Explanation:
According to your question,
no. a. ans would be like; chemical formula is defined as the an expression which determines no. and type of molecule of a compound.
b. no. ans; it tells that what type of compound is formed with the type and no. of atoms present in the atom.
c. no ans; the formulation of h2so4 states that it is acid named as hydrochloric acid which is formed by reacting of hydrogen (2 atoms ) ,sulpher (*1atom) and oxygen(4atoms).
Hope it helps...
For the reaction CO2(g) + H2(g)CO(g) + H20(g)
∆H°=41.2 kJ and ∆S°=42.1 J/K
The standard free energy change for the reaction of 1.96 moles of Co2(g) at 289 K, 1 atm would be_________KJ.
This reaction is (reactant, product)___________ favored under standard conditions at 289 K.
Assume that ∆H° and ∆S° are independent of temperature.
Answer:
The ΔG° is 29 kJ and the reaction is favored towards reactant.
Explanation:
Based on the given information, the ΔH°rxn or enthalpy change is 41.2 kJ, the ΔS°rxn or change in entropy is 42.1 J/K or 42.1 * 10⁻³ kJ/K. The temperature given is 289 K. Now the Gibbs Free energy change can be calculated by using the formula,
ΔG° = ΔH°rxn - TΔS°rxn
= 41.2 kJ - 289 K × 42.1 × 10⁻³ kJ/K
= 41.2 kJ - 12.2 kJ
= 29 kJ
As ΔG° of the reaction is positive, therefore, the reaction is favored towards reactant.
Why don't siblings look exactly alike
Answer:
Your genes play a big role in making you who you are. ... But brothers and sisters don't look exactly alike because everyone (including parents) actually has two copies of most of their genes. And these copies can be different. Parents pass one of their two copies of each of their genes to their kids.
Which of the following do we need to know in order to calculate pH during an acid-base titration of a strong monoprotic acid with a strong monoprotic base? Select all that apply
a. the concentration of the acid
b. the concentration of the base titrant
c. the initial volume of the acid solution
d. the volume of the titrant used
Answer:
the volume of the titrant used
Explanation:
Acid-base titrations are usually depicted on special graphs referred to as titration curve. A titration curve is a graph that contains a plot of the volume of the titrant as the independent variable and the pH of the system as the dependent variable.
Hence, a titration curve is a graphical plot showing the pH of the analyte solution plotted against the volume of the titrant as the reaction is in progress. The titration curve is drawn by plotting data obtained during a titration, that is, volume of the titrant added (plotted on the x-axis) and pH of the system (plotted on the y-axis).
Consider this synthesis of isoamyl acetate based on this week's experimental methods, after refluxing the reaction mixture for 25 minutes, what is likely present in solution
Answer:
acetic acid and phosphoric acid
Explanation:
After refluxing the reaction mixture ( synthesis of isoamyl acetate ) what is likely present in the solution is acetic acid and phosphoric acid, this due to the fact that if the reaction time between the reactants was less than the refluxing time which is 25 minutes,
there will be no reactant ( 3-methylbutanol )left in the reaction mixture
When salt is added to water, all of the following happens except? A. The salt breaks into positive chlorine ions and negative sodium icons B. the positive part of the water molecule is attracted to the negative ions C. The negative part of the water molecule is attracted to the positive ions D. The water molecules surround the dissociated ions
Answer:
The salt breaks into positive chlorine ions and negative sodium icons
Explanation:
The question requested for the wrong option in the list. If we look at the option selected, we will notice that sodium ions are positively charged ions since sodium is a metal. Metals produce cations (positive ions) because they loose electrons. Therefore, a sodium ion can never be negatively charged.
Similarly, chlorine is a highly electronegative nonmetal. It gains electrons in an ionic bond. Hence chlorine ions can not be positive.
Divers often inflate heavy duty balloons attached to salvage items on the sea floor. If a balloon is filled to a volume of 1.20 L at a pressure of 6.25 atm, what is the volume of the balloon when it reaches the surface?
Answer:
7.50 L
Explanation:
The balloon has a volume of 1.20 L (V₁) when the pressure at the sea floor is 6.25 atm (P₁). When it reaches the surface, the pressure is that of the atmosphere, that is, 1.00 atm (P₂). If we consider the gas to behave as an ideal gas and the temperature to be constant, we can calculate the final volume (V₂) using Boyle's law.
