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3 years ago

Hydrogen gas was collected by water displacement. what was pressure of the h2 collected if the temperature was 26°c?

1 answer:

4 0

The ideal gas law may be written as
$p= \frac{\rho R T}{M}$
where
p = pressure
ρ =density
T = temperature
M = molar mass
R = 8.314 J/(mol-K)

For the given problem,
ρ = 0.09 g/L = 0.09 kg/m³
T = 26°C = 26+273 K = 299 K
M = 1.008 g/mol = 1.008 x 10⁻³ kg/mol

Therefore
$p= \frac{(0.09 \, kg/m^{3})*(8.314 \, J/(mol-K))*(299 \, K)}{1.008 \times 10^{-3} \, kg/mol} =2.2195 \times 10^{5} \, Pa$

Note that 1 atm = 101325 Pa
Therefore
p = 2.2195 x 10⁵ Pa
= 221.95 kPa
= (2.295 x 10⁵)/101325 atm
= 2.19 atm

Answer:
2.2195 x 10⁵ Pa (or 221.95 kPa or 2.19 atm)

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There are 3 STOP codons in the genetic code - UAG, UAA, and UGA. These codons signal the end of the polypeptide chain during translation.

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3 years ago

Passing an electric current through a sample of water (H2O) can cause the water to decompose into hydrogen gas (H2) and oxygen g

Tema [17]

<u>Answer:</u> The mass of water that must be reacted is 56.28 grams.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ....(1)

Given mass of oxygen = 50 g

Molar mass of oxygen = 32.00 g/mol

Putting values in equation 1, we get:

$\text{Moles of oxygen}=\frac{50g}{32g/mol}=1.5625mol$

For the given chemical equation:

$2H_2O\rightarrow 2H_2+O_2$

By Stoichiometry of the reaction:

1 mole of oxygen is produced when 2 moles of water is reacted.

So, 1.5625 moles of oxygen is produced when = $\frac{2}{1}\times 1.5625=3.125mol$ of water is reacted.

To calculate the mass of water, we use equation 1:

Moles of water = 3.125 moles

Molar mass of water = 18.01 g/mol

Putting values in equation 1, we get:

$3.125mol=\frac{\text{Mass of water}}{18.01g/mol}\\\\\text{Mass of water}=56.28g$

Hence, the mass of water that must be reacted is 56.28 grams.

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4 years ago

An element 'X after reacting with acids liberates hydrogen gas and candisplace lead and mercury from their salt solutions. The m

adell [148]

The element X is Copper.

Explanation :

Reactive metals from 3d transition series like Cu are very reactive because their standard reduction potential values are low.

Hence, when treated with a acid it tends to liberate hydrogen gas.

Also, because of its low reduction potential value, it can easily displace lead and tin from their salt solutions.

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5 0

3 years ago

A vessel of volume 100 cm3 contains 0.25 mol o2 and 0.034 mol co2 at 10.0c. Calculate the partial pressure of each component and

Artyom0805 [142]

Answer: The partial pressure of oxygen is 76.56 atm , partial pressure of carbon dioxide is 10.44 atm and the total pressure is 87 atm.

Explanation:

According to the ideal gas equation:'

$PV=nRT$

P = Pressure of the gas = ?

V= Volume of the gas = $100cm^3=0.1L$ $1L=1000cm^3$

T= Temperature of the gas = 10°C = 373 K

R= Gas constant = 0.0821 atmL/K mol

n= moles of gas= 0.25 +0.034 = 0.284 moles

$P=\frac{nRT}{V}=\frac{0.284\times 0.0821\times 373}{0.1}=87atm$

$x_{O_2}$ = mole fraction of oxygen= $\frac{\text {moles of }O_2}{\text {moles of }O_2+\text{moles of }CO_2}=\frac{0.25}{0.25+0.034}=0.88$ ,

$x_{CO_2}$ =mole fraction of carbon dioxide= $\frac{\text {moles of }CO_2}{\text {moles of }O_2+\text{moles of }CO_2}=\frac{0.034}{0.25+0.034}=0.12$

partial pressure of oxygen = $p_{O_2}=x_{O_2}\times P=0.88\times 87=76.56atm$

partial pressure of carbon dioxide= $p_{CO_2}=x_{CO_2}\times P=0.12\times 87=10.44atmatm$

The partial pressure of oxygen is 76.56 atm , partial pressure of carbon dioxide is 10.44 atm and the total pressure is 87 atm.

8 0

3 years ago

A 5 g sample of lead (specific heat 0.129 /g˚C) is heated, then put in a calorimeter with 50 mL of water (specific heat 4.184 J/

Svetach [21]

Answer:

670.68°C

Explanation:

Given that:

volume of water = 50 ml but 1 g = 1 ml. Therefore the mass of water (m) = 50 ml × 1 g / ml = 50 g

specific heat (C) = 4.184 J/g˚C

Initial temperature = 20°C, final temperature = 22°C. Therefore the temperature change ΔT = final temperature - initial temperature = 22 - 20 = 2°C

The quantity of heat (Q) used to raise the temperature of a body is given by the equation:

Q = mCΔT

Substituting values:

Q = 50 g × 4.184 J/g˚C × 2°C = 418.4 J

Since the mass of lead = 5 g and specific heat = 0.129 J/g˚C. The heat used to raise the temperature of water is the same heat used to raise the temperature of lead.

-Q = mCΔT

-418.4 J = 5 g × 0.129 J/g˚C × ΔT

ΔT = -418.4 J / ( 5 g × 0.129 J/g˚C) = -648 .68°C

temperature change ΔT = final temperature - initial temperature

- 648 .68°C = 22°C - Initial Temperature

Initial Temperature = 22 + 648.68 = 670.68°C

4 0

3 years ago