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Veronika [31]
1 year ago
10

Perform the following conversions:

Chemistry
1 answer:
spayn [35]1 year ago
5 0

0 K in Celsius scale is -273 degree Celsius and in Fahrenheit scale 459 degree F.

The degree of a body's heat or cold is referred to as the gas's temperature. It is stated in temperature units such as, degree Celsius, Fahrenheit  and Kelvin these temperature units are interchangeable. To convert Celsius to Fahrenheit ° F = 9/5 ( ° C) + 32, kelvin to Fahrenheit= ° F = 9/5 (K - 273) + 32, Celsius to kelvin K = ° C + 273.  to convert Fahrenheit to Celsius ° C = 5/9 (° F - 32)  In this question,  to convert Fahrenheit to Celsius ° C = 5/9 (° F - 32) = 0-273 K (5/9 - F-32) =459 degree F. and kelvin to Celsius K - 273 = C = -273 degree Celsius

At 8^0C K= 8 + 273 =281 degree Celsius and at 8F , 8= 9/5(k-273) +32 =453 F

To learn more about Fahrenheit and Celsius:

brainly.com/question/1844449

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Oxygen gas is collected at a pressure of 1.21 atm in a container which has a volume of 10.0 L. What temperature (in Kelvin) must
stiv31 [10]

Answer : The temperature of gas is, 295 K

Explanation :

To calculate the temperature of gas we are using ideal gas equation:

PV=nRT

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n = number of moles = 0.500 mole

R = Gas constant = 0.0821L.atm/mol.K

T = Temperature of gas = ?

Now put all the given values in above equation, we get:

1.21atm\times 10.0L=0.500mole\times (0.0821L.atm/mol.K)\times T

T=294.8Kapprox 295K

Therefore, the temperature of gas is, 295 K

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3 years ago
A sample from a local stream is found to have 3.0 ppm nitrates (in the form of sodium nitrate). How many grams of sodium nitrate
maw [93]

Answer:

0.01028  grams of sodium nitrate would there be in 2.5 L of the stream.

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The ppm is the amount of solute (in milligrams) present in kilogram of a solvent. It is also known as parts-per million.

To calculate the ppm of oxygen in sea water, we use the equation:

\text{ppm}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 10^6

Both the masses are in grams.

We are given:

The ppm concentration of nitrates = 3.0 ppm

Mass of nitrates = x

Mass of steam= m

Volume of steam = V = 2.5 L = 2500 ml ( 1 L = 1000 mL)

Density of steam = d = 1.0 g/mL

M=d\times V=1.0 g/mL\times 2500 mL = 2500 g

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3.0=\frac{x}{2500 g}\times 10^6

x=0.0075 g

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Moles of nitrate = \frac{0.0075 g}{62 g/mol}=0.0001210 mol

1 mole of nitrate ion is present in 1 mole of sodium nitrate.

Then 0.0001210 moles of nitrate will be present in :

\frac{1}{1}\times 0.0001210 mol=0.0001210 mol of sodium nitrate;

Mass of 0.0001210 moles of sodium nitrate :

0.0001210 mol × 85 g/mol = 0.01028 g

0.01028  grams of sodium nitrate would there be in 2.5 L of the stream.

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