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3 years ago

-. You have 100 grams of methane. How many moles do you have?

Chemistry

1 answer:

Luba_88 [7]3 years ago

8 0

Answer:

6.25mol

Explanation:

Using the following formula;

Mole = mass ÷ molar mass

According to this question, there are 100 grams of methane (CH4)

Molar Mass of methane (CH4) = 12 + 1(4)

= 12 + 4

= 16g/mol

mole = 100/16

mole = 6.25moles

Therefore, there are 6.25moles of methane in 100g

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Does it take a lot of heat to melt a substance with a low melting point t

Helga [31]

low melting point means that a low amount of heat is required to melt said substance.

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3 years ago

Luis is helping his parents paint a border around the walls of a room. He uses a stencil to repeat the same design on each wall

iren [92.7K]

Answer:

Polymerization.

Explanation:

Polymerization can be defined as a type of chemical reaction in which molecules that are relatively small in size chemically combine to form a huge chain of molecules.

Simply stated, polymerization refers to a chemical reaction where two or more smaller molecules react to produce larger molecules of the same network or repetitive structural units.

In polymerization, the relatively small molecules are generally referred to as monomers while the larger molecules they produce are known as polymers.

Polymerization is given by the chemical formula;

nA -----> A(n).

In this scenario, Luis uses a stencil to repeat the same design on each wall to form one long grapevine with a bunch of grapes every foot along its length.

Hence, the type of chemical reaction this best model is polymerization because it involved repeating the same design (monomers) to form a long grapevine with a bunch of grapes (polymers).

5 0

3 years ago

What volume (in liters) of a solution contains 0.14 mol of KCl?

oksano4ka [1.4K]

Answer:

$\boxed {\boxed {\sf 0.078 \ L }}$

Explanation:

We are asked to find the volume of a solution given the moles of solute and molarity.

Molarity is a measure of concentration in moles per liter. It is calculated using the following formula:

$molarity= \frac{moles \ of \ solute}{liters \ of \ solution}$

We know there are 0.14 moles of potassium chloride (KCl), which is the solute. The molarity of the solution is 1.8 molar or 1.8 moles of potassium chloride per liter.

moles of solute = 0.14 mol KCl
molarity= 1.8 mol KCl/ L
liters of solution=x

Substitute these values/variables into the formula.

$1.8 \ mol \ KCl/ L = \frac { 0.14 \ mol \ KCl}{x}$

We are solving for x, so we must isolate the variable. First, cross multiply. Multiply the first numerator and second denominator, then the first denominator and second numerator.

$\frac {1.8 \ mol \ KCl/L}{1} = \frac{0.14 \ mol \ KCl}{x}$

$1.8 \ mol \ KCl/ L *x = 1*0.14 \ mol \ KCl$

$1.8 \ mol \ KCl/ L *x = 0.14 \ mol \ KCl$

Now x is being multiplied by 1.8 moles of potassium chloride per liter. The inverse operation of multiplication is division, so we divide both sides by 1.8 mol KCl/L.

$\frac {1.8 \ mol \ KCl/ L *x}{1.8 \ mol \ KCl/L} = \frac{0.14 \ mol \ KCl}{1.8 \ mol \ KCl/L}$

$x= \frac{0.14 \ mol \ KCl}{1.8 \ mol \ KCl/L}$

The units of moles of potassium chloride cancel.

$x= \frac{0.14 }{1.8 L}$

$x=0.07777777778 \ L$

The original measurements of moles and molarity have 2 significant figures, so our answer must have the same. For the number we found, that is the thousandth place. The 7 in the ten-thousandth place tells us to round the 7 up to a 8.

$x \approx 0.078 \ L$

There are approximately <u>0.078 liters of solution.</u>

5 0

3 years ago

Since both HBr and KOH are strong, we expect these reactions to occur fully. You mix 2 moles of HBr with 3 moles of KOH in enoug

AysviL [449]

Answer: pH = 14

Explanation: Please see the attachments below

8 0

3 years ago

. The freezing point of an aqueous solution containing a nonelectrolyte solute is – 2.79 °C. What is the boiling point of this s

Semmy [17]

Answer:

Boiling point of the solution is 100.78°C

Explanation:

This is about colligative properties.

First of all, we need to calculate molality from the freezing point depression.

ΔT = Kf . m . i

As the solute is nonelectrolyte, i = 1

0°C - (-2.79°C) = 1.86 °C/m . m . 1

2.79°C / 1.86 m/°C = 1.5 m

Now, we go to the boiling point elevation

ΔT = Kb . m . i

Final T° - 100°C = 0.52 °C/m . 1.5m . 1

Final T° = 0.52 °C/m . 1.5m . 1 + 100°C → 100.78°C

4 0

3 years ago