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miskamm [114]
2 years ago
9

PROBLEMS

Chemistry
1 answer:
AlekseyPX2 years ago
7 0

Answer:

1.) 0.1 M

2.) 0.2 M

3.) 1 M

4.) Solution #3 is the most concentrated because it has the highest molarity. This solution has the largest solute to solvent ratio. The more solvent there is, the lower the concentration and molarity.

Explanation:

To find the molarity, you need to (1) convert grams NaOH to moles (via molar mass from periodic table) and then (2) calculate the molarity (via the molarity equation). All of the answers should have 1 sig fig to match the given values.

Molar Mass (NaOH): 22.99 g/mol + 16.00 g/mol + 1.008 g/mol

Molar Mass (NaOH): 39.998 g/mol

4 grams NaOH           1 mole
----------------------  x  ------------------  = 0.1 moles NaOH
                                 39.998 g

1.)

Molarity = moles / volume (L)

Molarity = (0.1 moles) / (1 L)

Molarity = 0.1 M

2.)

Molarity = moles / volume (L)

Molarity = (0.1 moles) / (0.5 L)

Molarity = 0.2 M

3.)

Molarity = moles / volume (L)

Molarity = (0.1 moles) / (0.1 L)

Molarity = 1 M

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During a combustion reaction, 9.00 grams of oxygen reacted with 3.00 grams of CH4.
Monica [59]

Answer:

0.74 grams of methane

Explanation:

The balanced equation of the combustion reaction of methane with oxygen is:

  • CH₄ + 2 O₂ → CO₂ + 2 H₂O

it is clear that 1 mol of CH₄ reacts with 2 mol of O₂.

firstly, we need to calculate the number of moles of both

for CH₄:

number of moles = mass / molar mass = (3.00 g) /  (16.00 g/mol) = 0.1875 mol.

for O₂:

number of moles = mass / molar mass = (9.00 g) /  (32.00 g/mol) = 0.2812 mol.

  • it is clear that O₂ is the limiting reactant and methane will leftover.

using cross multiplication

1 mol of  CH₄ needs → 2 mol of O₂

???  mol of  CH₄  needs → 0.2812 mol of O₂

∴ the number of mol of CH₄ needed = (0.2812 * 1) / 2 = 0.1406 mol

so 0.14 mol will react and the remaining CH₄

mol of CH₄ left over = 0.1875 -0.1406 = 0.0469 mol

now we convert moles into grams

mass of CH₄ left over = no. of mol of CH₄ left over *  molar mass

                                    = 0.0469 mol * 16 g/mol = 0.7504 g

So, the right choice is 0.74 grams of methane

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3 years ago
aspirin C6H4(CO2)(CO2CH3),can be prepared in the chemistry laboratory by the reactions of salicylic acid, C6H4(CO2H)(OH),with ac
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262mol 1=kg
g=1000
from stoichoimetry

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x=102000/360
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Answer:

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