All categories

2 years ago

If I have 340 mL of a 1.5 M NaBr solution, what will the concentration be if I add 560 mL more water to it?

Chemistry

1 answer:

ipn [44]2 years ago

5 0

Answer:

0.5667 M ≅ 0.57 M.

Explanation:

It is known that the no. of millimoles of a solution before dilution is equal to the no. of millimoles of the solution after the dilution.

It can be expressed as:

(MV) before dilution = (MV) after dilution.

M before dilution = 1.5 M, V before dilution = 340 mL.

M after dilution = ??? M, V after dilution = 340 mL + 560 mL = 900 mL.

∴ M after dilution = (MV) before dilution/(V) after dilution = (1.5 M)(340 mL)/(900 mL) = 0.5667 M ≅ 0.57 M.

You might be interested in

The metal content of iron in ores can be determined by a redox procedure in which the sample is first oxidized with Br2 to conve

coldgirl [10]

Answer:

80.27%

Explanation:

Let's consider the following balanced equation.

2 Fe³⁺(aq) + Sn²⁺(aq) ⇒ 2Fe²⁺(aq) + Sn⁴⁺(aq)

First, we have to calculate the moles of Sn²⁺ that react.

$\frac{0.1015molSn^{2+} }{1L} .13.28 \times 10^{-3} L=1.348\times 10^{-3}molSn^{2+}$

We also know the following relations:

According to the balanced equation, 1 mole of Sn²⁺ reacts with 2 moles of Fe³⁺.
1 mole of Fe³⁺ is oxidized from 1 mole of Fe.
The molar mass of Fe is 55.84 g/mol.

Then, for 1.348 × 10⁻3 moles of Sn²⁺:

$1.348\times 10^{-3}molSn^{2+}.\frac{2molFe^{3+} }{1molSn^{2+} } .\frac{1molFe}{1molFe^{3+} } .\frac{55.84gFe}{1molFe} =0.1505gFe$

If there are 0.1505 g of Fe in a 0.1875 g sample, the mass percentage of Fe is:

$\frac{0.1505g}{0.1875g} \times 100 \% = 80.27\%$

5 0

4 years ago

Calculate the empirical formula of a compound that has a composition of 5.9% (by mass) hydrogen and 94.1% (by mass) oxygen.

adelina 88 [10]

Answer:

The empirical formula is the simplest form;

Given:

Oxygen O at 94.1% and

H at 5.9%

Assume 100grams.

94% = 0.941 x 100gm. = 94.1 gm x 1mole/16gm. = 5.88 moles of O

5.9% = 0.059 x 100gm. = 5.9gm. X 1moleH/1.002gm. = 5.88 moles of H

There is one mole of O for each mole of H so the empirical formula is $O_1H_1$

and written as OH.

8 0

3 years ago

Read 2 more answers

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

How many miles can you drive, if your car gets 21.3 miles per gallon and you have 12.0 gallons of gas?

Travka [436]

Answer:

255.6

Explanation:

If you have 12 gallons and get 21.3mpg,

-Multiply 21.3 by 12

-you can travel 255.6 miles before running out of gas.

-If you need to estimate, round up to 256 miles.

6 0

4 years ago

A chemist measures the energy change ?H during the following reaction:

Leno4ka [110]

Answer:

(A) endothermic

(A) Yes, absorbed

Explanation:

Let's consider the following thermochemical equation.

2 Fe₂O₃(s) ⇒ 4 FeO(s) + O₂(g) ΔH = 560 kJ

Since ΔH > 0, the reaction is endothermic.

We can establish the following relations:

560 kJ are absorbed when 2 moles of Fe₂O₃ react.
The molar mass of Fe₂O₃ is 160 g/mol.

Suppose 66.6 g of Fe₂O₃ react. The heat absorbed is:

$66.6g.\frac{1mol}{160g} .\frac{560kJ}{2mol} =117kJ$

5 0

3 years ago