P₁ × V₁ = P₂ × V₂
V₂ = P₁ × V₁ / P₂
V₂ = 6.25 atm × 1.20 L / 1.00 atm
V₂ = 7.50 L
Please help, Which type of molecule is shown below?
Answer:
Carbohydrate
Explanation:
A.pex
To refine aluminum from its ore, aluminum oxide is electrolyzed to form aluminum and oxygen. At which electrode does oxygen form? options: A) Both the anode and the cathode B) Cathode C) Neither electrode D) Anode
Answer:
im pretty sure its the anode
Explanation:
To solve such, we must know the concept of electrolysis reaction. The correct option is option D among all given options. At anode electrode oxygen forms.
What is chemical reaction?Chemical reaction is a process in which two or more than two molecules collide in right orientation and energy to form a new chemical compound. The mass of the overall reaction should be conserved. There are so many types of chemical reaction reaction like combination reaction, double displacement reaction.
Electrolysis is the process of passing an electric current through a material to cause a chemical change. A chemical change occurs when a material loses or acquires the electron. To refine aluminum from its ore, aluminum oxide is electrolyzed to form aluminum and oxygen. At anode electrode oxygen forms.
Therefore, the correct option is option D among all given options. At anode electrode oxygen forms.
Learn more about the chemical reactions, here:
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Suppose a student completes an experiment with an average value of 2.9 mL and a calculated standard deviation of 0.71 mL. What is the minimum value within a 1 SD range of the average
Answer:
The correct answer is 2.2 mL.
Explanation:
Given:
Average: 2.9 mL
SD: 0.71 mL
We can define a 1 SD range in which the value of volume (in mL) will be comprised:
Volume (mL) = Average ± SD = (2.9 ± 0.7) mL
Maximum value= Average + SD= 2.9 + 0.7 mL = 3.6 mL
Minimum value= Average - SD = 2.9 - 0.7 mL = 2.2 mL
Thus, the minimum value within a 1 SD range of the average is 2.2 mL
The minimum value within 1 SD is 2.19 mL
The z score is used to determine by how many standard deviations the raw score is above or below the mean. The z score is given by:
[tex]z=\frac{x-\mu}{\sigma} \\\\where\ x\ is\ raw\ score, \mu=mean,\sigma=standard\ deviation[/tex]
Given that μ = 2.9 mL, σ = 0.71 mL; hence:
The minimum value within 1 SD range = μ ± σ = 2.9 ± 0.71 = (2.19, 3.61)
Therefore the minimum value within 1 SD is 2.19 mL
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Using these metal ion/metal standard reduction potentials Cd2+(aq)|Cd(s) Zn2+(aq)|Zn(s) Ni2+(aq)|Ni(s) Cu2+(aq)/Cu(s) -0.40 V -0.76 V ‑0.25 V +0.34 V Calculate the standard cell potential for the cell whose reaction is Ni2+(aq) + Zn(s) →Zn2+(aq)+ Ni(s)
Answer: The standard cell potential for the cell is +0.51 V
Explanation:
Given : [tex]E^0_{Ni^{2+}/Ni}=-0.25V[/tex]
[tex]E^0_{Zn^{2+}/Zn}=-0.76V[/tex]
The given reaction is:
[tex]Ni^{2+}(aq)+Zn(s)\rightarrow Zn^{2+}(aq)+Ni(s)[/tex]
As nickel is undergoing reduction, it acts as cathode and Zinc is undergoing oxidation, so it acts as anode.
[tex]E^0_{cell}=E^0_{cathode}-E^0_{anode}[/tex]
where both [tex]E^0[/tex] are standard reduction potentials.
