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ycow [4]
3 years ago
11

a 17.0 g sample of ammonia contains 14.0 g of nitrogen and 3.0 g of hydrogen . what is the mass % of hydrogen in ammonia

Chemistry
1 answer:
lina2011 [118]3 years ago
4 0

Answer:

% age of hydrogen = 17.65%

Explanation:

Given data:

Mass of ammonia = 17.0 g

Mass of hydrogen = 3.0 g

Mass of nitrogen = 14.0 g

Percentage of hydrogen = ?

Solution:

Formula:

% age of hydrogen = mass of hydrogen / mass of ammonia × 100

Now we will put the values:

% age of hydrogen = 3.0 / 17.0 × 100

% age of hydrogen = 17.65%

For oxygen:

% age of hydrogen = mass of hydrogen / mass of ammonia × 100

Now we will put the values:

% age of hydrogen = 14/ 17.0 × 100

% age of hydrogen = 82.35%

Total percentage =  17.65% + 82.35%

Total percentage = 100%

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8 0
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If an ice cube weighing 25.0 g with an initial
riadik2000 [5.3K]

Answer:

11

∘

C

Explanation:

As far as solving this problem goes, it is very important that you do not forget to account for the phase change underwent by the solid water at

0

∘

C

to liquid at

0

∘

C

.

The heat needed to melt the solid at its melting point will come from the warmer water sample. This means that you have

q

1

+

q

2

=

−

q

3

(

1

)

, where

q

1

- the heat absorbed by the solid at

0

∘

C

q

2

- the heat absorbed by the liquid at

0

∘

C

q

3

- the heat lost by the warmer water sample

The two equations that you will use are

q

=

m

⋅

c

⋅

Δ

T

, where

q

- heat absorbed/lost

m

- the mass of the sample

c

- the specific heat of water, equal to

4.18

J

g

∘

C

Δ

T

- the change in temperature, defined as final temperature minus initial temperature

and

q

=

n

⋅

Δ

H

fus

, where

q

- heat absorbed

n

- the number of moles of water

Δ

H

fus

- the molar heat of fusion of water, equal to

6.01 kJ/mol

Use water's molar mass to find how many moles of water you have in the

100.0-g

sample

100.0

g

⋅

1 mole H

2

O

18.015

g

=

5.551 moles H

2

O

So, how much heat is needed to allow the sample to go from solid at

0

∘

C

to liquid at

0

∘

C

?

q

1

=

5.551

moles

⋅

6.01

kJ

mole

=

33.36 kJ

This means that equation

(

1

)

becomes

33.36 kJ

+

q

2

=

−

q

3

The minus sign for

q

3

is used because heat lost carries a negative sign.

So, if

T

f

is the final temperature of the water, you can say that

33.36 kJ

+

m

sample

⋅

c

⋅

Δ

T

sample

=

−

m

water

⋅

c

⋅

Δ

T

water

More specifically, you have

33.36 kJ

+

100.0

g

⋅

4.18

J

g

∘

C

⋅

(

T

f

−

0

)

∘

C

=

−

650

g

⋅

4.18

J

g

∘

C

⋅

(

T

f

−

25

)

∘

C

33.36 kJ

+

418 J

⋅

(

T

f

−

0

)

=

−

2717 J

⋅

(

T

f

−

25

)

Convert the joules to kilojoules to get

33.36

kJ

+

0.418

kJ

⋅

T

f

=

−

2.717

kJ

⋅

(

T

f

−

25

)

This is equivalent to

0.418

⋅

T

f

+

2.717

⋅

T

f

=

67.925

−

33.36

T

f

=

34.565

0.418

+

2.717

=

11.026

∘

C

Rounded to two sig figs, the number of sig figs you have for the mass of warmer water, the answer will be

T

f

=

11

∘

C

Explanation:

3 0
2 years ago
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