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3 years ago

Around four hundred million years after the formation of Earth, the planet's atmosphere would have been almost entirely _____.

Chemistry

2 answers:

Olegator [25]3 years ago

4 0

the answer is option d

posledela3 years ago

3 0

Answer:

D.) Carbon Dioxide

Explanation:

You welcome

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Since the number of atoms in a substance is so large , a unit to count them was created. This unit is the number of atoms in 12

larisa86 [58]

The number of atoms in one mole of any substance is measured by Avogadro's number. The value of Avogadro's number is 6.023 x 10 ^23. It is named after scientist Avogadro who proposed this number. 12 grams of carbon-12 represents 1 mole of carbon-12. For this reason, the number of atoms present in 1 mole of any substance is 6.023 x 10 ^23. Therefore, the number of atoms present in 1 mole carbon-12 is 6.023 x 10^23.

(Answer) This unit is the number of atoms in 12 grams of carbon-12 and known as Avogadro's number.

7 0

3 years ago

Am I incorrect ?<br> Or nah?

Mademuasel [1]

I think so... I'm currently learning this too but you should be correct

4 0

3 years ago

Read 2 more answers

What is the percent composition of nitrogen in a 2.57 g sample of Al(NO3)3?

Lisa [10]

Answer:

19.8% of Nitrogen

Explanation:

In the Al(NO₃)₃ there are:

1 atom of Al

3 atoms of N

And 9 atoms of O

The molar mass of Al(NO₃)₃ is:

1 Al * (26.98g/mol) = 26.98g/mol

3 N * (14g/mol) = 42g/mol

9 O * (16g/mol) = 144g/mol

26.98 + 42 + 144 = 212.98g/mol

We can do a conversion using these molar masses to find the mass of nitrogen is the sample, that is:

2.57g * (42g/mol / 212.98g/mol) =

0.51g N

Percent composition of nitrogen is:

0.51g N / 2.57g * 100

= 19.8% of Nitrogen

6 0

2 years ago

What mass of butane in grams is necessary to produce 1.5×103 kj of heat what mass of co2 is produced?

kari74 [83]

The heat of reaction (i.e. combustion) of butane ( $C_{4} H_{10}$ ) when reacted with oxygen ( $O_{2}$ )  is -2658 kJ/mol butane, and the chemical reaction is given by:

$C_{4} H_{10}$ + $\frac{13}{2} O_{2}$ ---> $4 CO_{2}$ + $5 H_{2}O$

The mass of butane required in the reaction is based on the heat produced by the reaction, which is given to be -1,500 kJ. The minus sign is added because the reaction releases heat (exothermic), which means that the products are in a "lower energy state" than the reactants.

Dividing this with the heat of reaction per mole of butane reacted would give the number of moles butane required. Then, multiplying the answer with the molar mass of butane which is 58 grams/mole, will give the mass of butane required.

Moles of butane = [(-1,500 kJ)/(-2658 kJ/mol butane)]
Moles of butane = 0.5643 moles butane

Mass of butane  = 0.5643 moles butane * 58 grams/mol butane
Mass of butane = 32.73 grams butane

The mass of carbon dioxide ( $CO_{2}$ ) can be determined by multiplying the moles of butane ( $C_{4} H_{10}$ ) with the mole ratio of ( $CO_{2}$ ) produced to the ( $C_{4} H_{10}$ ) reacted, and then with the molar mass of ( $CO_{2}$ ), which is 44 grams/mole.

Mass of carbon dioxide produced
= 0.5643 moles butane * [4 moles $CO_{2}$ / 1 mole $C_{4} H_{10}$ ] * 44 grams/mole $CO_{2}$

Mass of carbon dioxide produced
= 99.32 grams $CO_{2}$

Thus, the mass of butane required is 32.73 grams, and the mass of carbon dioxide produced from the reaction of this amount of butane is 99.32 grams.

4 0

3 years ago

Read 2 more answers

What is the percent yield if 3.47 g of copper is produced when 1.87 g of aluminium is reacted with excess of copper (II) sulfato

Serhud [2]

Answer:

copper is limiting

Explanation:

5 0

2 years ago