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Taya2010 [7]
3 years ago
10

To what volume should 5.06 mL of an 7.83 M sulfuric acid solution be diluted in order to obtain a final solution that is 2.36 M?

Chemistry
1 answer:
Akimi4 [234]3 years ago
4 0

the answer is 16.8 mL H2SO4 (aq).

the formula for dilation is M1*V1=M2*V2, where M is the molarity and V is the volume.

so plugging in...

1. (7.83)(5.06)=2.36*V2

2. multiply 7.83 and 5.06 then divide that by 2.36

3. V2= 16.788050...

4. to three sigfigs (because the problem has a minimum of 3) is 16.8 mL H2SO4 (aq).

i hope this is correct! i did it without pencil and paper and i’m currently learning this too! if it’s wrong i’m so truly sorry :(

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A 50/50 blend of engine coolant and water (by volume) is usually used in an automobile's engine cooling system. If a car's cooli
Diano4ka-milaya [45]

Answer:

\large \boxed{109.17 \, ^{\circ}\text{C}}

Explanation:

Data:

50/50 ethylene glycol (EG):water

V = 4.70 gal

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ρ(water) = 0.988 g/mL

Calculations:

The formula for the boiling point elevation ΔTb is

\Delta T_{b} = iK_{b}b

i is the van’t Hoff factor —  the number of moles of particles you get from 1 mol of solute. For EG, i = 1.

1. Moles of EG

\rm n = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1  gal}} \times \dfrac{\text{1000 mL}}{\text{1 L}} \times \dfrac{\text{1.11 g}}{\text{1 mL}} \times \dfrac{\text{1 mol}}{\text{62.07 g}} = \text{159 mol}

2. Kilograms of water

m = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1  gal}} \times \dfrac{\text{998 g}}{\text{1 L}} \times \dfrac{\text{1 kg}}{\text{1000 g}} = \text{8.88 kg}

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b =  \dfrac{\text{159 mol}}{\text{8.88 kg}} = \text{17.9 mol/kg}

4. Increase in boiling point

\rm \Delta T_{b} = iK_{b}b = 1 \times 0.512 \, \, ^{\circ}\text{C} \cdot kg \cdot mol^{-1} \, \times 17.9 \cdot mol \cdot kg^{-1} = 9.17 \, ^{\circ}\text{C}

5. Boiling point

\rm T_{b} = T_{b}^{\circ} + \Delta T_{b} = 100.00 \, ^{\circ}\text{C} + 9.17 \, ^{\circ}\text{C} = \mathbf{109.17 \, ^{\circ}C}\\\rm \text{The boiling point of the solution is $\large \boxed{\mathbf{109.17 \, ^{\circ}C}}$}

7 0
3 years ago
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You can detect it from the units of the different choices.

a- has the unit J/g.°C that is the unit of the specific heat capacity (c).

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c- has the unit g/mol which is the unit of the molar mass.

d- has the unit kJ/g which is the unit of the enthalpy divided by the no. of rams that is the specific entha;py of fusion.

<em>So, the right choice is: d-  334 kJ/g.</em>

5 0
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Answer:

think it helps you

<h2>Explanation:</h2>

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4 0
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Answer:

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Explanation:

Given:

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The electron configuration of Nitrogen and Iodine is given below;

N(7) = 1s^2,2s^22p^3\\I(53) = 1s^2,2s^22p^6,3s^23p^63d^{10},4s^24p^64d^{10},5s^25p^5

There are 5 electrons in valance shell of Nitrogen atom and 7 electrons in valance shell of Iodine atom.

So, 3 atom of Iodine shares 1 electron with 1 electrons of Nitrogen.

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Therefore, the option that isn't true is option B. "13C NMR spectra display peaks for only carbons that bear hydrogen atoms". This is false because 13C NMR will show every peak in the spectrum and it doesn't matter if it's only carbons that bear hydrogen atoms as everything will be shown.

4 0
2 years ago
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