Thus putting the values we get:
[tex]E^0_{cell}=-0.25-(-0.76)[/tex]
[tex]E^0_{cell}=0.51V[/tex]
Thus the standard cell potential for the cell is +0.51 V
Methanol liquid burns readily in air. One way to represent this equilibrium is: 2 CO2(g) + 4 H2O(g)2 CH3OH(l) + 3 O2(g) We could also write this reaction three other ways, listed below. The equilibrium constants for all of the reactions are related. Write the equilibrium constant for each new reaction in terms of K, the equilibrium constant for the reaction above. 1) CH3OH(l) + 3/2 O2(g) CO2(g) + 2 H2O(g) K1 = 2) CO2(g) + 2 H2O(g) CH3OH(l) + 3/2 O2(g) K2 = 3) 2 CH3OH(l) + 3 O2(g) 2 CO2(g) + 4 H2O(g)
Answer:
Answers are in the explanation
Explanation:
It is possible to obtain K of equilibrium of related reactions knowing the laws:
A + B ⇄ C K₁
C ⇄ A + B K = 1 /K₁
The inverse reaction has the inverse K equilibrium
2A + 2B ⇄ 2C K = K₁²
The multiplication of the coefficients of reaction produce a k powered to the number you are multiplying the coefficients
For the reaction:
2 CO2(g) + 4 H2O(g) ⇄ 2 CH3OH(l) + 3 O2(g) K
1) CH3OH(l) + 3/2 O2(g) ⇄ CO2(g) + 2 H2O(g)
This is the inverse reaction but also the coefficients are dividing in the half, that means:
[tex]K_1 = \frac{1}{k^{1/2}} = (1/K)^{1/2}[/tex]
2) CO2(g) + 2 H2O(g) ⇄ CH3OH(l) + 3/2 O2(g)
Here,the only change is the coefficients are the half of the original reaction:
[tex]K_2 = K^{1/2}[/tex]
3) 2CH3OH(l) + 3 O2(g) ⇄ 2 CO2(g) + 4 H2O(g)
This is the inverse reaction. Thus, you have the inverse K of equilibrium:
[tex]K_3 = \frac{1}{K}[/tex]
the pain reliever codeine is a weak base with a kb equal to 1.6 x 10^-6. what is the ph of a 0.05 m aqueous codeine solution
Answer:
[tex]pH=10.45[/tex]
Explanation:
Hello,
In this case, for the dissociation of the given base, we have:
[tex]base\rightleftharpoons OH^-+CA[/tex]
Whereas CA accounts for conjugated acid and OH⁻ for the conjugated base. In such a way, equilibrium expression is:
[tex]Kb=\frac{[OH^-][CA^+]}{[base]}[/tex]
And in terms of the reaction extent [tex]x[/tex] we can write:
[tex]1.6x10^{-6}=\frac{x*x}{0.05M-x}[/tex]
For which the roots are:
[tex]x_1=-0.000284M\\x_2=0.000282M[/tex]
For which clearly the result is the positive root which also equals the concentration of hydroxyl ions and we can compute the pOH:
[tex]pOH=-log([OH^-])=-log(0.000282)\\\\pOH=3.55[/tex]
And the pH:
[tex]pH=14-pOH=14-3.55\\\\pH=10.45[/tex]
Regards.
The pH of the solution is 10.45.
Let us represent codeine with the generic formula BH. We can set up the ICE table as follows;
:B(aq) + H2O(l) ⇄ BH(aq) + OH^-(aq)
I 0.05 0 0
C -x +x +x
E 0.05 - x x x
We know that the Kb of codeine is 1.6 x 10^-6, Hence;
1.6 x 10^-6 = x^2/0.05 - x
1.6 x 10^-6 (0.05 - x ) = x^2
8 x 10^-8 - 1.6 x 10^-6x = x^2
x^2 + 1.6 x 10^-6x - 8 x 10^-8 = 0
x = 0.00028 M
The concentration of hydroxide ions = 0.00028 M
Given that pOH = - log[0.00028 M]
pOH = 3.55
pH + pOH = 14
pH = 14 - 3.55
pH = 10.45
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Which ONE of these cations has the same number of unpaired electrons as Fe2+ ? A) Ni2+ B) Fe3+ C) Cr2+ D) Mn2+ E) Co2+
Answer:
Explanation:
Fe2+ Has 4 unpaired electrons.
By method of elimination;
Option A: Ni2+ has two unpaired electrons. so this option is wrong.
Option B: There are 5 unpaired electrons in the Fe3+ ion. so this option is wrong.
Option C: There are 4 unpaired electrons in the Cr2+ ion. so this option is correct.
Option D: There are 5 unpaired electrons in the Mn2+ ion. so this option is wrong.
Option E: There are 3 unpaired electrons in the Co2+ ion. so this option is wrong.
The solubility product for Ag3PO4 is 2.8 × 10‑18. What is the solubility of silver phosphate in a solution which also contains 0.10 moles of silver nitrate per liter?
Answer:
2.8x10⁻¹⁵ M.
