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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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Which type of mixture can be separated using magnetism?

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Answer: The correct option is heterogeneous mixture whose components are attracted differently to a magnet.

Explanation: There are two types of mixtures:

1) Homogeneous mixtures: In these mixtures, the particles are uniformly distributed throughout the mixture. These particles cannot be separated.

2) Heterogeneous Mixtures: These are the mixtures where the particles are visible separated and are not-uniformly distributed. These particles can be separated easily.

If magnet is used to separate the components of a mixture, the heterogeneous mixtures will only get separated.

To separate the components by a magnet, the components of a mixture should attract the magnet differently. One component should attract the magnet and another should not. Hence, they can be easily separated.

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3 years ago

a sample of compound determined to contain 1.71 g C and 0.287 g H. The corresponding empirical formula is

fiasKO [112]

Answer: The empirical formula is $CH_2$ .

Explanation:

Mass of C = 1.71 g

Mass of H = 0.287 g

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{1.71g}{12g/mole}=0.142moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.287g}{1g/mole}=0.287moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{0.142}{0.142}=1$

For H = $\frac{0.287}{0.142}=2$

The ratio of C: H = 1: 2

Hence the empirical formula is $CH_2$ .

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3 years ago

How many grams of lead(II) sulfate (303 g/mol) are needed to react with sodium chromate (162 g/mol) in order to produce 0.162 kg

Afina-wow [57]

Answer : The mass of $PbSO_4$ needed are, 1.515 grams.

Explanation :

First we have to calculate the mole of $PbCrO_4$ .

$\text{Moles of }PbCrO_4=\frac{\text{Mass of }PbCrO_4}{\text{Molar mass of }PbCrO_4}=\frac{0.162g}{323g/mole}=0.005mole$

Now we have to calculate the moles of $PbSO_4$ .

The balanced chemical reaction will be,

$PbSO_4+Na_2CrO_4\rightarrow PbCrO_4+Na_2SO_4[tex]From the balanced chemical reaction, we conclude thatAs, 1 mole of [tex]PbCrO_4$ produced from 1 mole of $PbSO_4$

So, 0.005 mole of $PbCrO_4$ produced from 0.005 mole of $PbSO_4$

Now we have to calculate the mass of $PbSO_4$

$\text{Mass of }PbSO_4=\text{Moles of }PbSO_4\times \text{Molar mass of }PbSO_4$

$\text{Mass of }PbSO_4=0.005mole\times 303g/mole=1.515g$

Therefore, the mass of $PbSO_4$ needed are, 1.515 grams.

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