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meriva
3 years ago
15

Two unknown molecular compounds were being studied. A solution containing 5.00 g of compound A in 100. g of water froze at a low

er temperature than a solution containing 5.00 g of compound B in 100. g of water. Which compound has the greater molar mass
Chemistry
1 answer:
LenaWriter [7]3 years ago
4 0

Answer:

Compound B has greater molar mass.

Explanation:

The depression in freezing point is given by ;

\Delta T_f=i\times k_f\times m..[1]

m=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Mass of solvent in kg}}

Where:

i = van't Hoff factor

k_f = Molal depression constant

m = molality of the solution

According to question , solution with 5.00 g of A in 100.0 grams of water froze at at lower temperature than solution with 5.00 g of B in 100.0 grams of water.

The depression in freezing point of solution with A solute: \Delta T_{f,A}

Molar mass of A = M_A

The depression in freezing point of solution with B solute: \Delta T_{f,B}

Molar mass of B = M_B

\Delta T_{f,A}>\Delta T_{f,B}

As we can see in [1] , that depression in freezing point is inversely related to molar mass of the solute.

\Delta T_f\propto \frac{1}{\text{Molar mass of solute}}

M_A

This means compound B has greater molar mass than compound A,

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Calculate the number of moles in a 2.98 g sample of aluminum.
Ksju [112]
Molar mass <span>Aluminium ( Al) = 26.98 g/mol

1 mole ------------ 26.98 g
? mole ----------- 2.98 g

moles Al = 2.98 x 1 / 26.98

moles Al = 2.98 / 26.98

= 0.110 moles

hope this helps!

</span>
3 0
3 years ago
Pentaborane-9, B5H9, is a colorless, highly reactive liquid that will burst into flame when exposed to oxygen. The reaction is 2
mina [271]

<u>Answer:</u> The amount of energy released per gram of B_5H_9 is -71.92 kJ

<u>Explanation:</u>

For the given chemical reaction:

2B_5H_9(l)+12O_2(g)\rightarrow 5B_2O_3(s)+9H_2O(l)

The equation used to calculate enthalpy change is of a reaction is:  

\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f_{(product)}]-\sum [n\times \Delta H^o_f_{(reactant)}]

The equation for the enthalpy change of the above reaction is:

\Delta H^o_{rxn}=[(5\times \Delta H^o_f_{(B_2O_3(s))})+(9\times \Delta H^o_f_{(H_2O(l))})]-[(2\times \Delta H^o_f_{(B_5H_9(l))})+(12\times \Delta H^o_f_{(O_2(g))})]

Taking the standard enthalpy of formation:

\Delta H^o_f_{(B_2O_3(s))}=-1271.94kJ/mol\\\Delta H^o_f_{(H_2O(l))}=-285.83kJ/mol\\\Delta H^o_f_{(B_5H_9(l))}=73.2kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol

Putting values in above equation, we get:

\Delta H^o_{rxn}=[(5\times (1271.94))+(9\times (-285.83))]-[(2\times (73.2))+(12\times (0))]\\\\\Delta H^o_{rxn}=-9078.57kJ

We know that:

Molar mass of pentaborane -9 = 63.12 g/mol

By Stoichiometry of the reaction:

If 2 moles of B_5H_9 produces -9078.57 kJ of energy.

Or,

If (2\times 63.12)g of B_5H_9 produces -9078.57 kJ of energy

Then, 1 gram of B_5H_9 will produce = \frac{-9078.57kJ}{(2\times 63.12)}\times 1g=-71.92kJ of energy.

Hence, the amount of energy released per gram of B_5H_9 is -71.92 kJ

8 0
3 years ago
Calculate the concentration of a solution that has 6.7 moles in a volume of 0.6 liters.
Yuri [45]

Answer:

11.2 M

Explanation:

Given data:

Number of moles = 6.7 mol

Volume of solution = 0.6 L

Concentration /Molarity = ?

Solution:

Molarity:

It is number of moles of solute in to per kg or litters of solution. It can be calculated by the following formula.

Molarity = number of moles / Volume in L

Now we will put the values in formula.

Molarity = 6.7 mol / 0.6 L

Molarity = 11.2 mol/L

Molarity = 11.2 M

6 0
4 years ago
If a force of 1.4N is applied to a block with a mass of 7kg, what is the acceleration of the block in meters per second squared?
Anton [14]

Answer:

<h2>0.2 m/s²</h2>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

a =  \frac{f}{m}  \\

f is the force

m is the mass

From the question we have

a =  \frac{1.4}{7} =  \frac{1}{5}   \\

We have the final answer as

<h3>0.2 m/s²</h3>

Hope this helps you

5 0
2 years ago
Read 2 more answers
Tap water used for drinking is considered in the DIL calculations.<br> A. True<br> B. False
guajiro [1.7K]

Tap water used for drinking is considered in the DIL calculations; true
7 0
4 years ago
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