Explanation:
Hello,
In this case, the dissociation reaction for silver phosphate is:
[tex]Ag_3PO_4(s)\rightleftharpoons 3Ag^+(aq)+PO_4(aq)[/tex]
Therefore, the equilibrium expression is:
[tex]Ksp=[Ag^+]^3[PO_4^-][/tex]
In such a way, since the initial solution contains an initial concentration of silver ions (from silver nitrate) of 0.10M, we can write the equilibrium expression in terms of the reaction extent [tex]x[/tex]:
[tex]2.8x10^{-18}=(0.10+3x)^3*(x)[/tex]
Thus, solving for [tex]x[/tex] we have:
[tex]x=2.8x10^{-15}M[/tex]
Thus, the molar solubility of silver phosphate is 2.8x10⁻¹⁵ M.
Regards.
Chlorine can be prepared in the laboratory by the reaction of manganese dioxide with hydrochloric acid, HCl(aq), as
described by the chemical equation
MnO,(s) + 4 HCl(aq)
MnCl(aq) + 2 H2O(l) + Cl (8)
How much MnO(s) should be added to excess HCl(aq) to obtain 175 mL C12(g) at 25 °C and 715 Torr?
mass of MnO2
Answer:
Explanation:
MnO₂(s) + 4 HCl(aq) = MnCl₂(aq) + 2 H₂O(l) + Cl₂
87 g 22.4 x 10³ mL
volume of given chlorine gas at NTP or at 760 Torr and 273 K
= 175 x ( 273 + 25 ) x 715 / (273 x 760 )
= 179.71 mL
22.4 x 10³ mL of chlorine requires 87 g of MnO₂
179.4 mL of chlorine will require 87 x 179.4 / 22.4 x 10³ g
= 696.77 x 10⁻³ g
= 696.77 mg .
21. What are the two main ways of working with clay?
Answer:
Diferentes tipos de arcilla
ARCILLA DE LADRILLOS. Contiene muchas impurezas. ...
ARCILLA DE ALFARERO. Llamada también barro rojo y utilizada en alfarería y para modelar. ...
ARCILLA DE GRES. Es una arcilla con gran contenido de feldespato. ...
ARCILLAS “BALL CLAY” O DE BOLA. ...
CAOLIN. ...
ARCILLA REFRACTARIA. ...
BENTONITA.
Explanation:
Answer:
Coil method and the slab method.
Explanation:
One hundred fifty joules of heat are removed from a heat reservoir at a temperature of 150 K. What is the entropy change of the reservoir (in J/K)?
Answer:
ΔS surrounding (entropy change of the reservoir) = -1 J/K
Explanation:
Given:
Change in heat (ΔH) = 150 joules
Temperature (T) = 150 K
Find:
ΔS surrounding (entropy change of the reservoir)
Computation:
ΔS surrounding (entropy change of the reservoir) = - ΔH / T
ΔS surrounding (entropy change of the reservoir) = - 150 / 150
ΔS surrounding (entropy change of the reservoir) = -1 J/K
A laboratory technician combines 35.9 mL of 0.258 M chromium(II) chloride with 35.8 mL 0.338 M potassium hydroxide. How many grams of chromium(II) hydroxide can precipitate
Answer:
0.52 g of chromium(II) hydroxide, Cr(OH)2.
Explanation:
We'll begin by calculating the number of mole of chromium (ii) chloride, CrCl2 in 35.9 mL of 0.258 M chromium(II) chloride solution.
This can be obtained as follow:
Molarity of CrCl2 = 0.258 M
Volume = 35.9 mL = 35.9/1000 = 0.0359 L
Mole of CrCl2 =?
Molarity = mole /Volume
0.258 = mole of CrCl2 /0.0359
Cross multiply
Mole of CrCl2 = 0.258 x 0.0359
Mole of CrCl2 = 0.0093 mole
Next, we shall determine the number of mole of potassium hydroxide, KOH in 35.8 mL 0.338 M potassium hydroxide solution.
This can be obtained as follow:
Molarity of KOH = 0.338 M
Volume = 35.8 mL = 35.8/1000 = 0.0358 L
Mole of KOH =.?
Molarity = mole /Volume
0.338 = mole of KOH /0.0358
Cross multiply
Mole of KOH = 0.338 x 0.0358
Mole of KOH = 0.0121 mole.
