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iris [78.8K]
3 years ago
15

If 15 grams of Carbon dioxide is produced in a chemical reaction, how many grams of Carbon must be consumed in the reaction if w

e know there were 11 grams of Oxygen on the reactants side of the equation?
Chemistry
1 answer:
guajiro [1.7K]3 years ago
5 0

Answer:

Quantity of Carbon is 4.09 gm

Explanation:

Equation of carbon reacting with oxygen to give carbon dioxide is given by

C + O_{2} ⇒ CO_{2}

One mole of carbon reacts with one mole of Oxygen in this reaction to give One mole of Carbon dioxide.

So, 12 gm of carbon reacts with 32 gm of Oxygen in this reaction to give 44 gm of carbon dioxide.

15 gm of CO_{2} was formed in this reaction

Oxygen used in this reaction = \frac{15}{44}×32 = 10.91 gm ,

Thus Oxygen is in sufficient quantity in the reaction.

Now,

Carbon that must be used = \frac{15}{44}×12  = 4.09 gm.

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A chemist weighed out 20.7 g of sodium . Calculate the number of moles of sodium she weighed out
Pachacha [2.7K]

Answer:

about 0.9 mol

Explanation:

there are 22.990 g/mol of Na

20.7/22.99 = 0.900391 mol

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8 0
3 years ago
PLEASE SOMEONE HELP ME WITH THIS!
Bad White [126]

Answer:

1) 1.202 L , 2) 1.291 dg , 3) 204.877  and 4) 1.04x10^{3\\}

Explanation:

You need to review about conversion factors and how to use them in the correct order. You can cancel the units and get the ones that you need if you use the appropriate conversion factors, remember is a number that you can use to multiply or divide.

For your exercise:

1) The conversion factor is: 1 L = 1000 mL

You will need to divide by 1000 mL to obtain liters L

1202.57120 mL x  \frac{1 L}{1000 mL} = 1.202 L

2) The conversion factor is: 1 g = 10 dg

0.1290743 g x \frac{10 dg}{ 1 g} = 1.291 dg

For the next exercises, you need to follow some rules:

1.  All numbers  that are different from Zero (non-zero digits) are significant figures.

2.The Zeros between non-zeros digits (Imbedded zeros) always are significant, 2007.

3. If you want to be specific and want some zeros to be significant you need to add a decimal point. For example 500. or 500.0

4. Leading zeros (to the left) are not significant.

5. Trailing zeros (zeros to the right) in a whole number without decimal point are not significant.

3) 843.062  - 638.1848  = 204.8772

Now if we round to 6 significant figures we get 204.877

4)123.0 x 8.43 = 1036.89

Now we round to 3 significant figures because 8.43 has the least significant figures.  

1.04x10^{3}

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Answer:

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