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inna [77]
3 years ago
8

Given the following chemical reaction: 2 O2 + CH4 CO2 + 2 H2O What mass of oxygen is required to completely react with 20 grams

of CH4? 6 C 12.01 Carbon 1 H 1.01 Hydrogen 8 O 16.00 Oxygen A. 10 grams O2 B. 20 grams O2 C. 80 grams O2 D. 144 grams O2
Chemistry
1 answer:
Jobisdone [24]3 years ago
4 0
The chemical equation is:
CH₄ + 2O₂ → CO₂ + 2H₂O

First, we calculate the moles of methane present using:
Moles = mass / molecular mass
Moles = 20 / 16
Moles = 1.25

Next, we may observe from the chemical equation that the molar ratio between methane and oxygen is 1 : 2
So the moles of oxygen required are 2 x 1.25
2.5 moles of oxygen required

Mass = moles * molecular mass
Mass = 2.5 * 32
Moles = 80

C. 80 grams O₂
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Calculate the molarity of a solution with 114.95 grams of sodium dissolved in 2 L of water. ​
koban [17]

Answer:

The molarity of a solution is 2.5 M

Explanation:

Molarity is a concentration unit that describes how much of a solution is dissolved in solution.

Molarity of a solution can found by using the formula,

Molarity (M) = (moles of solute)/(Liters of Solution).

Given, mass of Sodium = 114.95 grams.

Volume of water = 2 L.

Here, Sodium is solute as it is dissolved in water, which is the solvent.

Moles of Sodium(solute) can be found by using the formula,

Number of Moles = mass/Molecular weight.

mass of Sodium = 114.95 grams.

Molecular weight = 22.989 grams

Number of Moles of Sodium(solute) =114.95/22.989 = 5.

Substituting the values in the formula, we get,

M = 5/2 = 2.5 M

3 0
3 years ago
Please help me someone ASAP!!!!
vagabundo [1.1K]

Answer:

Explanation:

g

6 0
3 years ago
A chemist makes 940. mL of sodium carbonate (Na_2CO_3) working solution by adding distilled water to 200. mL of a 0.171 m stock
Alexxandr [17]

Answer:

The answer is 0.0698 M

Explanation:

The concentration was prepared by a serial dilution method.

The formula for the preparation I M1V1 = M2V2

M1= the concentration of the stock solution = 0.171 M

V1= volume of the stock solution taken = 200 mL

M2 = the concentration produced

V2 = the volume of the solution produced = 940 mL

Substitute these values in the formula

0.171 × 200 = 490 × M2

34.2 = 490 × M2

Make M2 the subject of the formula

M2 = 34.2/490

M2 = 0.069795

M2 = 0.0698 M ( 3 s.f)

The concentration of the Chemist's working solution to 3 significant figures is 0.0698M

6 0
3 years ago
What is chemistry n laboratory​
grin007 [14]
a laboratory for research in chemistry. chem lab, chemistry lab. lab, laboratory, research lab, research laboratory, science lab, science laboratory - a workplace for the conduct of scientific research.
3 0
3 years ago
How many moles of oxygen are in 25.45 g of caco3?
emmainna [20.7K]
From the periodic table:
mass of oxygen = 16 grams
mass of calcium = 40 grams
mass of carbon = 12 grams
mass of CaCO3 = 40 + 12 + 3(16) = 100 grams
Therefore, each 100 grams of CaCO3 contains 3 moles of oxygen
To know the number of oxygen moles in 25.45 grams, we will simply do cross multiplication as follows:
number of oxygen moles = (25.45 x 3) / 100 = 0.7636
3 0
3 years ago
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