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Brilliant_brown [7]

3 years ago

9

A mass of oil is 20 grams and the volume is 55m3. From this information, calculate the density of oil? Write your answer in 2 si

gnificant figures and then into scientific notation?

2 answers:

Vikentia [17]3 years ago

8 0

Answer:

0.36363636363m3 ??

Vika [28.1K]3 years ago

3 0

Mass=20g=0.02kg

Volume=55m^3

$\\ \sf\longmapsto Density=\dfrac{Mass}{Volume}$

$\\ \sf\longmapsto Density=\dfrac{0.02}{55}$

$\\ \sf\longmapsto Density=0.0003kg/m^3$

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A mole is __<br> - objects.<br> Answer here

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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 $(C_{14}H_{10})$ = $-6.931 *10^3 \ kJ/mol$

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3 years ago

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Consider the following equilibrium: H2CO3+H2O = H3O+HCO3^-1. What is the correct equilibrium expression?

mylen [45]

Equilibrium expression is $Keq = \frac{[H3O+][HCO3^-]}{[H2CO3]}\\$

<u>Explanation:</u>

Equilibrium expression is denoted by Keq.

Keq is the equilibrium constant that is defined as the ratio of concentration of products to the concentration of reactants each raised to the power its stoichiometric coefficients.

Example -

aA + bB = cC + dD

So, Keq = conc of product/ conc of reactant

$Keq = \frac{[C]^c [D]^d}{[A]^a [B]^b}$

So from the equation, H₂CO₃+H₂O = H₃O+HCO₃⁻¹

$Keq = \frac{[H3O^+]^1 [HCO3^-]^1}{[H2CO3]^1 [H2O]^1}$

The concentration of pure solid and liquid is considered as 1. Therefore, concentration of H2O is 1.

Thus,

$Keq = \frac{[H3O+][HCO3^-]}{[H2CO3]}\\$

Therefore, Equilibrium expression is $Keq = \frac{[H3O+][HCO3^-]}{[H2CO3]}\\$

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3 years ago