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3 years ago

How many kilojoules of heat are released when 0.72 mole of oxygen gas are used to combust methane?

Chemistry

1 answer:

KonstantinChe [14]3 years ago

6 0

0.72 mole of oxygen would produce 320.4 kJ of heat.

Explanation:

CH₄ (g) + 2O₂ (g) → CO₂ (g) + 2H₂O (ℓ) + 890kJ

According to the equation,

2 moles of O₂ produces 890 kJ of heat

So, 0.72 moles of O₂ will produce:

$= \frac{890}{2} X 0.72\\\\= 445 X 0.72\\\\= 320.4 kJ$

Therefore, 0.72 mole of oxygen would produce 320.4 kJ of heat.

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a 3.81 g sample of NaHCO3 was completely decomposed. After decomposition, Na2CO3 had mass of 2.86g. Determine mass of H2CO3 prod

qaws [65]

Answer:

1.67g H2CO3 are produced

Explanation:

Based on the reaction:

2NaHCO3 → Na2CO3 + H2CO3

2 moles of NaHCO3 produce 1 mole of Na2CO3 and 1 mole of H2CO3

To solve this question we need to find the moles of Na2CO3 = Moles of H2CO3. With their moles we can find the mass of H2CO3 as follows:

Moles Na2CO3 -Molar mass: 105.99g/mol-

2.86g Na2CO3 * (1mol/105.99g) = 0.02698 moles Na2CO3 = Moles H2CO3

Mass H2CO3 -Molar mass: 62.03g/mol-

0.02698 moles * (62.03g/mol) =

<h3>1.67g H2CO3 are produced</h3>

8 0

3 years ago

PLEASE HELP ME

BabaBlast [244]

Answer:47.05% is the percent yield of nitrogen in the reaction.

Explanation:

heoretical yield of nitrogen gas = x

Moles of ammonia =

According to reaction,2 moles of ammonia gives 1 mol of nitrogen gas.

Then 2.3529 mol of ammonia will give:

of nitrogen gas

Mass of 1.1764 moles of nitrogen gas,x = 1.1764 mol × 28 g/mol=32.94 g

Experiential yield of nitrogen gas = 15.5 g

Percentage yield:

hope that help

47.05% is the percent yield of nitrogen in the reaction.

3 0

3 years ago

What is the pressure in atm exerted by 2.48 moles of a gas in a 250.0 mL container at 58 degrees celsius

Dvinal [7]

Since the question manages to include moles, pressure, volume, and temperature, then it is evident that in order to find the answer we will have to use the Ideal Gas Equation: PV = nRT (where P = pressure; V = volume; n = number of moles; R = the Universal Constant [0.082 L·atm/mol·K]; and temperature.

First, in order to work out the questions, there is a need to convert the volume to Litres and the temperature to Kelvin based on the equation:
250 mL = 0.250 L
58 °C = 331 K

Also, based on the equation P = nRT ÷ V

⇒ P = (2.48 mol)(0.082 L · atm/mol · K)(331 K) ÷ 0.250 L
⇒ P = (67.31 L · atm) ÷ 0.250 L
⇒ P = 269.25 atm

Thus the pressure exerted by the gas in the container is 269.25 atm.

7 0

3 years ago

Determine the enthalpy for this reaction: Al(OH)3(s)+3HCl(g)→AlCl3(s)+3H2O(l)

ivanzaharov [21]

Important information to solve the exercise :
Substance ΔHf (kJ/mol):
HCl(g)= −92.0 kJ/mol
Al(OH)3(s)= −1277.0 kJ/mol
 H2O(l)= −285.8 kJ/mol
AlCl3(s) =−705.6 kJ/mol

Al(OH)3(s)+3HCl(g)→AlCl3(s)+3H2O(l)
reactants products

products- reactants:

(−705.6) + (3 x −285.8) - ( −1277.0 ) - (3 x −92.0 ) = - 10.0 kJ per mole at 25°C

6 0

3 years ago

What volume of stock solution and water must you add to prepare 36.25ml of a 1.25M solution

Serjik [45]

Given :

Number of moles , n = 36.25 mol .

Molarity , M = 1.25 M .

To Find :

The volume of water required .

Solution :

Moarity is given by :

$M=\dfrac{n}{V}$

So , $V=\dfrac{n}{M}$

Here , n is number of moles and M is molarity .

Putting all values in above equation , we get :

$V=\dfrac{36.25}{1.25}\\\\V=29\ L$

Therefore , volume of water required is 29 L .

5 0

3 years ago