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2 years ago

10

The charcoal that was burned in a grill left only a fine, light ash. The ash weighs significantly less than the charcoal before

being burned. What happened to the extra mass?

1 answer:

Musya8 [376]2 years ago

8 0

Answer:

When the charcoal burned, it turned into a gas which caused the mass to change.

Explanation:

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A student has a 2.19 L bottle that contains a mixture of O 2 , N 2 , and CO 2 with a total pressure of 5.57 bar at 298 K . She k

Sergeeva-Olga [200]

<u>Answer:</u> The partial pressure of oxygen gas is 2.76 bar

<u>Explanation:</u>

To calculate the number of moles, we use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the gas = 5.57 bar

V = Volume of the gas = 2.19 L

T = Temperature of the gas = 298 K

R = Gas constant = $0.0831\text{ L bar }mol^{-1}K^{-1}$

n = Total number of moles = ?

Putting values in above equation, we get:

$5.57bar\times 2.19L=n\times 0.0831\text{ L. bar }mol^{-1}K^{-1}\times 298K\\\\n=\frac{5.57\times 2.19}{0.0831\times 298}=0.493mol$

To calculate the mole fraction of carbon dioxide, we use the equation given by Raoult's law, which is:

$p_{A}=p_T\times \chi_{A}$ ........(1)

where,

$p_A$ = partial pressure of carbon dioxide = 0.318 bar

$p_T$ = total pressure = 5.57 bar

$\chi_A$ = mole fraction of carbon dioxide = ?

Putting values in above equation, we get:

$0.318bar=5.57bar\times \chi_{CO_2}\\\\\chi_{CO_2}=\frac{0.381}{5.57}=0.0571$

Mole fraction of a substance is given by:

$\chi_A=\frac{n_A}{n_A+n_B}$

We are given:

Moles of nitrogen gas = 0.221 moles

Mole fraction of nitrogen gas, $\chi_{N_2}=\frac{0.221}{0.493}=0.448$

Calculating the partial pressure of oxygen gas by using equation 1, we get:

Mole fraction of oxygen gas = (1 - 0.0571 - 0.448) = 0.4949

Total pressure of the system = 5.57 bar

Putting values in equation 1, we get:

$p_{O_2}=5.57bar\times 0.4949\\\\p_{O_2}=2.76bar$

Hence, the partial pressure of oxygen gas is 2.76 bar

6 0

3 years ago

How many grams of aluminum will be deposited by 0.1F? (Al=27) a.0.3g b. 0.9. c. 9.0g. d. 2.7g<br>

andrezito [222]

Based on the charge on the aluminium ion, 0.9 g of aluminium are deposited by 0.1 F of electricity.

<h3>What is electrolysis?</h3>

Electrolysis is the decomposition of a substance known as an electrolyte when electric current is passed through it.

The mass and hence moles an electrolyte deposited when current is passed through it depends on the charge on the ion.

Aluminium ion has a charge of +3 and requires 3F of electricity to deposit 1 mole or 27 g of aluminium

0.1 F will discharge = 0.1/3 × 27 g of aluminium

mass of aluminium deposited = 0.9 g of aluminium.

Therefore, 0.9 g of aluminium are deposited by 0.1 F of electricity.

Learn more about electrolysis at: brainly.com/question/26050361

5 0

2 years ago

When an atom undergoes beta decay, wich of the following is true?

tigry1 [53]

<span> The atomic number increases by one and the element becomes a different element. </span>

3 0

3 years ago

Which subatomic particles affect the mass of an atom

klasskru [66]

Answer:

protons and neutrons

Explanation:

those particles account for 99.99% of mass

6 0

3 years ago

How many mL of 0.715 M HCl is required to neutralize 1.25 grams of sodium carbonate? (producing carbonic acid)

jonny [76]

Answer:

34 mL

Explanation:

We'll begin by calculating the number of mole in 1.25 g of sodium carbonate, Na₂CO₃. This can be obtained as follow:

Mass of Na₂CO₃ = 1.25 g

Molar mass of Na₂CO₃ = (23×2) + 12 + (16×3)

= 46 + 12 + 48

= 106 g/mol

Mole of Na₂CO₃ =?

Mole = mass /molar mass

Mole of Na₂CO₃ = 1.25 / 106

Mole of Na₂CO₃ = 0.012 mole

Next, we shall determine the number of mole HCl needed to react with 0.012 mole of Na₂CO₃.

The equation for the reaction is given below:

Na₂CO₃ + 2HCl —> H₂CO₃ + 2NaCl

From the balanced equation above,

1 mole of Na₂CO₃ reacted with 2 moles of HCl.

Therefore, 0.012 mole of Na₂CO₃ will react with = 0.012 × 2 = 0.024 mole of HCl.

Next, we shall determine the volume of HCl required for the reaction. This is illustrated:

Mole of HCl = 0.024 mole

Molarity of HCl = 0.715 M

Volume of HCl =?

Molarity = mole /Volume

0.715 = 0.024 / volume of HCl

Cross multiply

0.715 × volume of HCl = 0.024

Divide both side by 0.715

Volume of HCl = 0.024 / 0.715

Volume of HCl = 0.034 L

Finally, we shall convert 0.034 L to mL

This can be obtained as follow:

1 L = 1000 mL

Therefore,

0.034 L = 0.034 L × 1000 mL / 1 L

0.034 L = 34 mL

Therefore, 34 mL of HCl is needed for the reaction.

6 0

2 years ago