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2 years ago

Balance this eqaution:

Chemistry

1 answer:

klasskru [66]2 years ago

5 0

Answer:

H₂ + CuO → Cu + H₂O

Explanation:

We want to balance the given equation;

H₂ + CuO → Cu(OH)₂ + H₂O

To balance this we have to make sure the number of each element on the left is equal to the number of each element on the right.

Thus, the correct balanced equation is;

H₂ + CuO → Cu + H₂O

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Your job is to determine the concentration of ammonia in a commercial window cleaner. In the titration of a 25.0 mL sample of th

ruslelena [56]

Answer:

The initial concentration of ammonia is 0.14 M and the pH of the solution at equivalence point is 5.20

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ .....(1)

Molarity of HCl solution = 0.164 M

Volume of solution = 23.8 mL = 0.0238 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

$0.164M=\frac{\text{Moles of HCl}}{0.0238L}\\\\\text{Moles of HCl}=(0.146mol/L\times 0.0238L)=0.0035mol$

The chemical equation for the reaction of ammonia and HCl follows:

$NH_3+HCl\rightarrow NH_4^++Cl^-$

By Stoichiometry of the reaction:

1 mole of HCl reacts with 1 mole of ammonia

So, 0.0035 moles of HCl will react with = $\frac{1}{1}\times 0.0035=0.0035mol$ of ammonia

Calculating the initial concentration of ammonia by using equation 1:

Moles of ammonia = 0.0035 moles

Volume of solution = 25 mL = 0.025 L

Putting values in equation 1, we get:

$\text{Initial concentration of ammonia}=\frac{0.0035mol}{0.025L}=0.14M$

By Stoichiometry of the reaction:

1 mole of ammonia produces 1 mole of ammonium ion

So, 0.0035 moles of ammonia will react with = $\frac{1}{1}\times 0.0035=0.0035mol$ of ammonium ion

Calculating the concentration of ammonium ion by using equation 1:

Moles of ammonium ion = 0.0035 moles

Volume of solution = [23.8 + 25] mL = 48.8 mL = 0.0488 L

Putting values in equation 1, we get:

$\text{Molarity of ammonium ion}=\frac{0.0035mol}{0.0488L}=0.072M$

To calculate the acid dissociation constant for the given base dissociation constant, we use the equation:

$K_w=K_b\times K_a$

where,

$K_w$ = Ionic product of water = $10^{-14}$

$K_a$ = Acid dissociation constant

$K_b$ = Base dissociation constant = $1.8\times 10^{-5}$

$10^{-14}=1.8\times 10^{-5}\times K_a\\\\K_a=\frac{10^{-14}}{1.8\times 10^{-5}}=5.55\times 10^{-10}$

The chemical equation for the dissociation of ammonium ion follows:

$NH_4^+\rightarrow NH_3+H^+$

The expression of $K_a$ for above equation follows:

$K_a=\frac{[NH_3][H^+]}{[NH_4^+]}$

We know that:

$[NH_3]=[H^+]=x$

$[NH_4^+]=0.072M$

Putting values in above expression, we get:

$5.55\times 10^{-10}=\frac{x\times x}{0.072}\\\\x=6.32\times 10^{-6}M$

To calculate the pH concentration, we use the equation:

$pH=-\log[H^+]$

We are given:

$[H^+]=6.32\times 10^{--6}M$

$pH=-\log (6.32\times 10^{-6})\\\\pH=5.20$

Hence, the initial concentration of ammonia is 0.14 M and the pH of the solution at equivalence point is 5.20

5 0

3 years ago

How many joules of heat are required to raise the temperature of 174g of gold from 22°C to 85°C? The specific heat of gold is 0.

Slav-nsk [51]

Remembering the equation Q=MCdeltaT where
q=is the amount of heat energy
M=mass
C=specific heat
deltaT= change in temperature

Therefore, using the equation we can substitute values and solve for q.
Q= (15 grams) (0.129 J/(gx°C))(85-22)
Q=(15) ((0.129 J/(gx°C)) (63)
Q=121.9 Joules

The energy needed to raise the temperature of 15 grams of gold from 22 degrees Celsius to 85 degrees Celsius is then 121.9 Joules or 122 Joules (if rounded up).

7 0

3 years ago

I need helps plss!!!!

Leni [432]

Answer:

Copper = 2 Boron = 12 (copper and boron combined = 14)

Explanation:

the number after the element is the amount of atoms there is

8 0

2 years ago

What new information can you add to your definition of chemical change?

ivolga24 [154]

Something that can not be undone. Like baking bread..cant be undone.

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3 years ago

The 10 short columns in the middle of the periodic table make up the

nekit [7.7K]

Answer:

The transition elements or transition metals occupy the short columns in the center of the periodic table, between Group 2A and Group 3A.Explanation:

6 0

2 years ago

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Balance this eqaution:​

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