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emmasim [6.3K]
2 years ago
12

!!HELP PLS!! A compound sample contains 0.783g of C, 0.196 g of H, 0.521 g 0 and the molecular formula molar mass is 184.27g/mol

. What is molecular formula of this
substance?
Chemistry
1 answer:
Delvig [45]2 years ago
4 0

Answer:

First

divide each element by its Molecular Mass to get their respective moles

Then Divide through by the lowest of the moles

You'll have the ratio of Carbon Hydrogen and Oxygen to be

C2H3O

Given Molecular Mass=184.27

C2H3On=184.27

n(12x2 + 1x3 + 16) =184.27

Evaluating this... You'll have n=4.3

Pls check if you assigned the correct value to each element

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How much 3.0 M NaOH is needed to neutralize 30. mL of 0.75 M H2SO4?
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Answer:

15.0 mL

Explanation:

Please see the step-by-step solution in the picture attached below.

Hope this answer can help you. Have a nice day!

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Gaseous methane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water. Suppose 0.802 g of methane i
Kipish [7]

Answer:

1.07g

Explanation:

Step 1:

We will begin by writing the balanced equation for the reaction. This is given below:

CH4 + 2O2 —> CO2 + 2H2O

Step 2:

Determination of the masses of CH4 and O2 that reacted and the mass of H2O produced from the balanced equation. This is illustrated below:

Molar Mass of CH4 = 12 + (4x1) = 12 + 4 = 16g/mol

Molar Mass of O2 = 16x2 = 32g/mol

Mass of O2 from the balanced equation = 2 x 32 = 64g

Molar Mass of H2O = (2x1) + 16 = 2 + 16 = 18g/mol

Mass of H2O from the balanced equation = 2 x 18 = 36g

Summary:

From the balanced equation above,

16g of CH4 reacted with 64g of O2 to produce 36g of H2O.

Step 3:

Determination of the limiting reactant.

We need to know which of the reactant is limiting the reaction in order to obtain the maximum mass of water.

This is illustrated below:

From the balanced equation above,

16g of CH4 reacted with 64g of O2.

Therefore, 0.802g of CH4 will react with = (0.802 x 64)/16 = 3.21g of O2.

From the above calculations, a higher mass of O2 is needed to react with 0.802g of CH4. Therefore, O2 is the limiting reactant.

Step 4:

Determination of the mass of H2O produced from the reaction.

To obtain the maximum mass of H2O produced, the limiting reactant will be used because it will generate the maximum yield of the product.

From the balanced equation above,

64g of O2 produce 36g of H2O.

Therefore, 1.9g of O2 will produce = (1.9 x 36)/64 = 1.07g of H2O.

The maximum mass of water (H2O) produced by the reaction is 1.07g

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Read 2 more answers
A sodium hydroxide solution is made by mixing 8.70 g NaOH with 100 g of water. The resulting solution has a density of 1.087 g/m
Ahat [919]

Answer:

Mass fraction = 0.08004

Mole fraction = 0.0377

Explanation:

Given, Mass of NaOH = 8.70 g

Mass of solution = 8.70 + 100 g = 108.70 g

Mass\ fraction\ of\ NaOH=\frac {8.70}{108.70} = 0.08004

Molar mass of NaOH = 39.997 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{8.70\ g}{39.997\ g/mol}

Moles\ of\ NaOH= 0.2175\ mol

Given, Mass of water = 100 g

Molar mass of water = 18.0153 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{100\ g}{18.0153\ g/mol}

Moles\ of\ water= 5.5508\ mol

So, according to definition of mole fraction:

Mole\ fraction\ of\ NaOH=\frac {n_{NaOH}}{n_{NaOH}+n_{water}}

Mole\ fraction\ of\ NaOH=\frac {0.2175}{0.2175+5.5508}=0.0377

3 0
3 years ago
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