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swat32
3 years ago
13

Choose the aqueous solution that has the highest boiling point.These are all solutions of nonvolatile solutes and you shouldassu

me ideal van't Hoff factors where applicable.
a. 0.100 m NaNO3
b. 0.100m Li2SO4
c. 0.200m C3H8O3
d. 0.060m Na3PO4
Chemistry
1 answer:
lesya [120]3 years ago
4 0

<u>Answer:</u> The correct answer is Option b.

<u>Explanation:</u>

The expression of elevation in boiling point is given as:

\Delta T_b=i\times k_b\times m

where,

\Delta T_b = Elevation in boiling point

i = Van't Hoff factor  

T_b = change in boiling point

k_b = boiling point constant

m = molality

For the given options:

<u>Option 1:</u>  0.100 m NaNO_3

Value of i = 2

So, molal concentration will be = (0.100)\times 2=0.200m

<u>Option 2:</u>  0.100 m Li_2SO_4

Value of i = 3

So, molal concentration will be = (0.100)\times 3=0.300m

<u>Option 3:</u>  0.200 m C_3H_8O_3

Value of i = 1  (for non-electrolytes)

So, molal concentration will be = (0.200)\times 1=0.200m

<u>Option 4:</u>  0.060 m Na_3PO_4

Value of i = 4

So, molal concentration will be = (0.060)\times 4=0.24m

As, the molal concentration of Li_2SO_4 is the highest, so its boiling point will be the highest.

Thus, the order of increasing boiling points follows:

NaNO_3=C_3H_8O_3

Hence, the correct answer is Option b.

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tigry1 [53]

Answer:

The volume will be 2600 L.

Explanation:

Initial temperature (T_{1}) = 25°C = (25 + 273) K = 298K

Final temperature (T_{2}) = 54°C = (54 + 273) K = 327 K

By using the combined gas law equation,

                                 \frac{P_{1} V_{1} }{T_{1} } = \frac{P_{2} V_{2} }{T_{2}  }

                       or, \frac{760* 1200}{298} = \frac{380* V_{2} }{327}

                      or, 3060 × 327 = 380 × V_{2}

                      or, V_{2} ≈ 2600 L

Hence the volume of the balloon will be about 2600 L.

6 0
3 years ago
Read 2 more answers
What type of reaction takes place when atoms or molecules rearrange to form new substances?
kap26 [50]
This is a chemical reaction.
7 0
4 years ago
Mg (50.0 g) and HCI (75.0 g) were added together to produce MgCl2 and H2
zimovet [89]

Answer:

Mass of hydrogen produced  = 2.1 g

Mass of excess reactant left = 25.2 g

Explanation:

Given data:

Mass of Mg = 50.0 g

Mass of HCl = 75.0 g

Mass of hydrogen produced = ?

Mass of excess reactant left = ?

Solution:

Chemical equation:

Mg + 2HCl      →       MgCl₂ + H₂

Number of moles of Mg:

Number of moles = mass/molar mass

Number of moles = 50 g/ 24 g/mol

Number of moles = 2.1 mol

Number of moles of HCl:

Number of moles = mass/molar mass

Number of moles = 75 g/ 36.5 g/mol

Number of moles = 2.1 mol

now we will compare the moles of hydrogen gas with both reactant.

                     Mg       :        H₂

                       1          :       1

                     2.1         :     2.1

                   HCl         :        H₂

                      2          :          1

                    2.1          :        1/2×2.1 = 1.05 mol

HCl is limiting reactant and will limit the yield of hydrogen gas.

Mass of hydrogen:

Mass = number of moles × molar mass

Mass=  1.05 mol ×2 g/mol

Mass = 2.1 g

Mg is present in excess.

Mass of Mg left:

         HCl            :           Mg

             2             :            1

            2.1            :         1/2×2.1 = 1.05

Out of 2.1 moles of Mg 1.05 react with HCl.

Moles of Mg left = 2.1 mol - 1.05 mol = 1.05 mol

Mass of Mg left:

Mass = number of moles × molar mass

Mass = 1.05 mol × 24 g/mol

Mass = 25.2 g

5 0
3 years ago
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Answer:

If there reacted 1.5 moles of O2, there will be produced 1.0 mol of Fe2O3

Explanation:

Step 1: Data given

Number of moles oxygen reacted = 1.5 moles

Step 2: The balanced equation

4Fe + 3O2 → 2Fe2O3

Step 3: Calculate moles of Fe2O3

For 4 moles Fe consumed, we need 3 moles of O2 to produce 2 moles of Fe2O3

For 1.5 moles O2 consumed, we'll have 2/3 * 1.5 = 1.0 mol of Fe2O3

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Answer:

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Explanation:

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3 years ago
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