Next, we shall write the balanced equation for the reaction. This is given below:
2KOH + CrCl2 → Cr(OH)2 + 2KCl
From the balanced equation above,
2 mole of KOH reacted with 1 mole of CrCl2 to produce 1 mole of Cr(OH)2.
Next, we shall determine the limiting reactant. This can be obtained as follow:
From the balanced equation above,
2 mole of KOH reacted with 1 mole of CrCl2.
Therefore, 0.0121 mole of KOH will react with = (0.0121 x 1)/2 = 0.00605 mole of CrCl2.
From the calculations made above, we can see that only 0.00605 mole out of 0.0093 mole of CrCl2 is needed to react completely with 0.0121 mole of KOH.
Therefore, KOH is the limiting reactant.
Next, we shall determine the number of mole of Cr(OH)2 produced from the reaction.
In this case, we shall be using the limiting reactant because it will give the maximum yield of Cr(OH)2.
The limiting reactant is KOH and the number of mole of Cr(OH)2 produced can be obtained as illustrated below:
From the balanced equation above,
2 mole of KOH reacted to produce 1 mole of Cr(OH)2.
Therefore, Therefore, 0.0121 mole of KOH will react with = (0.0121 x 1)/2 = 0.00605 mole of Cr(OH)2.
Finally, we shall convert 0.00605 mole of Cr(OH)2 to grams.
This is illustrated below:
Mole of Cr(OH)2 = 0.00605 mole
Molar mass of Cr(OH)2 = 52 + 2(16 + 1) = 52 + 2(17) = 86 g/mol
Mass of Cr(OH)2 =..?
Mole = mass /Molar mass
0.00605 = mass of Cr(OH)2/86
Cross multiply
Mass of Cr(OH)2 = 0.00605 x 86
Mass of Cr(OH)2 = 0.52 g
Therefore, 0.52 g of chromium(II) hydroxide, Cr(OH)2 was produced.
what js the percent yield of lithium hydroxide from a reaction of 7.40 g of lithium with 10.2 g of water? the actual yield was measured to be12.1 g
Answer:
89%.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
2Li + 2H2O —> 2LiOH + H2
Next, we shall determine the masses of Li and H2O that reacted and the mass of LiOH produced from the balanced equation.
This is illustrated below:
Molar mass of Li = 7 g/mol
Mass of Li from the balanced equation = 2 x 7 = 14 g
Molar mass of H2O = (2x1) + 16 = 18 g/mol
Mass of H2O from the balanced equation = 2 x 18 = 36 g
Molar mass of LiOH = 7 + 16 + 1 = 24 g/mol
Mass of LiOH from the balanced equation = 2 x 24 = 48 g
Summary:
From the balanced equation above,
14 g of Li reacted with 36 g of H2O to produce 48 g of LiOH.
Next, we shall determine the limiting reactant. This can be obtained as follow:
From the balanced equation above,
14 g of Li reacted with 36 g of H2O.
Therefore, 7.4 g of Li will react with = (7.4 x 36)/14 = 19.03 g of H2O.
From the calculation made above, we can see that it will take a higher amount i.e 19.03 g than what was given i.e 10.2 g of H2O to react completely with 7.4 g of Li.
Therefore, H2O is the limiting reactant and Li is the excess reactant.
Next, we shall determine the theoretical yield of LiOH.
In this case we shall use the limiting reactant.
The limiting reactant is H2O and the theoretical yield of LiOH can be obtained as follow:
From the balanced equation above,
36 g of H2O reacted to produce 48 g of LiOH.
Therefore, 10.2 g of H2O will react to produce = (10.2 x 48)/36 = 13.6 g of LiOH.
Therefore, the theoretical yield of LiOH is 13.6 g
Finally, we shall determine the percentage yield of LiOH. This can be obtained as follow:
Actual yield = 12.1 g
Theoretical yield = 13.6 g
Percentage yield =..?
Percentage yield = Actual yield /Theoretical yield x 100
Percentage yield = 12.1/ 13.6 x 100
Percentage yield = 89%
Therefore, the percentage yield LiOH is 89%.
A compound (C_9H_9BrO_2) gives the following NMR data. Draw the structure of the compound.
'1^H-NMR: 1.39 ppm, t(3H); 4.38 ppm, q(2H); 7.57 ppm, d(2H); 7.90 ppm, d(2H)
13^C-NMR: 165.73; 131.56; 131.01; 129.84; 127.81; 61.18; 14.18
You do not have to consider stereo chemistry.
