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Mars2501 [29]
3 years ago
14

Physical Science pre-Test 2020-2021

Chemistry
1 answer:
poizon [28]3 years ago
4 0

Answer:

exothermic I think

Explanation:

IM TWKIGN THE TEST

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When carbon is burned in air, it reacts with oxygen to form carbon dioxide. When 14.4 g of carbon were burned in the presence of
Natasha2012 [34]

When carbon reacts with oxygen it forms CO2. This can depicted by the below equation.

C + O2→ CO2

It has been mentioned that when 14.4 g of C reacts with 53.9 g of O2, then 15.5 g of O2 remains unreacted. <u>This indicates that Carbon is the limiting reagent and hence the amount of CO2 produced is based on the amount of Carbon burnt.</u>

C + O2→ CO2

In the above equation , 1 mole of carbon reacts with 1 mole of O2 to produce 1 mole of CO2.

In this case 14.4 g of Carbon reacts with 53.9 of O2 to produce "x"g of CO2.

<u>No of moles = mass of the substance÷molar mass of the substance</u>

No of moles of carbon = 14.4 /12= 1.2 moles

No of moles of O2 = Mass of reacted O2/Molar mass of O2.

No of moles of O2 = (Total mass of O2 burned - Mass of unreacted O2)/32

No of moles of O2 = (53.9-15.5) ÷ 32 = 1.2 moles.

Hence as already discussed 1 mole of Carbon reacts with 1 mole of O2 to produce 1 mole of CO2. In this case 1.2 moles of carbon reacts with 1.2 moles of O2 to produce 1.2 moles of CO2.

Moles of carbon dioxide = Mass of CO2 produced /Molar mass of CO2

Mass of CO2 produced(x) = Moles of CO2 ×Molar mass of CO2

Mass of CO2 produced(x) = 1.2 x 44 = 52.8 g

<u>Thus 52.8 g of CO2 is produced.</u>

5 0
3 years ago
What do you call a large body of air?<br> wind<br> thunderstorm<br> cold front<br> air mass
Keith_Richards [23]

Answer:

thunderstrom

Explanation:

it will be answer

4 0
2 years ago
Read 2 more answers
How many moles are in 4.85 x 10^25 atoms of carbon?
Serjik [45]

Answer:

4.85 x 10^25

Explanation:

thats what i was told by a calculator

6 0
3 years ago
How does a meteorologist use an understanding of what causes weather to predict a storm?
Svetllana [295]
Meteorologists also use satellites to observe cloud patterns around the world, and radar is used to measure precipitation. All of this data is then plugged into super computers, which use numerical forecast equations to create forecast models of the atmosphere.
8 0
3 years ago
2) 2KClO3 --&gt; 2KCl + 3O2
aleksandrvk [35]

2 \text{ KClO}_3 \to 2 \text{ KCl}+3\text{ O}_2

a)

2 \text{ mols of KClO}_3 \equiv 3  \text{ mols of O}_2

19 \text{ mols of KClO}_3 \equiv 3\cdot 9,5  \text{ mols of O}_2

\boxed{19 \text{ mols of KClO}_3 \equiv 28,5  \text{ mols of O}_2}

b)

2 \text{ mols of KClO}_3 \equiv 2  \text{ mols of KCl}

62 \text{ mol of KClO}_3 \equiv 62  \text{ mol of KCl}

Using the atomic mass given in the periodic table:

62\cdot(39+35,5+16\cdot3) \text{ g of KClO}_3 \equiv 62  \text{ mol of KCl}

62\cdot122,5 \text{ g of KClO}_3 \equiv 62  \text{ mol of KCl}

7595 \text{ g of KClO}_3 \equiv 62  \text{ mol of KCl}

\boxed{7,595 \text{ kg of KClO}_3 \equiv 62  \text{ mol of KCl}}

c)

2 \text{ KCl}+3\text{ O}_2\to 2 \text{ KClO}_3

3 \text{ mols of O}_2 \equiv 2 \text{ mols of KCl}

Using the atomic mass given in the periodic table:

3\cdot(2\cdot 16) \text{ g of O}_2 \equiv 2\cdot(39+35,5)  \text{ g of KCl}

96\text{ g of O}_2 \equiv 149\text{ g of KCl}

\dfrac{39}{149}\cdot 96\text{ g of O}_2 \equiv \dfrac{39}{149}\cdot 149\text{ g of KCl}

\boxed{25,13\text{ g of O}_2 \equiv 39\text{ g of KCl}}

This result is an aproximation.

8 0
3 years ago
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