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3 years ago

How many moles of n are in 0.165 g of n2o?

Chemistry

2 answers:

ipn [44]3 years ago

4 0

Answer:

$n_N=0.0075molN_2$

Explanation:

Hello,

In order to compute the required moles, one performs the following mole mass relationship in which it is seen there are two nitrogens into the $N_2O$ :

$n_N=0.165gN_2O*\frac{1molN_2O}{44gN_2O} *\frac{2molN}{1molN_2O}\\n_N=0.0075molN_2$

Best regards.

WINSTONCH [101]3 years ago

3 0

Answer:

0.045 moles of N

Solution:

Data Given:

Mass of N₂O = 0.165 g

M.Mass of N₂O = 44.02 g.mol⁻¹

Step 1: Calculate Moles of N₂O as,

Moles = Mass ÷ M.Mass

Putting values,

Moles = 0.165 g ÷ 44.02 g.mol⁻¹

Moles = 0.02266 mol

Step 2: Calculate Moles of N,

As,

1 mole of N₂O contains = 2 moles of N

So,

0.02266 mol of N₂O will contain = X moles of N

Solving for X,

X = (0.02266 mol × 2 mol) ÷ 1 mol

X = 0.045 moles of N

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A chemistry student needs 45.0mL of pentane for an experiment. By consulting the CRC Handbook of Chemistry and Physics, the stud

Masja [62]

The mass of pentane the student should weigh out is

The density of pentane is 0.626 gcm-3

To calculate the mass of pentane following expression is used,

(Density is defined as the mass divide by volume)

Density = mass / volume

mass of pentane = Density of pentane * Volume of pentane

mass of pentane = 0.626 gcm-3 * 45.0 mL

= 28.17 g

Here the unit of mass of pentane is g,

However the unit of density is gcm-3 and unit of volume is mL i.e. cm3

Hence, Mass = gcm-3 * cm3

Mass = g

The mass of pentane the student should weigh out is 28.17g

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5 0

2 years ago

How many grams of oxygen are produced when 7.65 moles of water is decomposed

maksim [4K]

Answer:

The answer to your question is 122.4 g of O₂

Explanation:

Data

mass of O₂ = ?

moles of H₂O = 7.65

Process

1.- Write the balanced chemical reaction

2H₂O ⇒ 2H₂ + O₂

2.- Convert the moles of H₂O to grams

molar mass of H₂O = 2 + 16 = 18 g

18 g of H₂O ---------------- 1 mol

x ----------------- 7.65 moles

x = (7.65 x 18) / 1

x = 137.7 g H₂O

3.- Calculate the grams of O₂

36 g of H₂O -------------------- 32 g of O₂

137.7 g of H₂O ------------------- x

x = (32 x 137.7) / 36

x = 122.4 g of O₂

6 0

3 years ago

1. The solubility of lead(II) chloride at some high temperature is 3.1 x 10-2 M. Find the Ksp of PbCl2 at this temperature.

solniwko [45]

Answer:

1) The solubility product of the lead(II) chloride is $1.2\times 10^{-4}$ .

2) The solubility of the aluminium hydroxide is $1.6\times 10^{-10} M$ .

3)The given statement is false.

Explanation:

Solubility of lead chloride = $S=3.1\times 10^-2M$

$PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)$

S 2S

The solubility product of the lead(II) chloride = $K_{sp}$

$K_{sp}=[Pb^{2+}][Cl^-]^2$

$K_{sp}=S\times (2S)^2=4S^3=4\times (3.1\times 10^{-2})^3=1.2\times 10^{-4}$

The solubility product of the lead(II) chloride is $1.2\times 10^{-4}$ .

Concentration of aluminium nitrate = 0.000010 M

Concentration of aluminum ion = $1\timed 0.000010 M=0.000010 M$

Solubility of aluminium hydroxide in aluminum nitrate solution = $S$

$Al(OH)_3(aq)\rightleftharpoons Al^{3+}(aq)+3OH^-(aq)$

S 3S

The solubility product of the aluminium nitrate = $K_{sp}=1.0\times 10^{-33}$

$K_{sp}=[Al^{3+}][OH^-]^3$

$1.0\times 10^{-33}=(0.000010+S)\times (3S)^3$

$S=1.6\times 10^{-10} M$

The solubility of the aluminium hydroxide is $1.6\times 10^{-10} M$ .

$Molarity=\frac{Moles}{Volume (L)}$

Mass of NaCl= 3.5 mg = 0.0035 g

1 mg = 0.001 g

Moles of NaCl = $\frac{0.0035 g}{58.5 g/mol}=6.0\times 10^{-5} mol$

Volume of the solution = 0.250 L

$[NaCl]=\frac{6.0\times 10^{-5} mol}{0.250 L}=0.00024 M$

1 mole of NaCl gives 1 mole of sodium ion and 1 mole of chloride ions.

$[Cl^-]=[NaCl]=0.00024 M$

Moles of lead (II) nitrate = n

Volume of the solution = 0.250 L

Molarity lead(II) nitrate = 0.12 M

$n=0.12 M]\times 0.250 L=0.030 mol$

1 mole of lead nitrate gives 1 mole of lead (II) ion and 2 moles of nitrate ions.

$[Pb^{2+}]=[Pb(NO_2)_3]=0.030 M$

$PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)$

Solubility of lead(II) chloride = $K_{sp}=1.2\times 10^{-4}$

Ionic product of the lead chloride in solution :

$Q_i=[Pb^{2+}][Cl^-]^2=0.030 M\times (0.00024 M)^2=1.7\times 10^{-9}$

$Q_i$ ( no precipitation)

The given statement is false.

3 0

3 years ago

Not enough air is pumped into an inflatable life raft to make completely filled out the atm pressure is 14.7psi what is the pres

sasho [114]

If you do not inflate the life raft to make completely filled out, as long as you do not press or squeeze the life raft, the air inside it will be in equilibrium with the air outside the raft, and so the pressure inside the life raft will be the same atmospheric pressure, 14.7 psi.

Note that when the raft is swollen, if you punch it, the air will leave from it which means that the pressure inside is greater than the atmospheric pressure.

5 0

3 years ago

In Figure 13-3, which spheres have the GREATEST density (top, middle, or

natulia [17]

Answer:

bottom

Explanation:

the ones with the greatest density will sink. if they have a lighter density than water, they will float. if they have about the same, they will stay in the middle. and if they have a higher density they will sink.

3 0

3 years ago

Read 2 more answers