How many moles of AlCl3 are in 2.25L of a 0.15 molar solution?

Chemistry

1 answer:

Darya [45]3 years ago

5 0

Answer:

Option C. 0.34 moles

Explanation:

Data obtained from the question include:

Molarity = 0.15M

Volume = 2.25L

Mole =

Molarity = mole /Volume

Mole = Molarity x Volume

Mole = 0.15 x 2.25

Mole = 0.34mole

The number of mole of AlCl3 present in the solution is 0.34mole.

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A party balloon contains 5.50 x 10 ^22 atoms of helium gas. What is the mass, in grams of the

eduard

Answer:

$\boxed {\boxed {\sf A. \ 0.365 \ g\ He}}$

Explanation:

1. Convert Atoms to Moles

We must use Avogadro's Number: 6.022*10²³. This is the number of particles (atoms, molecules, ions, etc.) in 1 mole of a substance. In this case, the particles are atoms of helium. We can create a ratio.

$\frac {6.022*10^{23} \ atoms \ He}{1 \ mol \ He}$

Multiply by the given number of helium atoms.

$5.50 *10^{22} \ atoms \ He *\frac {6.022*10^{23} \ atoms \ He}{1 \ mol \ He}$

Flip the fraction so the atoms of helium cancel.

$5.50 *10^{22} \ atoms \ He *\frac {1 \ mol \ He}{6.022*10^{23} \ atoms \ He}$

$5.50 *10^{22} *\frac {1 \ mol \ He}{6.022*10^{23} }$

$\frac {5.50 *10^{22} \ mol \ He}{6.022*10^{23} }= 0.09133178346 \ mol \ He$

2. Convert Moles to Grams

We must use the molar mass, which is found on the Periodic Table.

Helium (He): 4.00 g/mol

Use this as a ratio.

$\frac { 4.00 \ g \ He }{ 1 \ mol \ He}$

Multiply by the number of moles we calculated. The moles will then cancel.

$0.09133178346 \ mol \ He *\frac { 4.00 \ g \ He }{ 1 \ mol \ He}$

$0.09133178346*\frac { 4.00 \ g \ He }{ 1 }$

$0.3653271338 \ g\ He$

3. Round

The original measurement has 3 significant figures (5, 5, and 0). Our answer must have the same. For the number we calculated, it is thousandth place. The 3 in the ten thousandth place tells us to leave the 5.

$0.365 \ g\ He$