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nekit [7.7K]
3 years ago
7

Convert 64,567 milliliters into nanometers

Chemistry
1 answer:
LiRa [457]3 years ago
8 0

Answer:

64567000000 nanolitres

Explanation:

Base 10 decimal system: 1 milli = 1000000 nano

We simply multiply 64,567 millilitres by 1000000 to get our number in nanolitres:

64567(1000000) = 64567000000 nanolitres

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Which nitrogen base sequence is the partner of C-A-T-C-G-A?
marysya [2.9K]

Answer:

The answer should be B. G-T-A-G-C-T

A pairs with T and G pairs with C.

6 0
3 years ago
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Ira Lisetskai [31]
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3 years ago
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2C4H10 + 13O2 Right arrow. 8CO2 + 10H2O<br><br> What is the mole ratio of butane to carbon dioxide?
Zielflug [23.3K]

2C4H10 + 13O2 Right arrow. 8CO2 + 10H2O

What is the mole ratio of butane to carbon dioxide?

ANSWER:

Butane is an organic compound with the formula C4H10 and carbon dioxide is compound (gas) with the formula CO2. So, the balanced coefficient of butane is 2, and that of carbon dioxide is 8. According to thid, the mole ratio of butane to carbon dioxide is 2:8 or 1:4.

7 0
3 years ago
2. Answer the following questions about a sample of calcium phosphate:
weqwewe [10]

Answer:

a) <u>310.18 g/mol</u>

<u>b) 4.352 moles Ca3(PO4)2</u>

<u>c) 2.6 * 10^24 molecules</u>

<u>d) 5.24 * 10^24 P atoms</u>

<u>e)13.056 moles Ca</u>

<u>f)</u>10825.3 grams

Explanation:

Step 1: Data given

Atomic mass of Ca = 40.08 g/mol

Atomic mass of P = 30.97 g/mol

Atomic mass of O = 16.0 g/mol

Step 2: Calculate molecular weight of Ca3(PO4)2

Molecular weight of Ca3(PO4)2 = 3*atomic mass of Ca + 2* atomic mass of P and 8* atomic mass of O

Molecular weight of Ca3(PO4)2 = 3*40.08 + 2*30.97 + 8*16.0  =<u> 310.18 g/mol</u>

Step 3: Calculate moles of Ca3(PO4)2 in 1350 grams

Moles Ca3(PO4)2 = mass Ca3(PO4)2 /molar mass

Moles Ca3(PO4)2 = 1350 grams / 310.18 g/mol

Moles Ca3(PO4)2 = <u>4.352 moles</u>

Step 4: Calculate molecules in 1350 grams

Molecules = moles * number of Avogadro

Molecules = 4.352 moles * 6.02 * 10^23

Molecules = <u>2.6 *10^24 molecules</u>

<u />

Step 5: Calculate moles Phosphorus

For 1 mol Ca3(PO4)2 we need 2 moles P

For 4.352 moles Ca3(PO4)2 we have 2*4.352 = 8.704 moles

Step 6: Calculate P atoms

Atoms P = 8.704 moles * 6.02*10^23

Atoms P =<u> 5.24 * 10^24 P atoms</u>

<u />

Step 7: Calculate moles Calcium in 1350 grams

For 1 mol Ca3(PO4)2 we have 3 moles Ca

For 4.352 moles we have 3*4.352 = <u>13.056 moles Ca</u>

<u />

<u />

<u>Step 8:</u> Calculate mass of 2.1 * 10^25 molecules of Ca3(PO4)2

Moles Ca3(PO4)2 = 2.1 * 10^25 / 6.02 * 10^23

Moles Ca3(PO4)2 = 34.9 moles

Mass Ca3(PO4)2 = 34.9 moles * 310.18 g/mol

Mass Ca3(PO4)2 = 10825.3 grams

4 0
3 years ago
The substance fluorine has the following properties: normal melting point: 53.5 K normal boiling point: 85.0 K triple point: 1.6
mihalych1998 [28]

Answer:

The true statements include;

- The sample is initially a gas.

- The final state of the substance is a solid.

- One or more phase changes will occur.

The untrue/false statements include;

- The liquid initially present will vaporize.

- The final state of the substance is a liquid.

Explanation:

A couple pieces of informatton on Fluorine is imitially provided.

The substance fluorine has the following properties: normal melting point: 53.5 K normal boiling point: 85.0 K triple point: 1.6×10-4 atm, 53.4 K critical point: 55 atm, 144.1 K

So, a question is now attached about a sample of Fluorine. A sample of fluorine at a pressure of 1.00 atm and a temperature of 90.3 K is cooled at constant pressure to a temperature of 49.3 K.

We are then told to examine a group of options to find the ones that are correct/apply.

Taking the options one at a time

- The sample is initially a gas.

The initial state of the Fluorine sample has its temperature at 90.3 K, which is above the gas' boiling point. Hence, the sample can be concluded to initially be a gas.

- The liquid initially present will vaporize.

The sample doesn't initially contain liquid. And even of it did, the temperature is cooled, not heated , Hence, this statement is wrong.

- The final state of the substance is a solid.

The sample of Fluorine moves from a temperature higher than boiling point (85.0 K), with the sample in gaseous form, to one that is at a lower temperature (49.3 K) than the gas' normal melting point (53.5 K).

At temperatures lower than melting point, a substance exists in the solid form. Hence, this statement is true. The final state of the substance is solid.

- One or more phase changes will occur.

In moving from 90.3 K to 49.3 K for the sample and passing through the substance's boiling and melting points (85.0 K and 53.5 K respectively) along the way, it is logical to conclude that there would be one or more phase changes will occur. This statement is true.

- The final state of the substance is a liquid.

This is false as we already established that the final state of the substance is a solid. Hence, this statement is false.

Hope this Helps!!!

7 0
4 years ago
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