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3 years ago

How many moles of n are in 0.195 g of n2o?

Chemistry

2 answers:

Blababa [14]3 years ago

8 0

Explanation:

As molar mass of $N_{2}O$ is 44.013 g/mol and it is given that its mass is 0.195 g.

Therefore, calculate its number of moles as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

= $\frac{0.195 g}{44.013 g/mol}$

= 0.0044 mol

Here, in 1 mole of $N_{2}O$ there is 2 mole of N. Thus, moles of N present in it is calculated as follows.

$2 \times 0.0044 mol$

= 0.0088 mol

Thus, we can conclude that in 0.195 g of $N_{2}O$ there are 0.0088 mol of N.

ivann1987 [24]3 years ago

4 0

0.195g N2O x 2moles of N / 28g of N = 5.46 moles of N

2X14= 28g.

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How many atoms are there in 8.88 g Si?

Mariana [72]

Answer:

$\boxed {\boxed {\sf 1.90 \times 10^{23} \ atoms \ Si}}$

Explanation:

We are asked to find how many atoms are in 8.88 grams of silicon.

<h3>1. Grams to Moles </h3>

First, we convert grams to moles. We use the molar mass or the mass of 1 mole of a substance. These values are found on the Periodic Table as they are equal to the atomic masses, but the units are grams per mole instead of atomic mass units.

Look up silicon's molar mass.

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We will convert using dimensional analysis. Set up a conversion factor with the molar mass.

$\frac { 28.085 \ g \ Si}{1 \ mol \ Si}$

We are converting 8.88 grams of silicon to moles, so we multiply by this value.

$8.88 \ g \ Si *\frac { 28.085 \ g \ Si}{1 \ mol \ Si}$

Flip the fraction so the units of grams of silicon cancel.

$8.88 \ g \ Si *\frac{1 \ mol \ Si} { 28.085 \ g \ Si}$

$8.88 *\frac{1 \ mol \ Si} { 28.085 }$

$\frac {8.88} { 28.085 } \ mol \ Si$

$0.316183015845 \ mol \ Si$

<h3>2. Moles to Atoms </h3>

Next, we convert moles to atoms. We use Avogadro's Number or 6.022 × 10²³. This is the number of particles (atoms, molecules, formula units, etc.) in 1 mole of a substance. In this case, the particles are atoms of silicon.

Set up another conversion factor.

$\frac {6.022 \times 10^{23} \ atoms \ Si}{1 \ mol \ Si}$

Multiply by the number of moles we calculated.

$0.316183015845\ mol \ Si *\frac {6.022 \times 10^{23} \ atoms \ Si}{1 \ mol \ Si}$

The units of moles of silicon cancel.

$0.316183015845 * \frac {6.022 \times 10^{23} \ atoms \ Si}{1}$

$0.316183015845 * {{6.022 \times 10^{23} \ atoms \ Si}$

$1.90405412 \times 10^{23} \ atoms \ Si$

<h3>3. Significant Figures</h3>

The original measurement of 8.88 grams has 3 significant figures, so our answer must have the same.

For the number we calculated, that is the hundredth place. The 4 in the thousandth place tells us to leave the 0 in the hundredth place.

$1.90 \times 10^{23} \ atoms \ Si$

<u>8.88 grams of silicon contains 1.90 ×10²³ atoms of silicon.</u>

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3 years ago

Water is being pumped from the bottom of a well 150 feet deep at a rate of 200 gal/hour into a vented storage tank 30 feet above

KengaRu [80]

Explanation:

As the given data is as follows.

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Heat energy required to heat 700 gal can be calculated as follows:

Heat Required = $5810 lbs \times 1 BTU/lb^{o}F \times (120^{o}F - 35^{o}F)$

Thus, water rises till $120^{o}F$ .

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Answer:

<em>The solution expected to contain the greatest number of solute particles is: </em><u>A) 1 L of 1.0 M NaCl</u>

Explanation:

The number of particles is calculated as:

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molarity × volume in liters × number of ions per unit formula.

b) <u>For covalent compounds</u>:

molarity × volume in liters

The difference is a factor which is the number of particles resulting from the dissociation or ionization of one mole of the ionic compound.

So, calling M the molarity, you can write:

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This table show the calculations for the four solutions from the list of choices:

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A) NaCl ionic ions Na⁺ and Cl⁻ 1.0 1.0 × 1 × 2 = 2

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C) Glucose covalent molecules 0.5 0.5 × 1 × 1 = 0.5

D) Glucose covalent molecules 1.0 1.0 × 1 × 1 = 1

Therefore, the rank in increasing number of particles is for the list of solutions given is: C < B = D < A, which means that the solution expected to contain the greatest number of solute particles is the solution A) 1 L of 1.0 M NaCl.

8 0

3 years ago