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zvonat [6]
2 years ago
14

The electrolysis of water forms H2 and O2. 2H2O Right arrow. 2H2 O2 What is the percent yield of O2 if 10.2 g of O2 is produced

from the decomposition of 17.0 g of H2O
Chemistry
1 answer:
Mama L [17]2 years ago
8 0

Answer:

67.5%

Explanation:

Step 1: Write the balanced equation for the electrolysis of water

2 H₂O ⇒ 2 H₂ + O₂

Step 2: Calculate the theoretical yield of O₂ from 17.0 g of H₂O

According to the balanced equation, the mass ratio of H₂O to O₂ is 36.04:32.00.

17.0 g H₂O × 32.00 g O₂/36.04 g H₂O = 15.1 g O₂

Step 3: Calculate the percent yield of O₂

Given the experimental yield of O₂ is 10.2 g, we can calculate its percent yield using the following expression.

%yield = (exp yield / theoret yield) × 100%

%yield = (10.2 g / 15.1 g) × 100% = 67.5%

You might be interested in
How much heat is absorbed/released when 20.00 g of NH3(g) reacts in the presence of excess O2 (g) to produce NO (g) and H2O (l)
FrozenT [24]

Answer:

a. 342.9 kJ of heat are absorbed.

Explanation:

Calculation of the moles of NH_3 as:-

Mass = 20.00 g

Molar mass of NH_3 = 17.031 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{20.00\ g}{17.031\ g/mol}

Moles= 1.1743\ mol

Given that:- \Delta H=+1168\ kJ

It means that 1 mole of NH_3 undergoes reaction and absorbs 1168\ kJ of heat

So,

1168 mole of NH_3 undergoes reaction and absorbs 1168\times 1168\ kJ of heat

<u>Amount of heat absorbed = + 342.9 KJ</u>

7 0
3 years ago
How do metalloids affect modern technology?
Lady bird [3.3K]
<em><u>examples of metalloids :
</u>Boron,Silicon ,Germanium are some metalloids ...
<u>what they do:
</u>they are used to form the semiconductors and these semiconductors are used in modern computer technology like in circuits ,chips and computer based gadgets... :) <u>
</u></em>
7 0
3 years ago
It has been suggested that the surface melting of ice plays a role in enabling speed skaters to achieve peak performance. Carry
Fittoniya [83]

Answer:

a) Pf = 689.4 bar

b) P = 226.6 bar

c) T = 269.99 K

Explanation:

a)

The molar volume of ice is equal to:

Vi = m/p = (18.02x10^-3 kg H2O/1 mol H2O)*(1 m^3/920 kg) = 1.96x10^-5 m^3 mol^-1

The molar volume of liquid water is equal to:

Vl = (18.02x10^-3 kg/1 mol)*(1 m^3/997 kg) = 1.8x10^-5 m^3 mol^-1

The change in volume is equal to:

ΔVchange = Vi-Vl = 1.96x10^-5 - 1.8x10^-5 = 1.5x10^-6 m^3 mol^-1

using the Clapeyron equation:

Pf = Pi + ((ΔHf*ΔT)/(ΔVf*Ti)) = 1.013x10^5 Pa + ((6010 J mol^-1 * 4.7 K)/(1.5x10^-6 * 273.15 K)) = 6.89x10^7 Pa = 689.4 bar

b)

For the pressure we will use the equation:

P = (m*g)/A, where m is the mass, g is the acceleration of gravity and A is the area. Replacing values:

P = (79 kg * 9.81 m s^-2)/(1.9x10^-4 m * 0.18 m) = 2.26x10^7 Pa = 226.6 bar

c)

From Clapeyron´s expression we need to clear ΔT:

ΔT = ((Pf-Pi)*ΔV*Ti)/ΔHf = ((2.26x10^7 - 6.89x10^7)*1.5x10^-6*273.15)/6010 = -3.16 K

you can evaluate the new melting point of ice:

T = Ti + ΔT = 273.15 K - 3.16 K = 269.99 K

4 0
3 years ago
A 50.0 mL solution of 0.129 M KOH is titrated with 0.258 M HCl. Calculate the pH of the solution after the addition of each of t
kobusy [5.1K]

Answer:

A- pH = 13.12

B- pH = 12.91

C- pH = 12.71

D- pH = 12.43

E- pH = 11.55

F- pH = 7

G- pH = 2.46

H- pH = 1.88

Explanation:

This is a titration of a strong base with a strong acid. The neutralization reaction is: KOH (aq) + HCl (aq) →  H₂O(l) + KCl(aq)

Our pH at the equivalence point is 7, because we have made a neutral salt.

To determine the volume at that point we state the formula for titration:

mmoles of base = mmoles of acid

Volume of base  . M of base = Volume of acid . M of acid

50mL . 0.129M = 0.258 M . Volume of acid

Volume of acid = (50mL . 0.129M) / 0.258 M →  25 mL (Point <u>F</u>)

When we add 25 mL of HCl, our pH will be 7.

