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avanturin [10]
3 years ago
10

Consider two aqueous solutions of NaCl. Solution 1 is 4.00 M and solution 2 is 0.10 M. In what ratio (solution 1 to solution 2)

must these solutions be mixed in order to produce a 0.86 M solution of NaCl
Chemistry
1 answer:
Maru [420]3 years ago
6 0

Answer:

The ratio of solution 1 to solution 2 is 24.20 to 100.00.

Explanation:

We will mix V₁ (L) of solution 1 with V₂ (L) of solution 2 to get the final solution.

So the mole concentration in the final solution is calculated as below, note that C₁ is the concentration of solution 1, and C₂ is the concentration of solution 2

[M] = \frac{V_{1} C_{1} +V_{2}C_{2}}{V_{1}+V_{2}} = \frac{4 V_{1} + 0.1 V_{2}}_{V_{1}+V_{2}}}=0.86

Then we can calculate for the ratio

\frac{V_{1}}{V_{2}}=\frac{0.86-0.10}{4.00-0.86}  =\frac{0.76}{3.14} or \frac{24.20}{100.00}

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A container is filled with a mixture of gases:
strojnjashka [21]

Answer:

The partial pressure of chlorine gas in the mixture is 1.55 atm.

Explanation:

Partial pressure of oxygen gas = p_1=1.0 atm

Partial pressure of nitrogen gas = p_2=0.75 atm

Partial pressure of chlorine gas = p_3=?

Total pressure of the mixture of gases = P = 3.30 atm

Using Dalton's law of partial pressure:

P=p_1+p_2+p_3

3.30 atm=1.0 atm+ 0.75 atm+p_3

p_3=3.30atm - (1.0 atm+ 0.75 atm)=1.55 atm

The partial pressure of chlorine gas in the mixture is 1.55 atm.

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3 years ago
A runner runs for a distance of 350 yards and then runs for an additional 200 m. What is the total difference covered by the run
Rudiy27
Answer: 520.04 meters.

If you convert 350 yards to meters, you get 320.04.

320.04m + 200m = 520.04m.

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3 years ago
Information about how a chemical reacts under high pressure can be found in the ""reactivity"" section of a msds. True or false?
dezoksy [38]

Information about how a chemical reacts under high pressure can be found in the "reactivity" section of a MSDS: True.

<h3>What is MSDS?</h3>

MSDS is an abbreviation for Material Safety Data Sheet and it can be defined as a text-based document which is designed and developed to provide more information with respect to the chemical and physical properties of an equipment, chemical compound, or substance, based on the following:

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In this context, we can infer and logically deduce that it is true that the information about how a chemical react would under high pressure can be found in the "reactivity" section of a Material Safety Data Sheet (MSDS).

Read more on Material Safety Data Sheet here: brainly.com/question/3282390

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3 years ago
If 3.0 atm of pure HN3(g) is decomposed initially, what is the final total pressure in the reaction container? What are the part
lawyer [7]

This is an incomplete question, here is a complete question.

Hydrogen azide, HN₃, decomposes on heating by thefollowing unbalanced reaction:

HN_3(g)\rightarrow N_2(g)+H_2(g)

If 3.0 atm of pure HN₃ (g) is decomposed initially,what is the final total pressure in the reaction container? Whatare the partial pressures of nitrogen and hydrogen gas? Assume thatthe volume and temperature of the reaction container are constant.

Answer : The partial pressure of N_2 and H_2 gases are, 4.5 atm and 1.5 atm respectively.

Explanation :

The given unbalanced chemical reaction is:

HN_3(g)\rightarrow N_2(g)+H_2(g)

This reaction is an unbalanced chemical reaction because in this reaction number of hydrogen and nitrogen atoms are not balanced on both side of the reaction.

In order to balance the chemical equation, the coefficient '2' put before the HN_3 and the coefficient '3' put before the N_2 then we get the balanced chemical equation.

The balanced chemical reaction will be,

2HN_3(g)\rightarrow 3N_2(g)+H_2(g)

As we are given:

The pressure of pure HN_3 = 3.0 atm

p_{Total}=2\times p_{HN_3}=2\times 3.0atm=6.0atm

From the reaction we conclude that:

Number of moles of N_2 = 3 mol

Number of moles of H_2 = 1 mol

Now we have to calculate the mole fraction of N_2 and H_2

\text{Mole fraction of }N_2=\frac{\text{Moles of }N_2}{\text{Moles of }N_2+\text{Moles of }H_2}=\frac{3}{3+1}=0.75

and,

\text{Mole fraction of }H_2=\frac{\text{Moles of }H_2}{\text{Moles of }N_2+\text{Moles of }H_2}=\frac{1}{3+1}=0.25

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

According to the Raoult's law,

p_i=X_i\times p_T

where,

p_i = partial pressure of gas

p_T = total pressure of gas  = 6.0 atm

X_i = mole fraction of gas

p_{N_2}=X_{N_2}\times p_T

p_{N_2}=0.75\times 6.0atm=4.5atm

and,

p_{H_2}=X_{H_2}\times p_T

p_{H_2}=0.25\times 6.0atm=1.5atm

Thus, the partial pressure of N_2 and H_2 gases are, 4.5 atm and 1.5 atm respectively.

8 0
3 years ago
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