Question 3<br> 6.06 X 103 L<br> How many significant figures are there

PLEASE ASAP. CHM help!

vichka [17]

Answer:

34

Explanation:

6 0

4 years ago

QUESTION 20

Assoli18 [71]

Answer:

49.56

Explanation:

5 0

3 years ago

How many moles of water are in 1.23 x 10^18 water molecules?

TiliK225 [7]

<span>1.23x10^24 atoms/6.022x10^23 atom/mol = 2.04 mol H20 </span>

3 0

4 years ago

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A bottle of wine contains 9.81 grams of C2H5OH, dissolved in 87.5 grams of water. The final volume of the solution is 100.0 mL.

dimulka [17.4K]

Answer:

[EtOH] = 2.2M and Wt% EtOH = 10.1% (w/w)

Explanation:

1. Molarity = moles solute / Volume solution in Liters

=> moles solute = mass solute / formula weight of solute = 9.8g/46g·mol⁻¹ = 0.213mol EtOH

=> volume of solution (assuming density of final solution is 1.0g/ml) ...

volume solution = 9.81gEtOH + 87.5gH₂O = 97.31g solution x 1g/ml = 97.31ml = 0.09731 Liter solution

Concentration (Molarity) = moles/Liters = 0.213mol/0.09731L = 2.2M in EtOH

2. Weight Percent EtOH in solution (assuming density of final solution is 1.0g/ml)

From part 1 => [EtOH] = 2.2M in EtOH = 2.2moles EtOH/1.0L soln

= {(2.2mol)(46g/mol)]/1000g soln] x 100% = 10.1% (w/w) in EtOH.

3 0

3 years ago

In an experiment, a 0.5297 g sample of diphenylacetylene (C14H10) is burned completely in a bomb calorimeter. The calorimeter is

V125BC [204]

Answer:

the Molar heat of Combustion of diphenylacetylene $(C_{14}H_{10})$ = $-6.931 *10^3 \ kJ/mol$

Explanation:

Given that:

mass of diphenylacetylene $(C_{14}H_{10})$ = 0.5297 g

Molar Mass of diphenylacetylene $(C_{14}H_{10})$ = 178.21 g/mol

Then number of moles of diphenylacetylene $(C_{14}H_{10})$ = $\frac{mass}{molar \ mass}$

= $\frac{0.5297 \ g }{178.24 \ g/mol}$

= 0.002972 mol

By applying the law of calorimeter;

Heat liberated by 0.002972 mole of diphenylacetylene $(C_{14}H_{10})$ = Heat absorbed by $H_2O$ + Heat absorbed by the calorimeter

Heat liberated by 0.002972 mole of diphenylacetylene $(C_{14}H_{10})$ = msΔT + cΔT

= 1369 g × 4.184 J g⁻¹°C⁻¹ × (26.05 - 22.95)°C + 916.9 J/°C (26.05 - 22.95)°C

= 17756.48 J + 2842.39 J

= 20598.87 J

Heat liberated by 0.002972 mole of diphenylacetylene $(C_{14}H_{10})$ = 20598.87 J

Heat liberated by 1 mole of diphenylacetylene $(C_{14}H_{10})$ will be = $\frac{20598.87 \ J}{0.002972 \ mol}$

= 6930979.139 J/mol

= 6930.98 kJ/mol

Since heat is liberated ; Then, the Molar heat of Combustion of diphenylacetylene $(C_{14}H_{10})$ = $-6.931 *10^3 \ kJ/mol$

3 0

3 years ago

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Question 3 6.06 X 103 L How many significant figures are there