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Shtirlitz [24]
3 years ago
9

Hypothermic body temperature can range from 28.7 °C to 35.5 °C. What is this range in degrees Fahrenheit?

Chemistry
1 answer:
Amanda [17]3 years ago
5 0
Solution:

To convert Celsius to Fahrenheit, the formula to be used is (C * 9/5) + 32

So in our given, 28.7 degrees Celsius:

Farenheit = (28.7 * 9/5) + 32
= 51.66 + 32
= 83.66 degrees Fahrenheit

Next, 35.5 degrees Celsius:
Farenheit = (35.5 * 9/5) + 32
= 63.9 + 32
= 95.9 degrees Fahrenheit

So in Fahrenheit, the range would be 83.66 – 95.9 degrees Fahrenheit.
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Answer:

The volume of the gasoline in the tank is 18.85 cubic feet

Explanation:

The tank is a cylinder. The equation for calculating the cylinder volumen is πr²*h, r is radius and h is height. π is approximately 3.1416. So,  The tank is  resting horizontally on its side, its radius is 2 feet/2= 1 feet. h=a tank long= 6 feet.

With equation:

Vol gasoline=3.1416*1²*6=18.85 feet ³

I hope my answer helps you

6 0
3 years ago
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Answer:

Just here for the points sorry

Explanation:

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6 0
2 years ago
How would the boiling point of water be affected if you doubled the amount of water?
Flauer [41]

Answer:

The boiling point wouldn´t be effected. The amount of time it would take to get it to that point would increase

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8 0
3 years ago
How much heat, in joules, must be added to a 75.0-g iron block with a specific heat of 0.449 J/g °C to increase its temperature
svp [43]

Answer:

50,849.25 Joules

Explanation:

The amount of heat, Q, required to raise the temperature of a body with mass, m, and specific heat capacity, c is given by:

Q = mcΔT, where ΔT represents the change in temperature.

In the case of the iron block:

m = 75 g

c = 0.449 J/g °C

ΔT = 1535 - 25 = 1510 °C

Therefore,

Q = 75 g x 0.449 J/g °C x 1510 °C

     = 50,849.25 Joules

<em>Hence, </em><em>50,849.25 Joules </em><em> of heat must be added to  a 75.0-g iron block with a specific heat of 0.449 J/g °C to increase its temperature from 25 °C to its melting temperature of 1535 °C</em>

5 0
2 years ago
What is the pOH value of 0.0000877 M HCL
vredina [299]

Answer:

pOH=9.9

Explanation:

pH=-log[H+]= -log[0.0000877]

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pOH+ pH=14

pOH=14-4.06= 9.91

8 0
2 years ago
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