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2 years ago

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What is the total pressure (in atm) inside of a vessel containing N2 exerting a partial pressure of 0.256 atm, He exerting a par

tial pressure of 203 mmHg, and H2 exerting a partial pressure of 39.0 kPa?

2 answers:

Bezzdna [24]2 years ago

8 0

Answer: Total pressure inside of a vessel is 0.908 atm

Explanation:

According to Dalton's law, the total pressure is the sum of individual partial pressures. exerted by each gas alone.

$p_{total}=p_1+p_2+p_3$

$p_{N_2}$ = partial pressure of nitrogen = 0.256 atm

$p_{He}$ = partial pressure of helium = 203 mm Hg = 0.267 atm (760mmHg=1atm)

$p_{H_2}$ = partial pressure of hydrogen =39.0 kPa = 0.385 atm (1kPa=0.00987 atm)

Thus $p_{total}=p_{H_2}+p_{He}+p_{H_2}$

$p_{total}$ =0.256atm+0.267atm+0.385atm =0.908atm

Thus total pressure (in atm) inside of a vessel is 0.908

e-lub [12.9K]2 years ago

4 0

Answer:

The total pressure inside the vessel is 0.908 atm

Explanation:

Step 1: Data given

Partial pressure N2 = 0.256 atm

Partial pressure He = 203 mmHg = 0.267105 atm

Partial pressure H2 = 39.0 kPa = 0.3849 atm

Step 2: Calculate the total pressure

Total pressure = p(N2) + p(He) + p(H2)

Total pressure = 0.256 atm + 0.267105 atm + 0.3849 atm

Total pressure = 0.908 atm

The total pressure inside the vessel is 0.908 atm

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ZanzabumX [31]

Answer:

19,26 kJ

Explanation:

The work done when a gas expand with a constant atmospheric pressure is:

W = PΔV

Where P is pressure and ΔV is the change in volume of gas.

Assuming the initial volume is 0, the reaction of 500g of Zn with H⁺ (Zn(s) + 2H⁺(aq) → Zn²⁺(aq) + H₂(g)) produce:

500,0g Zn(s)× $\frac{1molZn}{65,38g}$ × $\frac{1molH_{2}(g)}{1molZn}$ = 7,648 moles of H₂

At 1,00atm and 303,15K (30°C), the volume of these moles of gas is:

V = nRT/P

V = 7,648mol×0,082atmL/molK×303,15K / 1,00atm

V = 190,1L

That means that ΔV is:

190,1L - 0L = <em>190,1L</em>

And the work done is:

W = 1atm×190,1L = 190,1atmL.

In joules:

190,1 atmL× $\frac{101,325}{1atmL}$ = <em>19,26 kJ</em>

I hope it helps!

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2 years ago

Three moles of helium gas (molar mass MM = 4.00 g/molg/mol) are in a rigid container that keeps the volume of the gas constant.

Anastaziya [24]

Answer:

The rms speed of the gas atoms after 3600 J of heat energy is added to the gas = 1150 m/s.

Explanation:

Mass of 3 moles of Helium = 3 moles × 4.00 g/mol = 12.00 g = 0.012 kg

The initial average kinetic energy of the helium atoms = (1/2)(m)(u²)

where u = initial rms speed of the gas = 850 m/s

Initial average kinetic energy of the gas = (1/2)(0.012)(850²) = 4335 J

Then, 3600 J is added to the gas,

New kinetic energy of the gas = 4335 + 3600 = 7935 J

New kinetic energy of Helium atoms = (1/2)(m)(v²)

where v = final rms speed of the gas = ?

7935 = (1/2)(0.012)(v²)

v² = (7935×2)/0.012

v² = 1,322,500

v = 1150 m/s

Hence, the rms speed of the gas atoms after 3600 J of heat energy is added to the gas = 1150 m/s.

Hope this Helps!!!

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1 year ago

If 4.9 kg of CO2 are produced during a combustion reaction, how many molecules of CO2 would be produced?

solmaris [256]

Answer:

6.7 x 10²⁶molecules

Explanation:

Given parameters

Mass of CO₂ = 4.9kg = 4900g

Unknown:

Number of molecules = ?

Solution:

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Number of moles = $\frac{mass}{molar mass}$

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Number of moles = $\frac{4900}{44}$ = 111.36mole

A mole of substance is the quantity of substance that contains the avogadro's number of particles.

1 mole = 6.02 x 10²³molecules

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andreev551 [17]

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LiRa [457]

Answer: 1.14

Explanation:

$HCl+NaOH\rightarrow NaCl+H_2O$

To calculate the molarity of acid, 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 $HCl$

are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=1\\M_1=?\\V_1=10.0mL\\n_2=1\\M_2=0.1M\\V_2=7.2mL$

Putting values in above equation, we get:

$1\times M_1\times 10.0=1\times 0.1\times 7.2\\\\M_1=0.072M$

To calculate pH of gastric juice:

molarity of $H^+$ = 0.072

$pH=-log[H^+]$

$pH=-log(0.072)=1.14$

Thus the pH of the gastric juice is 1.14

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2 years ago