Temperature has no effect on matter. A True B False

A 3L sample of an ideal gas at 178 K has a pressure of 0.3 atm. Assuming that the volume is constant, what is the approximate pr

harina [27]

Answer: The approximate pressure of the gas after it is heated to 278 K is 0.468 atm.

Explanation:

Given: $T_{1}$ = 178 K, $P_{1}$ = 0.3 atm

$T_{2}$ = 278 K, $P_{2}$ = ?

According to Gay Lussac law, at constant volume the pressure of a gas is directly proportional to the temperature.

Formula used to calculate the pressure is as follows.

$\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}\\$

Substitute the values into above formula is as follows.

$\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}\\\frac{0.3 atm}{178 K} = \frac{P_{2}}{278 K}\\P_{2} = 0.468 atm$

Thus, we can conclude that the approximate pressure of the gas after it is heated to 278 K is 0.468 atm.

3 0

3 years ago

The zinc within a copper-plated penny will dissolve in hydrochloric acid if the copper coating is filed down in several spots (s

Mrac [35]

Answer:

Mass of hydrogen gas evolved is 0.0749 grams.

Explanation:

Total pressure of the gases = p = 758 mmHg

Vapor pressure of water = 23.78 mmHg

Pressure of hydrogen gas ,P = p - 23.78 mmHg = 758 mmHg - 23.78 mmHg

P = 734.22 mmHg = $\frac{734.22}{760} atm=0.966 atm$

Temperature of of hydrogen gas ,T= 25°C =298.15 K

Volume of hydrogen gas = V = 0.949 L

Moles of hydrogen gas =n

PV = nRT (Ideal gas equation )

$n=\frac{PV}{RT}=\frac{0.966 atm\times 0.949 L}{0.0821 atm L/mol K\times 298.15 K}$

n = 0.03745 mol

Moles of hydrogen gas = 0.03745 mol

Mass of 0.03745 moles of hydrogen gas = 0.03745 mol × 2 g/mol = 0.0749 g

Mass of hydrogen gas evolved is 0.0749 grams.

3 0

3 years ago

A rigid tank contains 0.66 mol of oxygen (O2). Find the mass of oxygen that must be withdrawn from the tank to lower the pressur

dsp73

Answer:

12.8 g of $O_{2}$ must be withdrawn from tank

Explanation:

Let's assume $O_{2}$ gas inside tank behaves ideally.

According to ideal gas equation- $PV=nRT$

where P is pressure of $O_{2}$ , V is volume of $O_{2}$ , n is number of moles of $O_{2}$ , R is gas constant and T is temperature in kelvin scale.

We can also write, $\frac{V}{RT}=\frac{n}{P}$

Here V, T and R are constants.

So, $\frac{n}{P}$ ratio will also be constant before and after removal of $O_{2}$ from tank

Hence, $\frac{n_{before}}{P_{before}}=\frac{n_{after}}{P_{after}}$

Here, $\frac{n_{before}}{P_{before}}=\frac{0.66mol}{43atm}$ and $P_{after}=17atm$

So, $n_{after}=\frac{n_{before}}{P_{before}}\times P_{after}=\frac{0.66mol}{43atm}\times 17atm=0.26mol$

So, moles of $O_{2}$ must be withdrawn = (0.66 - 0.26) mol = 0.40 mol

Molar mass of $O_{2}$ = 32 g/mol

So, mass of $O_{2}$ must be withdrawn = $(32\times 0.40)g=12.8g$

7 0

3 years ago

What’s the answer to these 3 questions? thanks!

Citrus2011 [14]

Water has h bonding
H-H
Sodium fluoride
I think

4 0

2 years ago

Which of the following explains this observation?

NikAS [45]

It would be C i’m pretty sure

6 0

2 years ago