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svetlana [45]
3 years ago
11

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?
Sergeu [11.5K]

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.

5 0
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.

4 0
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.

6 0
2 years ago
Saftey goggles are worn in the lab- ​
Harrizon [31]

Answer:

Yes

Explanation:

Is this a question or what?

4 0
3 years ago
Which postmortem parameter reaches a maximum at 12 hours and is gone within 36 to 48 hours?1.rigor mortis2.cloudiness of the eye
Olenka [21]
The cloudiness of the eyes increases to a maximum and then decreases. This is because initially, after death, all the muscles relax, dilating the pupil. Some time later, rigor mortis sets in, contracting the pupil. Thus, the cloudiness fades.
3 0
3 years ago
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