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3 years ago

7

The volume of a sample of gas measured at 35 C and 1.00 atm pressure is 2.00 L. What must the final temperature be in order for

the gas to have a final volume of 3.00 L at 1.00 atm pressure?

1 answer:

____ [38]3 years ago

6 0

Answer:

the final temperature is 465.2k

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3 years ago

How many moles of NaCI are there in 1.14*1024 in particles?

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3 years ago

Which sequence of group 18 elements demonstrates a gradual decrease in forces of attraction

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3 years ago

A quantity of 0.225 g of a metal M (molar mass = 27.0 g/mol) liberated 0.303 L of molecular hydrogen (measured at 17°C and 741 m

Lana71 [14]

Answer:

Oxide of M is $M_2O_3$ and sulfate of $M_2(SO_4)_3$

Explanation:

0.303 L of molecular hydrogen gas measured at 17°C and 741 mmHg.

Let moles of hydrogen gas be n.

Temperature of the gas ,T= 17°C =290 K

Pressure of the gas ,P= 741 mmHg= 0.9633 atm

Volume occupied by gas , V = 0.303 L

Using an ideal gas equation:

$PV=nRT$

$n=\frac{PV}{RT}=\frac{0.9633 atm\times 0.303 L}{0.0821 atm L/mol K\times 290 K}=0.01225 mol$

Moles of hydrogen gas produced = 0.01225 mol

$2M+2xHCl\rightarrow 2MCl_x+xH_2$

Moles of metal = $\frac{0.225 g}{27.0 g/mol}=8.3333 mol$

So, 8.3333 mol of metal M gives 0.01225 mol of hydrogen gas.

$\frac{8.3333}{0.01225 mol}=\frac{2}{x}$

x = 2.9 ≈ 3

$2M+6HCl\rightarrow 2MCl_3+3H_2$

$MCl_3\rightarrow M^{3+}+Cl^-$

Formulas for the oxide and sulfate of M will be:

Oxide of M is $M_2O_3$ and sulfate of $M_2(SO_4)_3$ .

3 0

4 years ago

What volume of 3.00 M HCl will form a solution with an acidic pH when mixed with 100 mL of 3.00 M NaOH

Maksim231197 [3]

<u>Answer:</u> The volume of acid should be less than 100 mL for a solution to have acidic pH

<u>Explanation:</u>

To calculate the volume of acid needed to neutralize, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is HCl

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH

We are given:

$n_1=1\\M_1=3.00M\\V_1=?mL\\n_2=1\\M_2=3.00M\\V_2=100mL$

Putting values in above equation, we get:

$1\times 3.00\times V_1=1\times 3.00\times 100\\\\V_1=\frac{1\times 3.00\times 100}{1\times 3.00}=100mL$

For a solution to be acidic in nature, the pH should be less than the volume of acid needed to neutralize.

Hence, the volume of acid should be less than 100 mL for a solution to have acidic pH

7 0

4 years ago