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bearhunter [10]
3 years ago
15

How many moles of n are in 0.185 g of n2o?

Chemistry
2 answers:
pickupchik [31]3 years ago
8 0

The mass of N_{2}O is 0.185 g.

To calculate the number of moles first calculate the molar mass of N_{2}O .

M_{N_{2}O}=2\times M_{N}+M_{O}

Molar mass of nitrogen and oxygen is 14 g/mol and 16 g/mol respectively

Thus, M_{N_{2}O}=2\times 14 g/mol+18 g/mol=44 g/mol

Now, number of moles can be calculated from mass and molar mass as follows:

n=\frac{m}{M}=\frac{0.185 g}{44 g/mol}=0.0042 mol

Therefore, number of moles of 0.185 g N_{2}O is 0.0042 mol.

solong [7]3 years ago
3 0

Answer : The number of moles of nitrogen are 0.00840 moles.

Explanation : Given,

Mass of N_2O = 0.185 g

Moles of N_2O = 44 g/mole

First we have to calculate the moles of N_2O.

\text{ Moles of }N_2O=\frac{\text{ Mass of }N_2O}{\text{ Molar mass of }N_2O}=\frac{0.185g}{44g/mole}=0.00420moles

Now we have to calculate the moles of nitrogen (N).

In N_2O, there are 2 moles of nitrogen atom and 1 mole of oxygen atom.

As, 1 mole of N_2O contains 2 moles of nitrogen

So, 0.00420 mole of N_2O contains 2 × 0.00420 = 0.00840 moles of nitrogen

Therefore, the number of moles of nitrogen are 0.00840 moles.

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A 72.0 mL aliquot of a 1.40 M solution is diluted to a total volume of 248 mL. A 124 mL portion of that solution is diluted by a
Aliun [14]

Answer: 0.20 M

Explanation:

According to the dilution law,

M_1V_1=M_2V_2

where,

M_1 = molarity of stock solution = 1.40 M

V_1 = volume of stock solution = 72.0 ml

M_2 = molarity of diluted solution = m

V_2 = volume of diluted solution = 248 ml

1.40\times 72.0=m\times 248

m=0.41M

Now 124 mL portion of this prepared solution is diluted by adding 133 mL of water.

According to the dilution law,

M_1V_1=M_2V_2

where,

M_1 = molarity of stock solution = 0.41 M

V_1 = volume of stock solution = 124 ml

M_2 = molarity of diluted solution = m

V_2 = volume of diluted solution = (124 +133) ml = 257 ml

0.41\times 124=m\times 257

m=0.20M

Thus the final concentration of the solution is 0.20 M.

8 0
3 years ago
What type of forces which exist in liquid hydrogen fluoride ?
Misha Larkins [42]

Answer:

H-F is a polar covalent molecule in which dipole-dipole interactions exists

7 0
2 years ago
Need help please?!!!!!​
puteri [66]

(a) 33.6 L of oxygen would be produced.

(b) 106 grams of  Na_2CO_3 would be needed

<h3>Stoichiometric calculations</h3>

1 mole of gas = 22.4 L

(a) From the equation, 2 moles of KClO_3 produce 3 moles of O_2. 1 mole of  KClO_3 will, therefore, produce 1.5 moles of  O_2.

1.5 moles of oxygen = 22.4 x 1.5 = 33.6 L

(b) 22.4 L of CO_2 is produced at STP. This means that 1 mole of the gas is produced.

From the equation, 1 mole of  CO_2 requires 1 mole of Na_2CO_3.

Molar mass of  Na_2CO_3 = (23x2)+ (12)+(16x3) = 106 g/mol

Mass of 1 mole  Na_2CO_3 = 1 x 106 = 106 grams

More on stoichiometric calculations can be found here: brainly.com/question/27287858

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6 0
2 years ago
10.0 mL of a HF solution was titrated with a 0.120 N solution of KOH; 29.6 mL of
Ilia_Sergeevich [38]

Answer:

0.355 N of HF

Explanation:

The titration reaction of HF with KOH is:

HF + KOH → H₂O + KF

<em>Where 1 mole of HF reacts per mole of KOH</em>

<em />

Moles of KOH are:

0.0296L × (0.120 equivalents / L) = 3.552x10⁻³ equivalents of KOH = equivalents of HF.

As volume of the titrated solution was 10.0mL, normality of HF solution is:

3.552x10⁻³ equivalents of HF / 0.010L =<em> 0.355 N of HF</em>

3 0
3 years ago
A chemical engineer must calculate the maximum safe operating temperature of a high-pressure gas reaction vessel. The vessel is
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