Why are bottles made from amorphous solid's such as plastic and glass

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

Read 2 more answers

What is the molarity of a solution if you take 12.0 grams of ca(no3)2 and mix it with 0.105 l of water?

Lunna [17]

The molarity of a solution if it tale 12.0 grams of Ca(No3)2 is calculated as below

molarity = moles/volume in liters
moles = mass/molar mass = 12.0 g/ 164 g/mol = 0.073 moles

molarity is therefore = 0.073/0.105 = 0.7 M

8 0

3 years ago

A ball is dropped from a height of 1.20 m and hits the floor. The ball compresses and then reforms to spring upwards from the fl

diamong [38]

Answer:

$\large \boxed{\text{98 m$\cdot$s}^{-2}}$

Explanation:

The formula for the velocity of the ball is

$v = \sqrt{2gh}$

1. Velocity at time of impact

$v = -\sqrt{2 \times 9.807 \times 1.20} = -\sqrt{23.54} = -\textbf{4.85 m/s}$

2. Velocity on rebound

The ball has enough upward velocity to reach a height of 0.86 m.

$v = \sqrt{2 \times 9.807 \times 0.86} = \sqrt{16.87} =\textbf{4.11 m/s}$

3. Acceleration

$a = \dfrac{\Delta v}{\Delta t} = \dfrac{4.11 - (-4.85)}{ 0.091} = \dfrac{8.96 }{0.091} =\textbf{98 m$\cdot$s}^{\mathbf{-2}}\\\\\text{The acceleration while the ball is in contact with the floor is $\large \boxed{\textbf{98 m$\cdot$s}^{\mathbf{-2}}}$}$

5 0

3 years ago

The crystallization of salt from evaporating seawater a. is an example of precipitation from a solution. b. is, in mineralogic t

Ad libitum [116K]

Answer:

is an example of precipitation from a solution.

Explanation:

3 0

3 years ago

What super power would you want to have and why?<br><br> I will give Brainly to the best answer

Darya [45]

Answer:

If I could have a super power it would be invisibility. Sometimes you wish you weren't there and if you were hiding from someone then they couldn't find you.

3 0

2 years ago