suffix and prefixes to denote the substitution of the hydrogen atoms.
It is found that, when a dilute gas expands quasistatically from 0.40 to 5.0 L, it does 210 J of work. Assuming that the gas temperature remains constant at 300 K, how many moles of gas are present
Answer:
[tex]n=0.033mole[/tex]
Explanation:
From the question we are told that:
Initial volume [tex]V_1=0.40L[/tex]
Final Volume[tex]V_2=5.0L[/tex]
Work [tex]W=210J[/tex]
Temperature [tex]T=300k[/tex]
Generally the equation for Ideal gas is mathematically given by
[tex]W=nRTIn\frac{V_2}{V_1}[/tex]
[tex]n=\frac{W}{RTIn\frac{V_2}{V_1}}[/tex]
[tex]n=\frac{210}{8.32*300In\frac{5.0}{0.4}}[/tex]
[tex]n=0.033mole[/tex]
any two functions of crystals
Answer:
1. Participating in calcium homeostatis storage of calcium.
2. High capacity calcium (Ca) regulation and protection against herbivory
[tex]\large \boxed{\sf 2 \: functions \: of \: crystals \: are :- } [/tex]
_________________
⟹
[tex] \sf \: \underline{ Calcium \: oxalate \: (CaOx) \: crystals} \: are \: distributed \: \\\sf among \: all \: taxonomic \: levels \\ \sf\: of \: photosynthetic \: organisms \: from \\ \sf \: small \: algae \: to \: angiosperms \: and \: giant \: gymnosperms .[/tex]
__________________
⟹
[tex]\sf Bone \: is \: mostly \: made \: of \: \underline{mineral \: crystals} \: \\ \sf and \: the \: protein \: collagen. \: The \: mineral \: crystals \: bone \\ \sf\: provide \: strength \: and \: rigidity \: for \: the \: matrix \: upon \: \\ \sf \: and \: within \: which \: they \: are \: deposited.[/tex]
how many of the electrons in a molecule of ethane are not involved in bondind
Ethane consists of 6C−H bonds and 1C−C bond. Total number of bonds is 7. Each bond is made up of two electrons
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The decomposition of ethyl amine, C2H5NH2, occurs according to the reaction: C2H5NH2(g)⟶C2H4(g)+NH3(g) At 85∘C, the rate constant for the reaction is 2.5 x 10-1 s-1. What is the half-life (in sec) of this reaction?
Answer:
2.772 seconds
Explanation:
Given that;
t1/2 = 0.693/k
Where;
t1/2 = half life of the reaction
k= rate constant
Note that decomposition is a first order reaction since the rate of reaction depends on the concentration of one reactant
t1/2 = 0.693/2.5 x 10-1 s-1
t1/2= 2.772 seconds
A reaction at 5.0°C evolves 583.mmol of dinitrogen difluoride gas. Calculate the volume of dinitrogen difluoride gas that is collected. You can assume the pressure in the room is exactly 1atm . Round your answer to 3 significant digits.
Answer: The volume of dinitrogen difluoride gas collected is 13.31 L.
Explanation:
Given: Temperature = [tex]5.0^{o}C[/tex] = (5 + 273) K = 278 K
Moles = 583 mmol (1 mmol = 0.001 mol) = 0.583 mol
Pressure = 1 atm
Formula used to calculate volume is as follows.
PV = nRT
where,
P = pressure
V = volume
n = no. of moles
R = gas constant = 0.0821 L atm/mol K
T = temperature
Substitute the values into above formula as follows.
[tex]PV = nRT\\1 atm \times V = 0.583 \times 0.0821 L atm/mol K \times 278 K\\V = 13.31 L[/tex]
Thus, we can conclude that the volume of dinitrogen difluoride gas collected is 13.31 L.
How many grams of P4O10 (292.88 g/mol) form when phelpsphorous (P4, 125.52 g/mol) reacts with 16.2 L of O2 (33.472 g/mol) ) at standard temperature and pressure
Answer:
40.5 g of P₄O₁₀ are produced
Explanation:
We state the reaction:
P₄ + 5O₂ → P₄O₁₀
We do not have data from P₄ so we assume, it's the excess reactant.
We need to determine mass of oxygen and we only have volumne so we need to apply density.
Density = mass / volume, so Mass = density . volume
Denstiy of oxygen at STP is: 1.429 g/L
1.429 g/L . 16.2L = 23.15 g
We determine the moles: 23.15 g . 1mol / 33.472g = 0.692 moles
5 moles of O₂ can produce 1 mol of P₄O₁₀
Our 0.692 moles may produce (0.692 . 1)/ 5 = 0.138 moles
We determine the mass of product:
0.138 mol . 292.88 g/mol = 40.5 g
5.60g of glyceraldehydes was dissolved in 10ml of a solvent and placed in a 50mm cell if the rotation is 1.74 calculate the specific rotation?
Answer:
6.214 degrees-mL/gdm
Explanation:
The specific rotation α' = α/LC where α = observed rotation, L = length of tube and C = concentration of solution.