You do not have to explicitly draw H atoms.
Do not include lone pairs in your answer.
Answer:
ethyl 4-bromobenzoate
Explanation:
In this question, we can start with the Index of Hydrogen Deficiency (I.H.D):
[tex]I.H.D=\frac{2C+2+N-H-X}{2}=\frac{(2*9)+2+0-9-1}{2}~=~5[/tex]
This indicates, that we can have a benzene ring (I.H.D = 4) and a carbonyl group (I.H.D = 1), for a total of 5.
Additionally, in the 1H-NMR info, we have a triplet 1.39 (3H) followed by a doublet 4.38 (2H), this indicates the presence of an ethyl group ([tex]CH_3-CH_2-[/tex]). Also, in the formula, we have 2 oxygens if we have carbonyl group with 2 oxygens we have a high probability to have an ester group.
[tex]O=C-O-CH_2-CH_3[/tex]
Now, if we add this to the benzene ring and the "Br" atom that we have in the formula, we will have ethyl 4-bromobenzoate.
See figures 1 and 2 to further explanations.
I hope it helps!
If a sample of C-14 initially contains 1.6 mmol of C-14, how many millimoles will be left after 2250 years
Answer: 1.2 millimoles will be left after 2250 years
Explanation:
Expression for rate law for first order kinetics is given by:
[tex]t=\frac{2.303}{k}\log\frac{a}{a-x}[/tex]
where,
k = rate constant
t = age of sample
a = let initial amount of the reactant
a - x = amount left after decay process
a) for completion of half life:
Half life is the amount of time taken by a radioactive material to decay to half of its original value.
[tex]t_{\frac{1}{2}}=\frac{0.693}{k}[/tex]
[tex]k=\frac{0.693}{5730}=0.00012years^{-1}[/tex]
b) Amount left after 2250 years
[tex]2250=\frac{2.303}{k}\log\frac{1.6}{a-x}[/tex]
[tex]2250=\frac{2.303}{0.00012}\log\frac{1.6}{a-x}[/tex]
[tex]\log\frac{1.6}{a-x}=0.117[/tex]
[tex]\frac{1.6}{a-x}=1.31[/tex]
[tex]{a-x}=\frac{1.6}{1.31}=1.2[/tex]
Thus 1.2 millimoles will be left after 2250 years
What is buffers and mention its importance?
Answer:
Buffer is the chemical substance that addition of acids and bases, maintaining constant environment,its called Buffer.
Explanation:
Buffers are use in the system to maintain the value of pH, and the contain the pH value is not to change.Buffer maintain the body of pH for the optimal activity,and they are solution of pH constant.Buffer in used in the lab and that to maintain growth of the micro tissues and the culture media.Buffer are used in maintain necessary optimal reaction activity,determine the indicator of solution with pH.Buffer capacity is that concentration to the buffering agent, is the very small increase,buffer capacity to the pH is 32% , of the maximum value of pH.Buffers in a acid regions to the desired of that value to the particular buffer agent.Buffers can be made from that a mixture of the base and acid, buffer can be a wide range of the obtained.Buffers that the pH calculation and they required to performed in the critic acid that the overlap over the buffer range.The front curve of a spectacle lens is called?
Answer:
Corrective lense or just lens.
Explanation:
In which pair do both compounds exhibit predominantly ionic bonding? A) KCl and CO2 B) SO2 and BaF2 C) F2 and N2O D) N2O3 and Rb2O E) NaF and SrO
Answer:
E) NaF and SrO
Explanation:
The ionic bonding occurs between atoms with a great difference in electronegativity. This usually happens between a metal and a non-metal.
In which pair do both compounds exhibit predominantly ionic bonding?
A) KCl and CO₂. NO. C and O are non-metals and present covalent bonding.
B) SO₂ and BaF₂. NO. S and O are non-metals and present covalent bonding.
C) F₂ and N₂O. NO. Both compounds contain non-metals and present covalent bonding.
D) N₂O₃ and Rb₂O. NO. N and O are non-metals and present covalent bonding.
E) NaF and SrO. YES. Na and Sr are metals while F and O are non-metals.
