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1 year ago

What is the total pressure exerted by a mixture of 48.0 grams of CH4 and 56.0 grams of

Chemistry

1 answer:

Oksi-84 [34.3K]1 year ago

4 0

The pressure of the gas is obtained as 48 atm.

<h3>What is the total pressure?</h3>

Now we know that;

Number of moles of CH4 = 48.0 grams /16 g/mol = 3 moles

Number of moles of H2 = 56.0 grams/2 g/mol = 28 moles

Total number of moles present = 3 moles + 28 moles = 31 moles

Using;

PV =nRT

P = total pressure

V = total volume

n = total number of moles

R = gas constant

T = temperature

P = nRT/V

P = 31 * 0.082 * 286/15

P = 48 atm

Learn more about pressure of a gas:brainly.com/question/18124975

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WILL MARK BRAINIEST <br><br> Can someone please help me understand this?

Mila [183]

To convert from moles to grams you divide by the molar mass of the element. To convert from grams to moles you X by the molar mass element

5 0

3 years ago

Which of the following statements is true with regard to transverse and longitudinal waves?

Ipatiy [6.2K]

Answer:

D (The last answer)

Explanation:

In a transverse wave, particles oscillate perpendicular to the direction of wave motion.

In a longitudinal wave, the oscillations of particles are parallel to the direction of propagation.

5 0

3 years ago

Read 2 more answers

Use bond energies from Table 10.3 in the textbook to estimate the enthalpy change (ΔH) for the following reaction. C2H2(g)+H2(g)

Digiron [165]

Answer: $=176.6kJmol^{-1}$

Explanation:Bond energy of H-H is 436.4 kJ/mole

Bond energy of C-H is 414 kJ/mol

Bond energy of C=C is 620 kJ/mol

Bond energy of C≡C is 835 kJ/mol

$\Delta H= {\text {sum of bond energies of reactants}}- {\text {sum of bond energies of products}}$

$\Delta H$ = {1B.E(C≡C)+2B.E(C-H) +1B.E(H-H)} - {1B.E(C=C)+4B.E(C-H)}

$\Delta H= {1B.E(835kJmole^{-1})+2B.E(414kJmole^{-1}) +1B.E(436.4kJmole^{-1})} - {1B.E(620kJmole^{-1})+4B.E(414kjmole^{-1})}$

$=176.6kJmol^{-1}$

7 0

3 years ago

A lead ball is added to a graduated cylinder containing 50.6 ml of water, causing the level of the water to increase to 93.0 mL.

Kamila [148]

42.4 ml is the volume in milliliters of the lead ball if a lead ball is added to a graduated cylinder containing 50.6 ml of water.

<h3>What is a graduated cylinder?</h3>

A tall narrow container with a volume scale is used especially for measuring liquids.

The graduated cylinder contains water

mL is a volume unit.

Water volume = 50.6 ml

The lead ball caused an increase in volume from 50.6 ml to 93.0 mL.

The new volume is the lead ball volume plus the original water volume :

Final volume = Vlead ball+ Water original volume

$93.0 mL= V_(lead ball) +50.6 ml$

$V_(lead ball) = 93.0 mL - 50.6 ml$

$V_(lead ball) = 42.4 ml$

Hence, 42.4 ml is the volume in milliliters of the lead ball.

Learn more about the graduated cylinder here:

brainly.com/question/13386106

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4 0

2 years ago

At 25°C and constant pressure, carbon monoxide gas combines with oxygen gas to give carbon dioxide gas with the evolution of 10.

stealth61 [152]

Answer : The value of $\Delta H$ for the reaction is, -565.6 kJ

Explanation :

First we have to calculate the molar mass of CO.

Molar mass CO = Atomic mass of C + Atomic mass of O = 12 + 16 = 28 g/mole

Now we have to calculate the moles of CO.

$\text{Moles of }CO=\frac{\text{Mass of }CO}{\text{Molar mass of }CO}=\frac{1g}{28g/mole}=\frac{1}{28}mole$

Now we have to calculate the value of $\Delta H$ for the reaction.

The balanced equation will be,

$2CO(g)+O_2(g)\rightarrow 2CO_2(g)$

From the balanced chemical reaction we conclude that,

As, $\frac{1}{28}mole$ of CO release heat = 10.1 kJ

So, 2 mole of CO release heat = $2\times 28\times 10.1=565.6kJ$

Therefore, the value of $\Delta H$ for the reaction is, -565.6 kJ (The negative sign indicates the amount of energy is released)

4 0

3 years ago