Calculate the pressure using the formula P=1.03+ mass on syringe/area of top of syringe

Use the combined gas law to calculate the "corrected" volume of hydrogen at STP

solong [7]

Answer:

We will derive the combined gas equation from the law of Gases

Boyle's law

P∝1/V

PV=Constant

Charles law

V∝T

Avogadro's Law

V∝n

By combining these three equations we get the combined gas equation

PV=nRT

V=nRT/P

If n=1 mole

V=RT/P

By putting the value of R, T, and P in the above equation we can calculate the volume of the gas at STP.

4 0

3 years ago

Type the correct answer in the box. Express your answer to three significant figures. A balloon is filled with 0.250 mole of air

katrin2010 [14]

Answer:

∴ The absolute pressure of the air in the balloon in kPa = 102.69 kPa.

Explanation:

We can solve this problem using the general gas law:

PV = nRT, where,

P is the pressure of the gas (atm),

V is the volume of the gas in L (V of air = 6.23 L),

n is the no. of moles of gas (n of air = 0.25 mole),

R is the general gas constant (R = 0.082 L.atm/mol.K),

T is the temperature of gas in K (T = 35 °C + 273 = 308 K).

∴ P = nRT / V = (0.25 mole)(0.082 L.atm/mol.K)(308 K) / (6.23 L) = 1.0135 atm.

Now, we should convert the pressure from (atm) to (kPa).

1.0 atm → 101.325 kPa,

1.0135 atm → ??? kPa.

∴ The absolute pressure of the air in the balloon in kPa = (101.325 kPa)(1.0135 atm) / (1.0 atm) = 102.69 kPa.

4 0

3 years ago

Read 2 more answers

A sample of Strontium-90 is found to have decayed to one-fourth of its original amount after 58.2 years. What is the half-life o

mafiozo [28]

The half-life of strontium-90 is 28.8 years.

3 0

4 years ago

Read 2 more answers

What law of motion is a small car is eaiser to start than moving a larger car

ASHA 777 [7]

Answer:

Second Law

Explanation:

5 0

3 years ago

A 0.875-g sample of anthracite coal was burned in a bomb calorimeter. The temperature rose from 22.50 to 23.80°C. The heat capac

kherson [118]

Answer:

a) 26.65 kJ was the heat evolved by the reaction.

b) $3.046\times 10^7 kJ$ is the energy released on burning 1 metric ton of this type of coal

Explanation:

Heat capacity of the calorimeter = C = 20.5 kJ/°C

Initial temperature of the calorimeter, $T_1$ = 22.50°C

Final temperature of the calorimeter, $T_2$ = 23.80°C

The heat evolved by the reaction = Q

$Q=C(T_2-T_1)$

$Q=20.5 kJ/^oC\times (23.80^oC-22.50^oC)$

$Q=26.65 kJ$

26.65 kJ was the heat evolved by the reaction.

0.875 g sample of anthracite coal was burned in a bomb calorimeter

0.875 g sample of anthracite coal gives 26.65 kJ of heat.

1 metric ton= 1000 kg

1000 kg = 1000 × 1000 g = 1,000,000 (1 kg =1000 g)

Then burning 1,000,000 g coal will give:

$=\frac{26.65 kJ }{0.875}\times 1,000,000 g=3.046\times 10^7 kJ$

$3.046\times 10^7 kJ$ is the energy released on burning 1 metric ton of this type of coal

8 0

3 years ago