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ANEK [815]
2 years ago
6

If 50.0 g of oxygen and 50.0 g of hydrogen react to produce water.

Chemistry
1 answer:
Elan Coil [88]2 years ago
6 0

m(O2)=50g

m(H2)=50g

m(H2O)-?

n2(H2)-?

2H2 + O2 = 2H2O

n(O2)= m (O2)/M(O2) =50g / 32 g/mol= 1,56 mol.

n(H2)= m (H2)/M(H2) =50g / 2 g/mol= 25 mol.

Since oxygen gas is the limiting reactant,

n(O2)< 2 n(H2) from reaction.

n(H2O)= 2n(O2)= 2n (H2 reac.)=2*1,56 mol=3,12mol.

m(H2O)=n(H2O)*M(H2O)= 3,12mol* 18 g/mol.

n2(H2)= n(H2) - n (H2 reac.)=25mol - 3,12mol=21,88mol.

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A sample of nitric acid has a mass of 8.2g. It is dissolved in 1L of water. A 25mL aliquot of this acid is titrated with NaOH. T
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Answer:

18.075 mL of NaOH solution was added to achieve neutralization

Explanation:

First, let's formulate the chemical reaction between nitric acid and sodium hydroxide:

NaOH + HNO3 → NaNO3 + H2O

From this balanced equation we know that 1 mole of NaOH reacts with 1 mole of HNO3 to achieve neutralization. Let's calculate how many moles we have in the 25 mL aliquot to be titrated:

63.01 g of HNO3 ----- 1 mole

8.2 g of HNO3 ----- x = (8.2 g × 1 mole)/63.01 g = 0.13014 moles of HNO3

So far we added 8.2 grams of nitric acid (0.13014 moles) in 1 L of water.

1000 mL solution ---- 0.13014 moles of HNO3

25 mL (aliquot) ---- x = (25 mL× 0.13014 moles)/1000 mL = 0.0032535 moles

So, we now know that in the 25 mL aliquot to be titrated we have 0.0032535 moles of HNO3. As we stated before, 1 mole of NaOH will react with 1 mole of HNO3, hence 0.0032535 moles of HNO3 have to react with 0.0032535 moles of NaOH to achieve neutralization. Let's calculate then, in which volume of the given NaOH solution we have 0.0032535 moles:

0.18 moles of NaOH ----- 1000 mL Solution

0.0032535 moles---- x=(0.0032535moles×1000 mL)/0.18 moles = 18.075mL

As we can see, we need 18.075 mL of a 0.18 M NaOH solution to titrate a 25 mL aliquot of the prepared HNO3 solution.

4 0
3 years ago
Which of the following is not apart of daltons atomic theory
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3. Atoms change into other atoms in chemical reactions.

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dem82 [27]

Answer:

Rate = k [HO2]

Rate constant = 0.8456us-1

Explanation:

Time(us) 0 0.6 1.0 1.4 1.8 2.4

[HO2](uM) 8.5 5.1 3.6 2.6 1.1 1.1

The rate law is given as;

Rate = k [HO2]^x

Where x signify the order of reaction.

For an order of reaction, the rate constant is constant for all concentrations. We are going to use this to obtain the order of reaction.

Zero Order:

[A] = [A]o -kt

5.1 = 8.5 - k(0.6)

-k (0.6) = 5.1 - 8.5

k = 5.67

3.6 = 5.1 - k(0.4)

-k (0.4) = 3.6 - 5.1

k = 3.75

The fact that the rate constant was not constant means the reaction is not a zero order reaction.

First Order:

ln[A] = ln[A]o -kt

(5.1) = ln(8.5) - k(0.6)

-k (0.6) = ln(5.1) - ln(8.5)

k = 0.8524

ln(3.6) = ln(5.1) - k(0.4)

-k (0.4) = ln(3.6) - ln(5.1)

k = 0.8708

ln(2.6) = ln(3.6) - k (0.4)

-k (0.4) = ln(2.6) - ln(3.6)

k = 0.8136

From the three calculations we see that the value of the rate constant is fairly constant in the range of 0.8 This means our reaction is a first order reaction.

The rate law is given as;

Rate = k [HO2]

We can represent the rate constant as the average of the three rate constants calculated above;

Rate constant = (0.8136 + 0.8708 + 0.8524 / 3)

Rate constant = 0.8456us-1

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