Given that α = 1.74, L = length of cell = 50 mm = 0.50 dm and C = m/V where m = mass of glyceraldehyde = 5.60 g and V = volume = 10 ml
So, C = m/V = 5.60 g/10 ml = 0.560 g/ml
Since α' = α/LC
substituting the values of the variables into the equation, we have
α' = α/LC
α' = 1.74/(0.50 dm × 0.560 g/ml)
α' = 1.74/(0.28 gdm/l)
α' = 0.006214 °mL/gdm
α' = 6.214 °mL/gdm
α' = 6.214 degrees-mL/gdm
How many moles in 3.30g of iron
The answer below is correct but to give you the process, here it is:
Molar mass of iron, Fe = 55.85 g/mol
3.30g/(55.85 g/mol) = 0.0591 mol
Some organic solvents do not work well in liquid-liquid aqueous extractions. Ethanol (HOCH2CH3) is a common inexpensive solvent, but is a poor solvent for extractions. In ten or fewer words, provide an explanation for why ethanol is a poor solvent selection for extraction.
Answer:
See explanation
Explanation:
Extraction has to do with the separation of the components of a mixture by dissolving the mixture in a set up involving two phases. One phase is the aqueous phase (beneath) while the other is the organic phase (on top). The solvents used for the two phases must not be miscible. Water commonly is used for the aqueous phase.
Ethanol is an important solvent in chemistry but the solvent is miscible with water in all proportions. As a result of this, ethanol is a poor solvent for carrying out extraction.
Solid aluminum (AI) and oxygen (0) gas react to form solid aluminum oxide (AIO). Suppose you have 7.0 mol of Al and 13.0 mol of o, in a reactor. Suppose as much as possible of the Al reacts. How much will be left? Round your answer to the nearest 0.1 mol mol 0.
Answer:
[tex]n_{O_2}^{leftover}=7.7mol[/tex]
Explanation:
Hello there!
In this case, according to the given information, it turns out firstly necessary for us to set up the corresponding chemical equation:
[tex]4Al+3O_2\rightarrow 2Al_2O_3[/tex]
In such a way, we calculate the moles of aluminum consumed by 13.0 moles of oxygen in the reaction, by applying the 4:3 mole ratio between them:
[tex]n_{Al}=13.0molO_2*\frac{4molAl}{3molO_2} =17.3molAl[/tex]
This means that Al is actually the limiting reactant and oxygen is in excess, for that reason we calculate the moles of oxygen consumed by 7.0 moles of aluminum:
[tex]n_{O_2}=7.0molAl*\frac{3molO_2}{4molAl} =5.3molO_2[/tex]
Thus, the leftover of oxygen is:
[tex]n_{O_2}^{leftover}=13.0mol-5.3mol\\\\n_{O_2}^{leftover}=7.7mol[/tex]
Whereas all the aluminum is assumed to be consumed.
Regards!
Please explain how to do it as well!
Write a complete, balanced equation for the following reactions:
a) The combustion of C₆H₁₂O (teachers note: You figure out products).
b) Aqueous ferric iron (III) sulfate plus barium hydroxide (teachers note: You figure out the products).
Answer:
a) C₆H₁₂O + 8.5 O₂ ⇒ 6 CO₂ + 6 H₂O
b) Fe₂(SO₄)₃ + 3 Ba(OH)₂ ⇒ 3 BaSO₄ + 2 Fe(OH)₃
Explanation:
a) A combustion is a reaction of a compound with oxygen to produce carbon dioxide and water. The corresponding equation is:
C₆H₁₂O + O₂ ⇒ CO₂ + H₂O
We will start balancing C atoms by multiplying CO₂ by 6 and H atoms by multiplying H₂O by 6.
C₆H₁₂O + O₂ ⇒ 6 CO₂ + 6 H₂O
Then, we get the balanced equation by multiplying O₂ by 8.5.
C₆H₁₂O + 8.5 O₂ ⇒ 6 CO₂ + 6 H₂O
b) This is a double displacement reaction of the general structure:
Salt 1 + Base 1 = Salt 2 + Base 2
The corresponding equation is:
Fe₂(SO₄)₃ + Ba(OH)₂ ⇒ BaSO₄ + Fe(OH)₃
First, we will balance Fe atoms by multiplying Fe(OH)₃ by 2 and S atoms by multiplying BaSO₄ by 3.
Fe₂(SO₄)₃ + Ba(OH)₂ ⇒ 3 BaSO₄ + 2 Fe(OH)₃
Then, we will get the balanced equation by multiplying Ba(OH)₂ by 3.
Fe₂(SO₄)₃ + 3 Ba(OH)₂ ⇒ 3 BaSO₄ + 2 Fe(OH)₃
Answer:
a.[tex]\bold{C_{6}H_{12}O_{6}\rightarrow 6CO_{2}+6H_{2}O}[/tex]
△H=−72 kcal
The energy required for production of 1.6 g of glucose is [molecular mass of glucose is 180 gm]
b.
[tex]\bold{Fe_{2}(SO_{4})_{3}+3Ba(OH)_{2}\rightarrow 3BaSO_{4}+2Fe(OH)_{3}}[/tex]
The iron(III) ions and chloride ions remain aqueous and are spectator ions in a reaction that produces solid barium sulfate.
