All categories

2 years ago

A student wants to make a fair comparison of the reactivity of the metals with

Chemistry

1 answer:

Otrada [13]2 years ago

7 0

Answer:

the concentration of hydrochloric acid

the temperature in heating

You might be interested in

A. 1

Lubov Fominskaja [6]

I think it’s B. I’m not 100% sure but I believe it is B! Srry if this is late btw.

8 0

2 years ago

The best reading for the volume of the gas in the syringe in the picture shown right is

spayn [35]

Answer:

48.5 mL , just did it on edge

6 0

3 years ago

What is the role of roughages in the balanced diet

ira [324]

Roughage, also known as Dietary fibre is thought to play an important role in digestive health, heart health, preventing Type 2 diabetes, as well as maintaining energy levels.

5 0

2 years ago

Which of these groups CANNOT harvest energy from the sun?

Tomtit [17]

Viruses..................

8 0

3 years ago

Read 2 more answers

The constant volume heat capacity of a gas can be measured by observing the decrease in temperature when it expands adiabaticall

miv72 [106K]

Answer:

The value is $C_p = 42. 8 J/K\cdot mol$

Explanation:

From the question we are told that

$\gamma = \frac{C_p }{C_v}$

The initial volume of the fluorocarbon gas is $V_1 = V$

The final volume of the fluorocarbon gas is $V_2 = 2V$

The initial temperature of the fluorocarbon gas is $T_1 = 298.15 K$

The final temperature of the fluorocarbon gas is $T_2 = 248.44 K$

The initial pressure is $P_1 = 202.94\ kPa$

The final pressure is $P_2 = 81.840\ kPa$

Generally the equation for adiabatically reversible expansion is mathematically represented as

$T_2 = T_1 * [ \frac{V_1}{V_2} ]^{\frac{R}{C_v} }$

Here R is the ideal gas constant with the value

$R = 8.314\ J/K \cdot mol$

$248.44 = 298.15 * [ \frac{V}{2V} ]^{\frac{8.314}{C_v} }$

=> $C_v = 31.54 J/K\cdot mol$

Generally adiabatic reversible expansion can also be mathematically expressed as

$P_2 V_2^{\gamma} = P_1 V_1^{\gamma}$

=> $[ 81.840 *10^3] [2V]^{\gamma} = [202.94 *10^3] V^{\gamma}$

=> $2^{\gamma} = 2.56$

=> $\gamma = 1.356$

$\gamma = \frac{C_p}{C_v} \equiv 1.356 = \frac{C_p}{31.54}$

=> $C_p = 42. 8 J/K\cdot mol$

3 0

3 years ago