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inysia [295]
2 years ago
8

The freezing point of water in degrees Celsius is [ Select ] __________. The freezing point of water in degrees Fahrenheit is [

Select ] __________. The freezing point of water in Kelvin is [ Select ] __________. (Use whole numerals/numbers for all choices — no letters or symbols)
Chemistry
1 answer:
ElenaW [278]2 years ago
4 0

Answer: 0, 32 , 273

Explanation:

The freezing point of water in degree Celsius, Fahrenheit and Kelvin are in the following order;   0  , 32  , 273

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Determine the mass of a gold bar that has a density of 19.3 g/cm3 and is 4.72 cm high by 8.21 cm long by 3.98 cm deep.
TEA [102]
The density of the gold is 19.3 grams/cc so each cc weighs 19.3grams. Now we can obtain the volume of gold from the given dimensions ie 4.72x8.21x3.98= 154.23 cc. So for the answer, just multiply the volume or 154.23 x 19.3= 2976.6 grams is the answer.
5 0
3 years ago
Carbon forms a great variety of stable compounds. Which of the following is not a reason for this variety?
katen-ka-za [31]

Answer:

D.

Explanation:

Carbon atoms are not strongly electronegative  and  tend to form  covalent bonds.

4 0
2 years ago
What compound do cells break down for energy in the process of cellular respiration?
arsen [322]

Answer:

Glucose

Explanation:

Glucose is broken down into water and carbon dioxide.

hope this helps! :)

6 0
3 years ago
I need the answer so much
Mariulka [41]
I don't know sorry just google it
4 0
2 years ago
Methane, the principal component of natural gas, is used for heating and cooking. The combustion process is CH4(g) + 2 O2(g) → C
antoniya [11.8K]

Answer:

392.97 litres

Explanation:

From the equation of reaction, we can see that 1 mole of methane yielded 1 mole of carbon iv oxide. Hence, 15.9 moles of methane will yield 15.9 moles of carbon iv oxide.

At s.t.p one mole of a gas occupies a volume of 22.4L ,hence 15.9 moles of a gas will occupy a volume of 22.4 × 15.9 which equals

356.16L.

Now, we can use the general gas equation to get the volume produced at the values given.

We have the following values;

V1 = 356.16L P1= 1 atm ( standard pressure) T1 = 273K ( standard temperature) V2 = ? T2 = 23.7 + 273 = 296.7K P2 = 0.985 atm

The general form of the general gas equation is given as :

(P1V1)T1 = (P2V2)/T2

After substituting the values , we get V2 to be 392.97Litres

8 0
2 years ago
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