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nikklg [1K]
2 years ago
11

You decide to establish a new temperature scale on which the melting point of ammonia (-77.75 ∘c) is 0∘a, and the boiling point

of ammonia (-33.35∘c) is 100 ∘a. part a what would be the boiling point of water in ∘a?
Chemistry
2 answers:
Hoochie [10]2 years ago
7 0
First we need to know that the boiling point of water in C is 100 and we just need to solve for x in the equation:

-33.75-(-77.75) / 100 = 100-(-77.75) / x
44.4/100 = 177.75 / x
 x = 177.75*100/44.4 = 400.33

The boiling point of water in ∘a would be 400.33∘a.

AveGali [126]2 years ago
3 0

<u>Answer:</u> The boiling point of water is 400.34^oA

<u>Explanation:</u>

To convert the units of temperature, we use the equation:

\frac{\text{X-LFP}}{\text{UFP-LFP}}=\frac{\text{Y-LFP}}{\text{UFP-LFP}}

where,

X = temperature in ^oA

Y = temperature in ^oC

LFP = lower fixed point

UFP = upper fixed point

LFP of ^oA is 0°A

UFP of ^oA is 100°A

LFP of ^oC is -77.75°A

UFP of ^oC is -33.35°A

Normal boiling point of water = 100^oC

Putting values in above equation, we get:

\frac{T(^oA)-0^oA}{100^oA-0^oA}=\frac{100^oC-(-77.75^oC)}{-33.35^oC-(-77.75^oC)}\\\\\frac{T(^oA)}{100^oA}=\frac{177.75^oC}{44.4^oC}\\\\T(^oA)=400.34^oA

Hence, the boiling point of water is 400.34^oA

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