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3 years ago

Make the following conversions. Show your calculations.

Chemistry

1 answer:

Natalija [7]3 years ago

5 0

Answer:

1. 500 cm = 5 m

2. 1000 g = 1 kg

3. 455 L = 45, 500 cL

4. 0.865 m = 0.00865 m or 8.65 * 10^-3 n

5. 33.5 cm = 335 mm

6. 0.0198 m = 19800 micrometers

7. 57.65 cg = 5.765 * 10^8 nanograms

8. 1000 L = 1 kl

9. 99 degrees F = 37.222 degrees C

Explanation:

1. One meter = 100 centimeters

500 centimeters/ 100 centimeters = 5 meters

2. One kilogram = 1000 grams.

1000 grams/1000 grams = 1 kilogram

3. One Liter = 100 centiliters

455 liters * 100 centiliters = 45, 500 centiliters

4. One meter = 100 centimeters

0.865 centimeters/100 centimeters = 0.00865 meters

5. One centimeter = 10 millimeters

33.5 centimeters * 10 millimeters = 335 millimeters

6. One meter = 1.0 * 10^6 micrometers

0.0198 meters * 1 * 10^6 micrometers = 19800 micrometers.

7. One centigram = 1.0 * 10^7 nanograms

57.65 centigrams * 1 * 10^7 nanograms = 5.765 * 10^8 nanograms.

8. One kiloliter = 1000 L

1000 L/ 1000 L = 1 kiloliter

9. The formula for finding Celsius is 5/9(f - 32)

So 5/9(99 - 32) = x

5/9(67) = x

x = 335/9 or 37.222 degrees Celsius.

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A quantity of CO gas occupies a volume of 0.64 L at 0.90 atm and 307 K . The pressure of the gas is lowered and its temperature

levacccp [35]

This is an incomplete question, here is a complete question.

A quantity of CO gas occupies a volume of 0.48 L at 1.6 atm and 282 K. The pressure of the gas is lowered and its temperature is raised until its volume is 2.1 L. Find the density of the CO under the new conditions?

Answer : The density of the CO under the new conditions is, 0.213 g/L

Explanation :

First we have to determine the moles of CO gas by using ideal gas equation.

$PV=nRT$

where,

P = pressure of gas = 0.90 atm

V = volume of gas = 0.64 L

T = temperature of gas = 3 07 K

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

n = number of moles of gas = ?

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

$(0.90atm)\times (0.64L)=n\times (0.0821L.atm/mol.K)\times (307K)$

$n=0.0228mol$

Now we have to calculate the mass of CO gas.

$\text{ Mass of }CO=\text{ Moles of }CO\times \text{ Molar mass of }CO$

Molar mass of $CO$ = 28g/mole

$\text{ Mass of }CO=(0.0228moles)\times (28g/mole)=0.638g$

Now we have to calculate the density of CO gas.

$\text{Density of CO gas}=\frac{\text{Mass of CO gas}}{\text{Volume of CO gas}}$

$\text{Density of CO gas}=\frac{0.638g}{3.0L}=0.213g/L$

Therefore, the density of the CO under the new conditions is, 0.213 g/L

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She was expecting 122.7 grams of lead for her lesson that day but only got out 107.9 grams of lead. What was the percent yield o

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As such, the theoretical yield or expected yield is, in this case, 122.7g of Pb, the actual in this case is 107.9g so we have
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