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AfilCa [17]
2 years ago
13

What is the molarity of a solution that contains 3.4 moles of NaCL in 2.2 L of solution ?

Chemistry
1 answer:
jeka942 years ago
5 0

Answer:

what is the molarity of a solution that contains 3.4 moles of NaCL in 2.2 L of solution ?

Explanation:

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What is the percent by mass of potassium in KFe(CN)?
vladimir1956 [14]

Answer:

Explanation:

1. Add the atomic mass of all the elements.

39+55.8+12+14= 120.8

2.Divide atomic mass of potassium by total atomic mass

39/120.8= .323

2. Multiply by 100

.323*100= %32.3 %32.3

8 0
4 years ago
Which property of a substance is not altered by a physical change?
antiseptic1488 [7]
C. Composition

is the answer
3 0
4 years ago
How many grams are in 1.23 x 1020 atoms of arsenic?
Alina [70]

Given the data from the question, the mass of arsenic that contains 1.23×10²⁰ atoms is 0.0153 g

<h3>Avogadro's hypothesis </h3>

6.02×10²³ atoms = 1 mole of arsenic

But

1 mole of arsenic = 75 g

Thus, we can say that:

6.02×10²³ atoms = 75 g of arsenic

<h3>How to determine the mass that contains 1.23×10²⁰ atoms</h3>

6.02×10²³ atoms = 75 g of arsenic

Therefore,

1.23×10²⁰ atoms = (1.23×10²⁰ × 75) / 6.02×10²³ atoms)

1.23×10²⁰ atoms = 0.0153 g of arsenic

Thus, 1.23×10²⁰ atoms is present in 0.0153 g of arsenic

Learn more about Avogadro's number:

brainly.com/question/26141731

6 0
2 years ago
A sample of ammonia ^NH3h gas is completely decomposed to nitrogen and hydrogen gases over heated iron wool. If the total pressu
icang [17]

Answer : The partial pressure of N_2 and H_2 is, 216.5 mmHg and 649.5 mmHg

Explanation :

According to the Dalton's Law, the partial pressure exerted by component 'i' in a gas mixture is equal to the product of the mole fraction of the component and the total pressure.

Formula used :

p_i=X_i\times p_T

X_i=\frac{n_i}{n_T}

So,

p_i=\frac{n_i}{n_T}\times p_T

where,

p_i = partial pressure of gas

X_i = mole fraction of gas

p_T = total pressure of gas

n_i = moles of gas

n_T = total moles of gas

The balanced decomposition of ammonia reaction will be:

2NH_3\rightarrow N_2+3H_2

Now we have to determine the partial pressure of N_2 and H_2

p_{N_2}=\frac{n_{N_2}}{n_T}\times p_T

Given:

n_{N_2}=1\\\\n_{H_2}=3\\\\n_{T}=4\\\\p_T=866mmHg

p_{N_2}=\frac{1}{4}\times (866mmHg)=216.5mmHg

and,

p_{H_2}=\frac{n_{H_2}}{n_T}\times p_T

Given:

n_{H_2}=1\\\\n_{H_2}=3\\\\n_{T}=4\\\\p_T=866mmHg

p_{H_2}=\frac{3}{4}\times (866mmHg)=649.5mmHg

Thus, the partial pressure of N_2 and H_2 is, 216.5 mmHg and 649.5 mmHg

5 0
4 years ago
What organelles are involved in the production of proteins
DanielleElmas [232]

Answer:

endoplasmic reticulum (ER)

7 0
3 years ago
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