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Brut [27]
3 years ago
10

What is the difference between elements and compounds? *

Chemistry
1 answer:
scoundrel [369]3 years ago
8 0
Hi I need help pls pls pls pls pls pls pls so,s
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Calculate the grams of C6H6 needed to produce 25 g of CO2<br> Show your work
Anna007 [38]

Answer:

The given reaction is a combustion reaction of benzene,

C

6

H

6

. From its balanced chemical equation,

2

C

6

H

6

+

15

O

2

→

12

C

O

2

+

6

H

2

O

,

the mass of carbon dioxide

(

C

O

2

)

produced from 20 grams (g) of

C

6

H

6

is determined through the molar mass of the two compounds, given by,

M

M

C

O

2

=

44.01

g

/

m

o

l

M

M

C

6

H

6

=

78.11

g

/

m

o

l

and their mole ratio:

12

m

o

l

C

O

2

2

m

o

l

C

6

H

6

→

6

m

o

l

C

O

2

1

m

o

l

C

6

H

6

With this,

m

a

s

s

o

f

C

O

2

=

(

20

g

C

6

H

6

)

(

1

m

o

l

C

6

H

6

78.11

g

C

6

H

6

)

(

6

m

o

l

C

O

2

1

m

o

l

C

6

H

6

)

(

44.01

g

C

O

2

1

m

o

l

C

O

2

)

=

(

20

)

(

6

)

(

44.01

)

g

C

O

2

78.11

=

5281.2

g

C

O

2

78.11

m

a

s

s

o

f

C

O

2

=

67.6

g

C

O

2

Therefore, the mass in grams of

C

O

2

formed from 20 grams of

C

6

H

6

is

67.6

g

C

O

2

.

it is a problem of app

3 0
3 years ago
A soccer ball is moving to the right. how can you increase the velocity of the ball?
Fantom [35]

Answer: The air will move more quickly around one side, making less pressure on that side of the ball

Explanation:

3 0
2 years ago
Determine the reducing agent in the following reaction. Explain your answer. 2 Li(s) + Fe(C2H3O2)2(aq) → 2 LiC2H3O2(aq) + Fe(s)
Brilliant_brown [7]

The reducing agent in the reaction 2Li(s) + Fe(CH₃COO)₂(aq) → 2LiCH₃COO(aq) + Fe(s) is lithium (Li).

The general reaction is:

2Li(s) + Fe(CH₃COO)₂(aq) → 2LiCH₃COO(aq) + Fe(s)   (1)

We can write the above reaction in <u>two reactions</u>, one for oxidation and the other for reduction:

  • Oxidation reaction

Li⁰(s) → Li⁺(aq) + e⁻   (2)

  • Reduction reaction

Fe²⁺(aq) + 2e⁻ → Fe⁰(s)    (3)

We can see that Li⁰ is oxidizing to Li⁺ (by <u>losing</u> one electron) in the lithium acetate (<em>reaction 2</em>) and that Fe²⁺ in iron(II) acetate is reducing to Fe⁰ (by <u>gaining</u> two <em>electrons</em>) (<em>reaction 3</em>).  

We must remember that the reducing agent is the one that will be oxidized by <u>reducing another element</u> and that the oxidizing agent is the one that will be reduced by <u>oxidizing another species</u>.

In reaction (1), the<em> reducing agent</em> is <em>Li</em> (it is oxidizing to Li⁺), and the <em>oxidizing agent </em>is<em> Fe(CH₃COO)₂</em> (it is reducing to Fe⁰).  

Therefore, the reducing agent in reaction (1) is lithium (Li).  

 

Learn more here:

  • brainly.com/question/10547418?referrer=searchResults
  • brainly.com/question/14096111?referrer=searchResults

I hope it helps you!

3 0
3 years ago
If the solubility of a gas is 10.5 g/L at 525 kPa pressure, what is the solubility of the gas when the pressure is 225 kPa? Show
Talja [164]

Answer:

4.5 g/L.

Explanation:

  • To solve this problem, we must mention Henry's law.
  • Henry's law states that at a constant temperature, the amount of a given gas dissolved in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid.
  • It can be expressed as: P = KS,

P is the partial pressure of the gas above the solution.

K is the Henry's law constant,

S is the solubility of the gas.

  • At two different pressures, we have two different solubilities of the gas.

<em>∴ P₁S₂ = P₂S₁.</em>

P₁ = 525.0 kPa & S₁ = 10.5 g/L.

P₂ = 225.0 kPa & S₂ = ??? g/L.

∴ S₂ = P₂S₁/P₁ = (225.0 kPa)(10.5 g/L) / (525.0 kPa) = 4.5 g/L.

8 0
3 years ago
When heat is applied to 80 grams of CaCO3, it yields 39 grams of B. Determine the percentage of the yield.
EleoNora [17]

The question is incomplete, the complete question is:

When heat is applied to 80 grams of CaCO3, it yields 39 grams of CaO Determine the percentage of the yield.

CaCO3→CaO + CO2

<u>Answer:</u> The % yield of the product is 87.05 %

<u>Explanation:</u>

The number of moles is defined as the ratio of the mass of a substance to its molar mass.

The equation used is:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}} ......(1)

We are given:

Given mass of CaCO_3 = 80 g

Molar mass of CaCO_3 = 100 g/mol

Putting values in equation 1, we get:

\text{Moles of }CaCO_3=\frac{80g}{100g/mol}=0.8mol

For the given chemical reaction:

CaCO_3\rightarrow CaO+CO_2

By stoichiometry of the reaction:

If 1 mole of CaCO_3 produces 1 mole of CaO

So, 0.8 moles of CaCO_3 will produce = \frac{1}{1}\times 0.8=0.8mol of CaO

We know, molar mass of CaO = 56 g/mol

Putting values in above equation, we get:

\text{Mass of CaO}=(0.8mol\times 56g/mol)=44.8g

The percent yield of a reaction is calculated by using an equation:

\% \text{yield}=\frac{\text{Actual value}}{\text{Theoretical value}}\times 100 ......(2)

Given values:

Actual value of the product = 39 g

Theoretical value of the product = 44.8 g

Plugging values in equation 2:

\% \text{yield}=\frac{39 g}{44.8g}\times 100\\\\\% \text{yield}=87.05\%

Hence, the % yield of the product is 87.05 %

6 0
3 years ago
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