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Ksenya-84 [330]

2 years ago

8

A sample of gas is enclosed in a container of fixed volume. Identify which of the following statements are true. Check all that

apply.If the container is heated, the gas particles will lose kinetic energy and temperature will increase.

1 answer:

daser333 [38]2 years ago

7 0

Explanation:

if container is heated, temperature increases,gas particles gain kinetic energy, collision increases, pressure increases

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Assuming ideal solution behavior, what is the freezing point of a solution of 9.04 g of I2 in 75.5 g of benzene?

cluponka [151]

Answer:

3.1°C

Explanation:

Using freezing point depression expression:

ΔT = Kf×m×i

<em>Where ΔT is change in freezing point, Kf is freezing point depression constant (5.12°c×m⁻¹), m is molality of the solution and i is Van't Hoff factor constant (1 For I₂ because doesn't dissociate in benzene).</em>

Molality of 9.04g I₂ (Molar mass: 253.8g/mol) in 75.5g of benzene (0.0755kg) is:

9.04g ₓ (1mol / 253.8g) = 0.0356mol I₂ / 0.0755kg = 0.472m

Replacing in freezing point depression formula:

ΔT = 5.12°cm⁻¹×0.472m×1

ΔT = 2.4°C

As freezing point of benzene is 5.5°C, the new freezing point of the solution is:

5.5°C - 2.4°C =

<h3>3.1°C</h3>

<em />

3 0

2 years ago

Ammonia gas is compressed from 21°C and 200 kPa to 1000 kPa in an adiabatic compressor with an efficiency of 0.82. Estimate the

Evgen [1.6K]

Explanation:

It is known that efficiency is denoted by $\eta$ .

The given data is as follows.

$\eta$ = 0.82, $T_{1}$ = (21 + 273) K = 294 K

$P_{1}$ = 200 kPa, $P_{2}$ = 1000 kPa

Therefore, calculate the final temperature as follows.

$\eta = \frac{T_{2} - T_{1}}{T_{2}}$

0.82 = $\frac{T_{2} - 294 K}{T_{2}}$

$T_{2}$ = 1633 K

Final temperature in degree celsius = $(1633 - 273)^{o}C$

= $1360^{o}C$

Now, we will calculate the entropy as follows.

$\Delta S = nC_{v} ln \frac{T_{2}}{T_{1}} + nR ln \frac{P_{1}}{P_{2}}$

For 1 mole, $\Delta S = C_{v} ln \frac{T_{2}}{T_{1}} + R ln \frac{P_{1}}{P_{2}}$

It is known that for $NH_{3}$ the value of $C_{v}$ = 0.028 kJ/mol.

Therefore, putting the given values into the above formula as follows.

$\Delta S = C_{v} ln \frac{T_{2}}{T_{1}} + R ln \frac{P_{1}}{P_{2}}$

= $0.028 kJ/mol \times ln \frac{1633}{294} + 8.314 \times 10^{-3} kJ \times ln \frac{200}{1000}$

= 0.0346 kJ/mol

or, = 34.6 J/mol (as 1 kJ = 1000 J)

Therefore, entropy change of ammonia is 34.6 J/mol.

3 0

2 years ago

What is the molarity of a naoh solution if 11.9 ml of a 0.220 m h2so4 solution is required to neutralize a 25.0-ml sample of the

bulgar [2K]

0.355M x 0.0282L= 0.01 moles of H2SO4. Remember sulphuric acid is diprotic so it will release 2 from each molecule.
<span>So moles of protons = 0.01 x 2 = 0.02 moles of H+ </span>
<span>For neutralization: moles H+ = moles OH- </span>
<span>Therefore moles of NaOH = 0.02 </span>
<span>conc = moles / volume </span>
<span>Conc NaOH = 0.02 / 0.025L = 0.8M </span>

4 0

2 years ago

Read 2 more answers

The double bond between carbon and oxygen is similar to an alkene C-C, except that C o is: a) shorter and weaker. b) shorter and

mina [271]

Answer:

the double bond between c and o is shorter and weaker

Explanation:

this is because the bond between c and o involves unequal sharing of electrons whole c and c involves hybridization sp2 of orbitals and also catenation phenomenon in which carbon could form long chain with it's other carbon

5 0

2 years ago

One liter of ocean water contains 35.06 g of salt. What volume of ocean water would contain 1.00 kg of salt? Express your answer

sweet [91]

Answer:

28.52 L

Explanation:

First, let's calculate the density of the ocean, which is the mass divided by the volume:

d = m/V

d = 35.06/1

d = 35.06 g/L

So, for a mass of 1.00 kg = 1000.00 g

d = m/V

35.06 = 1000.00/V

V = 1000.00/35.06

V = 28.52 L

How all the data are expressed with two significant figures, the volume must also be expressed with two.

7 0

2 years ago