A- At 0 mL of acid, we only have base.

KOH → K⁺ + OH⁻

[OH⁻] = 0.129 M

To make more easy the operations we will use, mmol.

mol . 1000 = mmoles → mmoles / mL = M

- log 0.129 = 0.889

14 - 0.889 = 13.12

B-  In this case we are adding, (7 mL . 0.258M) = 1.81 mmoles of H⁺

Initially we have  0.129 M . 50 mL = 6.45 mmoles of OH⁻

1.81 mmoles of H⁺ will neutralize, the 6.45 mmoles of OH⁻ so:

6.45 mmol - 1.81 = 4.64 mmoles of OH⁻

This mmoles of OH⁻ are not at 50 mL anymore, because our volume has changed. (Now, we have 50 mL of base + 7 mL of acid) = 57 mL of total volume.

[OH⁻] = 4.64 mmoles / 57 mL = 0.0815 M

- log 0.0815 M = 1.09 → pOH

pH = 14 - pOH → 14 - 1.09 = 12.91

C- In this case we add (12.5 mL . 0.258M) = 3.22 mmoles of H⁺

<em>Our initial mmoles of OH⁻ would not change through all the titration. </em>

Then 6.45 mmoles of OH⁻ are neutralized by 3.22 mmoles of H⁺.

6.45 mmoles of OH⁻ - 3.22 mmoles of H⁺ = 3.23 mmoles of OH⁻

Total volume is: 50 mL of base + 12.5 mL = 62.5 mL

[OH⁻] = 3.23 mmol / 62.5 mL = 0.0517 M

- log  0.0517 = 1.29 → pOH

14 - 1.11 = 12.71

D- We add (18 mL . 0.258M) = 4.64 mmoles of H⁺

6.45 mmoles of OH⁻ are neutralized by 4.64 mmoles of H⁺.

6.45 mmoles of OH⁻ - 4.64 mmoles of H⁺ = 1.81 mmoles of OH⁻

Total volume is: 50 mL of base + 18 mL = 68 mL

[OH⁻] = 1.81 mmol / 68 mL = 0.0265 M

- log  0.0265 = 1.57 → pOH

14 - 1.57 = 12.43

E- We add (24 mL . 0.258M) = 6.19 mmoles of H⁺

6.45 mmoles of OH⁻ are neutralized by 6.19 mmoles of H⁺.

6.45 mmoles of OH⁻ - 6.19 mmoles of H⁺ = 0.26 mmoles of OH⁻

Total volume is: 50 mL of base + 24 mL = 74 mL

[OH⁻] = 0.26 mmol / 74 mL = 3.51×10⁻³ M

- log  3.51×10⁻³  = 2.45 → pOH

14 - 2.45 = 11.55

F- This the equivalence point.

mmoles of OH⁻ = mmoles of H⁺

We add (25 mL . 0.258M) = 6.45 mmoles of H⁺

All the OH⁻ are neutralized.

OH⁻  +  H⁺  ⇄   H₂O              Kw

[OH⁻] = √1×10⁻¹⁴   →  1×10⁻⁷  →  pOH = 7

pH → 14 - 7 = 7

G- In this case we have an excess of H⁻

We add (26 mL . 0.258M ) = 6.71 mmoles of H⁺

We neutralized all the OH⁻ but some H⁺ remain after the equilibrium

6.71 mmoles of H⁺ - 6.45 mmoles of OH⁻ = 0.26 mmoles of H⁺

[H⁺] = 0.26 mmol / Total volume

Total volume is: 50 mL + 26 mL → 76 mL

[H⁺] = 0.26 mmol / 76 mL → 3.42×10⁻³ M

- log 3.42×10⁻³ = 2.46 → pH

H- Now we add (29 mL . 0.258M) = 7.48 mmoles of H⁺

We neutralized all the OH⁻ but some H⁺ remain after the equilibrium

7.48 mmoles of H⁺ - 6.45 mmoles of OH⁻ = 1.03 mmoles of protons

Total volume is 50 mL + 29 mL = 79 mL

[H⁺] = 1.03 mmol / 79 mL → 0.0130 M

- log 0.0130 = 1.88 → pH

After equivalence point, pH will be totally acid, because we always have an excess of protons. Before the equivalence point, pH is basic, because we still have OH⁻ and these hydroxides, will be neutralized through the titration, as we add acid.

5 0
3 years ago
Please help!!!!!<br> Please help!!!!!
lesantik [10]

Answer:

1.B

2.A

3. B

Explanation:

1. A chemical bond is the physical phenomenon of chemical substances being held together by attraction of atoms to each other through sharing

2. so 2 is electrons of one atom are transferred permanently to another atom.meaning the answer is A

3. is When atoms combine by forming covalent bonds, the resulting collection of atoms is called a molecule. We can therefore say that a molecule is the simplest unit of a covalent compound.

3 0
3 years ago
Read 2 more answers
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