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2 years ago

Which statement defines specific heat capacity for given sample?

Chemistry

2 answers:

ludmilkaskok [199]2 years ago

7 0

Answer:

D. the quantity of heat that is required to raise 1 g of the sample by 1*C (kelvin) at a constant pressure.

Explanation:

took the test

Aleks04 [339]2 years ago

6 0

Answer:

D. the quantity of heat that is required to raise 1 g of the sample by 1*C (kelvin) at a constant pressure.

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Consider the reaction 2CuCl2 + 4K - 2Cul + 4KCI + 12. If 4 moles of CuCl2 react with 4 moles of KI, what is the limiting reactan

Free_Kalibri [48]

Haha i’m trying to do the same one i’ll make sure if i find out how too to get back to you!

8 0

2 years ago

Using the balanced equation for the combustion of ethane: 2C2H6 + 7O2 → 4CO2 + 6H2O, how many moles of O2 needed to produce 12 m

pickupchik [31]

Answer:

14 moles of oxygen needed to produce 12 moles of H2O.

Explanation:

We are given that balance eqaution

$2C_2H_6+7O_2\rightarrow 4CO_2+6H_2O$

We have to find number of moles of O2 needed to produce 12 moles of H2O.

From given equation

We can see that

6 moles of H2O produced by Oxygen =7 moles

1 mole of H2O produced by Oxygen= $\frac{7}{6}$ moles

12 moles of H2O produced by Oxygen= $\frac{7}{6}\times 12$ moles

12 moles of H2O produced by Oxygen= $7\times 2$ moles

12 moles of H2O produced by Oxygen=14 moles

Hence, 14 moles of oxygen needed to produce 12 moles of H2O.

3 0

2 years ago

An aqueous CsCl solution is 8.00 wt% CsCl and has a density of 1.0643 g/mL at 20°C. What is the boiling point of this solution?

umka2103 [35]

<u>Answer:</u> The boiling point of solution is 100.53

<u>Explanation:</u>

We are given:

8.00 wt % of CsCl

This means that 8.00 grams of CsCl is present in 100 grams of solution

Mass of solvent = (100 - 8) g = 92 grams

The equation used to calculate elevation in boiling point follows:

$\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of pure solution}$

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

$\Delta T_b=iK_bm$

Or,

$\text{Boiling point of solution}-\text{Boiling point of pure solution}=i\times K_b\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

Boiling point of pure solution = 100°C

i = Vant hoff factor = 2 (For CsCl)

$K_b$ = molal boiling point elevation constant = 0.51°C/m

$m_{solute}$ = Given mass of solute (CsCl) = 8.00 g

$M_{solute}$ = Molar mass of solute (CsCl) = 168.4 g/mol

$W_{solvent}$ = Mass of solvent (water) = 92 g

Putting values in above equation, we get:

$\text{Boiling point of solution}-100=2\times 0.51^oC/m\times \frac{8.00\times 1000}{168.4g/mol\times 92}\\\\\text{Boiling point of solution}=100.53^oC$