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solniwko [45]
3 years ago
15

If you had a 100 mL of a solution of 0.01 M NaF, how many moles would that solution contain?

Chemistry
1 answer:
Daniel [21]3 years ago
5 0

Answer:

0.001 mole of NaF.

Explanation:

From the question given above, the following data were obtained:

Volume of solution = 100 mL

Molarity = 0.01 M

Mole of NaF =?

Next, we shall convert 100 mL to litre (L). This can be obtained as follow:

1000 mL = 1 L

Therefore,

100 mL = 100 mL × 1 L / 1000 ml

100 mL = 0.1 L

Thus, 100 mL is equivalent to 0.1 L.

Finally, we shall determine the number of mole of NaF present in the solution. This can be obtained as follow:

Volume of solution = 0.1 L

Molarity = 0.01 M

Mole of NaF =?

Molarity =mole /Volume

0.01 = mole of NaF / 0.1

Cross multiply

Mole of NaF = 0.01 × 0.1

Mole of NaF = 0.001 mole.

Thus, 0.001 mole of NaF is present in 100 mL of the solution.

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3 years ago
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Which chemical equation is unbalanced? c o2 right arrow. co2 sr o2 right arrow. 2sro 6h2 3o2 right arrow. 6h2o h2 h2 o2 right ar
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The unbalanced equation is one in which the moles of atoms are not equal on both sides of the reaction.

<h3>What is a balanced chemical equation?</h3>

A balanced chemical equation is one in wgich the moles of atoms in the reactants side is equal to the moles of atoms on the product side.

The given equations of reaction is not clearly stated.

Therefore, the unbalanced equation is one in which the moles of atoms are not equal on both sides of the reaction.

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6 0
2 years ago
An unknown gaseous substance has a density of 1.06 g/L at 31 °C and 371 torr. If the substance has the following percent composi
Anarel [89]

Answer:

C) C4H6 - Right answer

Explanation:

Let's combine the Ideal Gases Law with density to get the molecular formula for the unknown gas.

Density = mass / volume

1.06 g /L means that 1.06 grams of compound occupy 1 liter of volume.

P . V = n . R . T

Pressure in Torr must be converted to atm

760 Torr are 1 atm

371 Torr  are __ (371 .1)/760 = 0.488 atm

0.488 atm . 1L = 1.06g/MM . 0.082 . 304K

(0.488 atm . 1L) / 0.082 . 304K = 1.06g/ MM

Mass / Molar mass = Moles → That's why the 1.06 g / MM

0.0195 mol = 1.06g / MM

1.06g/0.0195 mol = MM →  54.3 g/m

Now, let's use the composition

100 g of compound have 88.8 g of C

54.3 g of compound have ___ (54.3  . 88.8) /100 = 48 g of C

100 g of compound have 11.2 g of H

54.3 g of compound have __ (54.3  .  11.2)/100 = 6 g of H

48 g of C are included un 4 atoms

6 g of H are included in 6 atoms

4 0
3 years ago
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GenaCL600 [577]

Answer:

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Explanation:

5 0
3 years ago
Show your work. 37.2 moles CO2 to oxygen atoms.
Rufina [12.5K]
4 moles of oxygen (6.0zzx10

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