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kotegsom [21]
3 years ago
12

The molality of a solution that is prepared by dissolving 64.2 mL of methanol (d = 0.791 g/mL, MM = 32 g/mol) in 137.6 g ethanol

is ______ m. Report your answer with three significant figures. Do not include units.
Chemistry
1 answer:
elena-s [515]3 years ago
3 0

Answer:

The molality is 11.5

Explanation:

Molality of solution = number of moles of methanol ÷ mass of ethanol in kilograms

Number of moles of methanol = density × volume/MM = 0.791×64.2/32 = 1.587 moles

Mass of ethanol = 137.6 g = 137.6/1000 = 0.1376 kg

Molality of solution = 1.587 ÷ 0.1376 = 11.5 (to 3 significant figures)

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So, a gas, and an
valina [46]

Answer:

d) cut the large sized Cu solid into smaller sized pieces

Explanation:

The aim of the question is to select the right condition for that would increases the rate of the reaction.

a) use a large sized piece of the solid Cu

This option is wrong. Reducing the surface area decreases the reaction rate.

b) lower the initial temperature below 25 °C for the liquid reactant, HNO3

Hugher temperatures leads to faster reactions hence this option is wrong.

c) use a 0.5 M HNO3 instead of 2.0 M HNO3

Higher concentration leads to increased rate of reaction. Hence this option is wrong.

d) cut the large sized Cu solid into smaller sized pieces

This leads to an increased surface area of the reactants, which leads to an increased rate of the reaction. This is the correct option.

5 0
3 years ago
The mole is defined as the amount of a substance containing the same number of particles as exactly 12 g of C-12. The amu is def
Pavlova-9 [17]

Answer:

Because it wouldn't make any sense

Explanation:

First of all, I think it's important to highlight the definition of isotope.

Isotope (Wikipedia): Variants of a particular chemical element <u>which differ in neutron number</u>...

This means that two isotopes are the same element but have different net mass per atom, due to the different number of neutrons.

Therefore, it's important to make the definition on the same isotope so that the proportion is equal. If the definition would be made on different isotopes, the proportion wouldn't have any sense. Let me be clear with this example:

mass in grams of a C-12 atom = 1.9944235 × 10 ^ -23 g  --> this is the mass of a single C atom.

By definition --> 1 mol of anything = 6.02 x 10 ^ 23 particles of anything

Therefore, we know how much a single C atom weights. How many grams do you think that 6.02 x 10 ^ 23 atoms of C (i.e a mol of C) could weight??

1 single C atom ----------------------------- 1.9944235 × 10 ^ -23 g

6.02 x 10 ^ 23 atoms of C   ------------ <u>12.006 g  !!</u>

These 12 g is the same quantity than above! Therefore, 1 mol of C weights 12 g. If the definition were made with 13 g of C-13 (the other C isotope), these numbers will not be the same --> There would be a contradiction.

Regarding the second question, we need to search Ne-20 atomic mass in grams -->  3,3509177 × 10 ^ -23 g

Hence, if we follow the same rule, the amu would be 1/12 of Ne-20.

[ 3,3509177 × 10 ^ -23 g ] / 12 = 2.79 ^ -24 g

3 0
3 years ago
Difference between shell and subshell?
juin [17]
The "sub shells" are the orientations and shapes for your orbitals, going in order by Shells are a collection of subshells with the same principle quantum number, and subshells are a collection of orbitals with the same principle quantum number and angular momentum quantum number. Hope this helps :)
3 0
3 years ago
Type the correct answer in the box. Express your answer to three significant figures.
VladimirAG [237]

<u>Given:</u>

Mass of calcium nitrate (Ca(NO3)2) = 96.1 g

<u>To determine:</u>

Theoretical yield of calcium phosphate, Ca3(PO4)2

<u>Explanation:</u>

Balanced Chemical reaction-

3Ca(NO3)2 + 2Na3PO4 → 6NaNO3 + Ca3(PO4)2

Based on the reaction stoichiometry:

3 moles of Ca(NO3)2 produces 1 mole of Ca3(PO4)2

Now,

Given mass of Ca(NO3)2 =  96.1 g

Molar mass of Ca(NO3)2 =  164 g/mol

# moles of ca(NO3)2 = 96.1/164 = 0.5859 moles

Therefore, # moles of Ca3(PO4)2 produced = 0.0589 * 1/3 = 0.0196 moles

Molar mass of Ca3(PO4)2 = 310 g/mol

Mass of Ca3(PO4)2 produced = 0.0196 * 310 = 6.076 g

Ans: Theoretical yield of Ca3(PO4)2 = 6.08 g



7 0
3 years ago
Read 2 more answers
A solution was prepared by dissolving 0.800 g of sulfur S8, in 100.0 g of acetic acid, HC2H3O2. Calculate the freezing point and
sammy [17]

<u>Answer:</u> The freezing point of solution is 16.5°C and the boiling point of solution is 118.2°C

<u>Explanation:</u>

To calculate the molality of solution, we use the equation:

Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}

Where,

m_{solute} = Given mass of solute (S_8) = 0.800 g

M_{solute} = Molar mass of solute (S-8) = 256.52 g/mol

W_{solvent} = Mass of solvent (acetic acid) = 100.0 g

Putting values in above equation, we get:

\text{Molality of solution}=\frac{0.800\times 1000}{256.52\times 100.0}\\\\\text{Molality of solution}=0.0312m

  • <u>Calculation for freezing point of solution:</u>

Depression in freezing point is defined as the difference in the freezing point of water and freezing point of solution.

\Delta T_f=\text{freezing point of acetic acid}-\text{Freezing point of solution}

To calculate the depression in freezing point, we use the equation:

\Delta T_f=iK_fm

or,

\text{Freezing point of acetic acid}-\text{Freezing point of solution}=iK_fm

where,

Freezing point of acetic acid = 16.6°C

i = Vant hoff factor = 1 (for non-electrolyte)

K_f = molal freezing point depression constant = 3.59°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

16.6^oC-\text{freezing point of solution}=1\times 3.59^oC/m\times 0.0312m\\\\\text{Freezing point of solution}=16.5^oC

Hence, the freezing point of solution is 16.5°C

  • <u>Calculation for boiling point of solution:</u>

Elevation in boiling point is defined as the difference in the boiling point of solution and freezing point of pure solution.

The equation used to calculate elevation in boiling point follows:

\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of acetic acid}

To calculate the elevation in boiling point, we use the equation:

\Delta T_b=iK_bm

or,

\text{Boiling point of solution}-\text{Boiling point of acetic acid}=iK_fm

where,

Boiling point of acetic acid = 118.1°C

i = Vant hoff factor = 1 (for non-electrolyte)

K_f = molal boiling point elevation constant = 3.08°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

\text{Boiling point of solution}-118.1^oC=1\times 3.08^oC/m\times 0.0312m\\\\\text{Boiling point of solution}=118.2^oC

Hence, the boiling point of solution is 118.2°C

5 0
3 years ago
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