Complete the sentences describing the cell.
a. In the nickel-aluminum galvanic cell, the cathode is ____ .
b. Therefore electrons flow from___ to ____.
c. The ____ electrode loses mass, while the ____ electrode gains mass.
Answer:
a. In the nickel-aluminum galvanic cell, the cathode is nickel electrode.
b. Therefore electrons flow from the aluminium electrode to the nickel electrode.
c. The aluminium electrode loses mass, while the nickel electrode electrode gains mass.
Explanation:
Voltaic or galvanic cells are electrochemical cells in which spontaneous oxidation-reduction reactions. The two halves of the redox reaction are separate and electron transfer is required to occur through an external circuit for the redox reaction to take place. That is, one of the metals in one of the half cells is oxidized while the metal of the other half cell is reduced, producing an exchange of electrons through an external circuit. This makes it possible to take advantage of the electric current.
Given:
E ⁰N i ⁺² = − 0.23 V is the standard reduction potential for the nickel ion
E ⁰ A l ⁺³ = − 1.66 V is the standard reduction potential for the aluminum ion
The most negative potentials correspond to more reducing substances. In this case, the aluminum ion is the reducing agent, where oxidation takes place. In the anodic half cell oxidations occur, while in the cathode half cell reductions occur. So the aluminum cell acts as the anode while the nickel cell acts as the cathode.
So a. In the nickel-aluminum galvanic cell, the cathode is nickel electrode.
The metal that is oxidized gives electrons to the metal that is reduced through the outer conductor. Then the electrons flow spontaneously from the anode to the cathode.
Then b. Therefore electrons flow from the aluminium electrode to the nickel electrode.
Ni⁺², being the cathode, accepts electrons, becoming Ni (s) and depositing on the Ni electrodes.
So, c. The aluminium electrode loses mass, while the nickel electrode electrode gains mass.
Which of the following compounds is more soluble in a 0.10 M NaCN solution than in pure neutral water? Ca3(PO4)2 AgBr CaCO3 Mg(OH)2 NH4ClO4
Answer:
AgBr
Explanation:
Silver bromide has a very low solubility product constant of about 7.7 ×10^-13 in pure water hence it is not quite soluble in pure water.
However, with NaCN, the AgBr forms the complex [Ag(CN)2]^2- which has a formation constant of about 5.6 ×10^8. This very high formation constant implies that the complex is easily formed leading to the dissolution of AgBr in NaCN.
The equation for the dissolution of AgBr in cyanide is shown below;
AgBr(s) + 2CN^-(aq) ----> [Ag(CN)2]^2-(aq) + Br^-(aq)
11. The mass (in grams) of FeSO4.7H2O required for preparation of 125 mL of 0.90 M
solution is:
(a) 16 g
(b) 25 g
(c) 13 g
(d) 31 g
(e) 43 g
Answer:
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Taking into account the definition of molarity and the molar mass of the compound, the correct answer is option (d): the mass of FeSO₄.7H₂O required for preparation of 125 mL of 0.90 M solution is 31 g.
In first place, you have to know tha molarity is a measure of the concentration of that substance that indicates the number of moles of solute present in the solution.
The molarity of a solution is calculated by dividing the moles of the solute by the volume of the solution.
[tex]molarity=\frac{number of moles of solute}{volume of solution}[/tex]
Molarity is expressed in units [tex]\frac{moles}{liter}[/tex].
In this case, you know:
molarity= 0.90 Mnumber of moles of solute= ?volume= 125 mL= 0.125 L (being 1000 mL=1 L)So, by definition of molarity, the number of moles is calculated as:
[tex]0.90 M=\frac{number of moles of solute}{0.125 L}[/tex]
Solving:
number of moles of solute= 0.90 M× 0.125 L
number of moles of solute= 0.1125 moles
On the other side, molar mass is the mass of one mole of a substance, which can be an element or a compound. In this case, the molar mass of FeSO₄.7H₂O is 277.85 [tex]\frac{g}{mole}[/tex].
Then you can apply the following rule of three: if by definition of molar mass, 1 mole of the compound contains 277.85 g, 0.1125 mole contains how much mass?
[tex]mass=\frac{0.1125 moles*277.85 g}{1 mole}[/tex]
Solving:
mass ≅ 31 g
Finally, the correct answer is option (d): the mass of FeSO₄.7H₂O required for preparation of 125 mL of 0.90 M solution is 31 g.
