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Vanyuwa [196]
3 years ago
14

What is the heat required to vaporize 650 grams of a liquid with a heat of vaporization of 723 joules/grams?

Chemistry
1 answer:
wolverine [178]3 years ago
3 0

Your answer would be E. 470,000 J

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65. Write an equation for the combustion of octane.
siniylev [52]

Answer:

2C8H18+25O2→16CO2+18H2O.



4 0
2 years ago
Which cannot be chemically broken down into simpler substances?
Anna71 [15]

Answer:

B.

Explanation:

elements

6 0
2 years ago
Read 2 more answers
The Nutrition Facts label for crackers states that one serving contains 19 g of carbohydrate, 4 g of fat, and 2 g of protein. Wh
sp2606 [1]

<u>Answer:</u> The amount of kilocalories contained in the given serving of crackes is 0.120 kCal

<u>Explanation:</u>

<u>As per the USDA:</u>

Carbohydrates provide 4 calories per gram, protein provides 4 calories per gram, and fat provides 9 calories per gram

We are given:

Mass of fat in the crackers = 4 g

Mass of carbohydrates in the crackers = 19 g

Mass of protein in the crackers = 2 g

Conversion factor used:  1 kCal = 1000 Cal

Applying unitary method:

  • <u>For Fat:</u>

1 gram of fat provides 9 calories

So, 4 gram of fat will provide = \frac{9}{1}\times 4=36Cal=0.036kCal

  • <u>For Carbohydrates:</u>

1 gram of carbohydrates provides 4 calories

So, 19 gram of carbohydrates will provide = \frac{4}{1}\times 19=76Cal=0.076kCal

  • <u>For Proteins:</u>

1 gram of proteins provides 4 calories

So, 2 gram of fat will provide = \frac{4}{1}\times 2=8Cal=0.008kCal

Total kilocalories per serving = [0.036 + 0.076 + 0.008] kCal = 0.120 kCal

Hence, the amount of kilocalories contained in the given serving of crackes is 0.120 kCal

8 0
3 years ago
A functional group introduces heteroatoms into a carbon chain to increase
ivanzaharov [21]

Answer:

reactivaty

Explanation:

here you go for the answer

5 0
3 years ago
If 60. liters of hydrogen gas at 546 K is cooled to 273 K at constant pressure, the new volume of the gas would be
Korvikt [17]

Answer:30 L

Explanation:

Initial Volume

=

V

1

=

60

l

i

t

e

r

Initial Temperature

=

T

1

=

546

K

Final Temperature

=

T

2

=

273

K

Final Vloume

=

V

2

=

?

?

Sol:-

Since the pressure is constant and the question is asking about temperature and volume, i.e,

V

1

T

1

=

V

2

T

2

⇒

V

2

=

V

1

⋅

T

2

T

1

=

60

⋅

273

546

=

60

2

=

30

l

i

t

e

r

⇒

V

2

=

30

l

i

t

e

r

Hence the new volume of the gas is

30

l

i

t

e

r

6 0
3 years ago
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