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3 years ago

How many moles of N are in 0.223 g of N2O

1 answer:

4 0

Answer:
There are twice the molar quantity of nitrogen atoms in nitrous oxide, i.e.
1×10−2⋅mol
Moles of nitrous oxide

=

0.217
⋅
g
44.013
⋅
g
⋅
m
o
l
−
1

=

5
×
10
−
3
⋅
m
o
l
.
Given the composition of
N
2
O
, there are this
2
×
5
×
10
−
3
⋅
m
o
l
of nitrogen atoms.

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The energy transfer between the system and the surroundings, in the form of heat at constant pressure is described by which of t

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3 years ago

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Answer:

C2H2O4

Explanation:

To get the molecular formula, we first get the empirical formula. This can be done by dividing the percentage compositions by the atomic masses. The percentage compositions are shown as follows :

C = 26.86%

H = 2.239%

O = 100 - ( 26.86 + 2.239) = 70.901%

We then proceed to divide by their atomic masses. Atomic mass of carbon is 12 a.m.u , H = 1 a.m.u , O = 16 a.m.u

The division is as follows:

C = 26.86/12 = 2.2383

H = 2.239/1 = 2.239

O = 70.901/16 = 4.4313

We now divide each by the smallest number I.e 2.2383

C = 2.2383/2.2383 = 1

H = 2.239/2.2383 = 1

O = 4.4313/2.2383 = 1.98 = 2

Thus, the empirical formula is CHO2.

To get the molecular formula, we use the molar mass .

(CHO2)n = 90

We add the atomic masses multiplied by n.

(12 + 1 + 2(16))n = 90

45n = 90

n = 90/45 = 2.

Thus , the molecular formula is C2H2O4

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3 years ago

What is the temperature of 0.47 mol of gas at a pressure of 1.5 atm and a volume of 10.5 l ?

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The temp is 0.002448 of the equation

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3 years ago

In the simulation, open the Custom mode. The beaker will be filled to the 0.50 L mark with a neutral solution. Set the pH to 2.8

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3 years ago

Under certain conditions, the equilibrium constant of the reaction below is Kc=1.7×10−3. If the reaction begins with a concentra

svetoff [14.1K]

Answer:

[Cl2] equilibrium = 0.0089 M

Explanation:

Given:

[SbCl5] = 0 M

[SbCl3] = [Cl2] = 0.0546 M

Kc = 1.7*10^-3

To determine:

The equilibrium concentration of Cl2

Calculation:

Set-up an ICE table for the given reaction:

$SbCl5(g)\rightleftharpoons SbCl3(g)+Cl2(g)$

I 0 0.0546 0.0546

C +x -x -x

E x (0.0546-x) (0.0546-x)

$Kc = \frac{[SbCl3][Cl2]}{[SbCl5]}\\\\1.7*10^{-3} =\frac{(0.0546-x)^{2} }{x} \\\\x = 0.0457 M$

The equilibrium concentration of Cl2 is:

= 0.0546-x = 0.0546-0.0457 = 0.0089 M

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3 years ago

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