Learn more about molarity with this example: brainly.com/question/15406534?referrer=searchResults
A solution contains 2.2 × 10-3 M in Cu2+ and 0.33 M in LiCN. If the Kf for Cu(CN)42- is 1.0 × 1025, how much copper ion remains at equilibrium?
Answer:
[Cu²⁺] = 2.01x10⁻²⁶
Explanation:
The equilibrium of Cu(CN)₄²⁻ is:
Cu²⁺ + 4CN⁻ ⇄ Cu(CN)₄²⁻
And Kf is defined as:
Kf = 1.0x10²⁵ = [Cu(CN)₄²⁻] / [Cu²⁺] [CN⁻]⁴
As Kf is too high you can assume all Cu²⁺ is converted in Cu(CN)₄²⁻ -Cu²⁺ is limiting reactant-, the new concentrations will be:
[Cu²⁺] = 0
[CN⁻] = 0.33M - 4×2.2x10⁻³ = 0.3212M
[Cu(CN)₄²⁻] = 2.2x10⁻³
Some [Cu²⁺] will be formed and equilibrium concentrations will be:
[Cu²⁺] = X
[CN⁻] = 0.3212M + 4X
[Cu(CN)₄²⁻] = 2.2x10⁻³ - X
Where X is reaction coordinate
Replacing in Kf equation:
1.0x10²⁵ = [2.2x10⁻³ - X] / [X] [0.3212M +4X]⁴
1.0x10²⁵ = [2.2x10⁻³ - X] / 0.0104858X + 0.524288 X² + 9.8304 X³ + 81.92 X⁴ + 256 X⁵
1.04858x10²³X + 5.24288x10²⁴ X² + 9.8304x10²⁵ X³ + 8.192x10²⁶ X⁴ + 2.56x10²⁷ X⁵ = 2.2x10⁻³ - X
1.04858x10²³X + 5.24288x10²⁴ X² + 9.8304x10²⁵ X³ + 8.192x10²⁶ X⁴ + 2.56x10²⁷ X⁵ - 2.2x10⁻³ = 0
Solving for X:
X = 2.01x10⁻²⁶
As
[Cu²⁺] = X
[Cu²⁺] = 2.01x10⁻²⁶How many moles of gold are equivalent to 1.204 × 1024 atoms? 0.2 0.5 2 5
Answer:
c
Explanation:
How many moles of gold are equivalent to 1.204 × 1024 atoms?
0.2
0.5
2
5
C) 2 Is the correct answer, I took the test and it was correct.
According to the concept of Avogadro's number, there are 2 moles of gold which are equivalent to 1.204×10²⁴ atoms.
What is Avogadro's number?
Avogadro's number is defined as a proportionality factor which relates number of constituent particles with the amount of substance which is present in the sample.
It has a SI unit of reciprocal mole whose numeric value is expressed in reciprocal mole which is a dimensionless number and is called as Avogadro's constant.It relates the volume of a substance with it's average volume occupied by one of it's particles .
According to the definitions, Avogadro's number depend on determined value of mass of one atom of those elements.It bridges the gap between macroscopic and microscopic world by relating amount of substance with number of particles.
Number of atoms can be calculated using Avogadro's number as follows: mass/molar mass×Avogadro's number.Number of moles=number of atoms/Avogadro's number=1.204×10²⁴ /6.023×10²³=1.999≅2
Thus, there are 2 moles of gold which are equivalent to 1.204×10²⁴ atoms.
Learn more about Avogadro's number,here:
https://brainly.com/question/11907018
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Emission of which one of the following leaves both atomic number and mass number unchanged?
(a) positron
(b) neutron
(c) alpha particle
(d) gamma radiation
(e) beta particle
Answer: Gamma Radiation
Explanation:
The emission of Gamma rays does not cause a change in both the atomic and mass number. They are electromagnetic radiation.
The radiations that leaves without changing the atomic mass and atomic number of the particle have been gamma radiations. Thus, option D is correct.
Radiations have been the energy that has been evolved by the particles during energy transitions. The nuclear decay results with the release of the energy from the particle resulting in the change in the atomic mass.
The electromagnetic radiations have been capable of emitting the radiation without changing the mass and atomic number of the element. The gamma radiations have been the electromagnetic radiations. Thus, option D is correct.
For more information about the emissions, refer to the link:
https://brainly.com/question/517329
