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Naily [24]
3 years ago
15

Calculate the potential for each cell as shown below and indicate whether the metal electrode in the half-reaction opposite the

standard hydrogen electrode, s.h.e., would be the anode or the cathode if the cell was shorted and electrons were allowed to flow freely at 25 °c.
a.s.h.e. || zn2 (aq, 0.0250 m) | zn(s)
Chemistry
1 answer:
Lesechka [4]3 years ago
8 0
.086567 is what I think as being a math matichian
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Balance the following equations. Do not include the states of matter.<br><br> (a) C + O2 → CO
krek1111 [17]

Answer:

C + O2 → CO2

Explanation:

C + O2 → CO ----------------- (1)

from equ (1) on reactant side, C has 1 mole, O has 2 moles

from equ (1) on product side, C has 1 mole, O has 1 mole

Thus, to balance the equation, O should have 2 moles

C + O2 → CO2

7 0
3 years ago
Read 2 more answers
opper(I) ions in aqueous solution react with NH 3 ( aq ) according to Cu + ( aq ) + 2 NH 3 ( aq ) ⟶ Cu ( NH 3 ) + 2 ( aq ) K f =
Y_Kistochka [10]

Answer:

55.373g/l

Explanation:

The dissolved amount of sparingly soluble salts is interlinked with a unitlesss quantity called as solubility product. It is a fixed quantity that only increases with the rise in temperatures and is used to predict the salting out of compounds. If the value of ionic product (Q) is larger than (Ksp), precipitation of compound occurs.

Given:

The solubility product of CuBr is 6.3×10−9.

The concentration of NH3 is 0.10 M.

Formula and Calculations:

The dissolution reaction (I) of CuBr is shown below.

The reaction showing dissolution of CuBr in NH3 is shown below.

The above reaction can be obtained by adding reaction (I) and (II) as shown below.

The equilibrium constants will get multiplied.

Suppose the solubility of CuBr is “s”.

It is given that concentration of NH3 is 0.10 M.

The equilibrium constant expression for the above reaction is as follows,

Here,

The concentration of pure solids is 1 M. Thus, the concentration of CuBr is 1 M.

As calculated, the value of Ksp is 396.9.

Substitute all the required values in above formula.  

On further solving above equation,

Therefore, the solubility of CuBr in ammonia is 0.386 M.

The formula to calculate solubility

Solubuility (g/l)= Molarity(M) x Molarmass

Chemistry homework question answer, step 2, image 10

The molar mass of CuBr is 143.45 g/mol.

The formula to calculate solubility in g/L is given below.

The molar mass of CuBr is 143.45 g/mol.

therefore,

solubility = 0.386M x 143.45g/mol

where (M = mol/l)

solubility = 55.375g/l

5 0
3 years ago
Which relationship or statement best describes delta s for the following reaction:
zalisa [80]

Answer:

b) Delta S < 0

Explanation:

The change in the entropy (ΔS) is related to the change in the number of gaseous moles of the reaction: Δn(g) = n(g, products) - n(g, reactants).

  • If Δn(g) > 0, the entropy increases (ΔS > 0).
  • If Δn(g) < 0, the entropy decreases (ΔS < 0).
  • If Δn(g) = 0, there is little or no change in the entropy

Let's consider the following equation.

2 H₂S(g) + 3 O₂(g) → 2 H₂O(g)

Δn(g) = 2 - 5 = - 3. Since Δn(g) < 0, the entropy decreases and ΔS < 0.

4 0
3 years ago
Write the net ionic equation for the reaction of aqueous copper(II) chloride with aqueous sodium phosphate. Remember to include
sergeinik [125]

Answer:

3 Cu²⁺(aq) + 2 PO₄³⁻(aq) ⇒ Cu₃(PO₄)₂(s)

Explanation:

Let's consider the molecular equation between aqueous copper(II) chloride and aqueous sodium phosphate.

3 CuCl₂(aq) + 2 Na₃PO₄(aq) ⇒ 6 NaCl(aq) + Cu₃(PO₄)₂(s)

The complete ionic equation includes all the ions and insoluble species.

3 Cu²⁺(aq) + 6 Cl⁻(aq) + 6 Na⁺(aq) + 2 PO₄³⁻(aq) ⇒ 6 Na⁺(aq) + 6 Cl⁻(aq) + Cu₃(PO₄)₂(s)

The net ionic equation includes only the ions that participate in the reaction (not spectator ions) and insoluble species.

3 Cu²⁺(aq) + 2 PO₄³⁻(aq) ⇒ Cu₃(PO₄)₂(s)

7 0
3 years ago
Valproic acid, used to treat seizures and bipolar disorder, is composed of C, H, and O. A 0.165-g sample is combusted to produce
sergeinik [125]

Answer:

The empirical formula is = C_4H_8O

The formula of Valproic acid = C_8H_{16}O_2

Explanation:

Mass of water obtained = 0.166 g

Molar mass of water = 18 g/mol

Moles of H_2O = 0.166 g /18 g/mol = 0.00922 moles

2 moles of hydrogen atoms are present in 1 mole of water. So,

<u>Moles of H = 2 x 0.00922 = 0.01844 moles </u>

Molar mass of H atom = 1.008 g/mol

<u>Mass of H in molecule = 0.01844 x 1.008 = 0.018588 g </u>

Mass of carbon dioxide obtained = 0.403 g

Molar mass of carbon dioxide = 44.01 g/mol

Moles of CO_2 = 0.403 g  /44.01 g/mol = 0.009157 moles

1 mole of carbon atoms are present in 1 mole of carbon dioxide. So,

<u>Moles of C = 0.009157 moles </u>

Molar mass of C atom = 12.0107 g/mol

<u>Mass of C in molecule = 0.009157 x 12.0107 = 0.11 g </u>

<u>Given that the Valproic acid only contains hydrogen, oxygen and carbon. </u>So,

Mass of O in the sample = Total mass - Mass of C  - Mass of H

Mass of the sample = 0.165 g

<u>Mass of O in sample = 0.165 - 0.11 - 0.018588 = 0.036412 g  </u>

Molar mass of O = 15.999 g/mol

<u>Moles of O  = 0.036412  / 15.999  = 0.002276 moles</u>

<u></u>

<u>Taking the simplest ratio for H, O and C as: </u>

<u>0.01844 : 0.002276 : 0.009157</u>

<u> = 8 : 1 : 4</u>

The empirical formula is = C_4H_8O

Molecular formulas is the actual number of atoms of each element in the compound while empirical formulas is the simplest or reduced ratio of the elements in the compound.

Thus,  

Molecular mass = n × Empirical mass

Where, n is any positive number from 1, 2, 3...

Mass from the Empirical formula = 4×12 + 8×1 + 16= 72 g/mol

Molar mass = 144 g/mol

So,  

Molecular mass = n × Empirical mass

144 = n × 72

<u>⇒ n = 2</u>

The formula of Valproic acid = C_8H_{16}O_2

7 0
3 years ago
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