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NikAS [45]
2 years ago
15

Help me pls!!!!!!!!!!!

Chemistry
1 answer:
Ludmilka [50]2 years ago
7 0

Answer:

i disagree with seths friend

Explanation:

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Over the years, the thermite reaction has been used for years to welding railroad nails, in cendiary bombs, and to ignite solid
koban [17]

Answer:

A. 109.61 g of Fe₂O₃

B. 36.99 g of Al

C. Maximum mass of Al₂O₃ produced is 69.90 g

Explanation:

The balanced equation for the reaction is given below:

Fe₂O₃ (s) + 2Al (s) → 2Fe (l) + Al₂O₃ (s)

Next, we shall determine the masses of Fe₂O₃ and Al that reacted and the masses of Fe and Al₂O₃ produced from the balanced equation. This can be obtained as follow:

Molar mass of Fe₂O₃ = (2×56) + (3×16)

= 112 + 48 = 160 g/mol

Mass of Fe₂O₃ from the balanced equation = 1 × 160 = 160 g

Molar mass of Al = 27 g/mol

Mass of Al from the balanced equation = 2 × 27 = 54 g

Molar mass of Fe = 56 g/mol

Mass of Fe from the balanced equation = 2 × 56 = 112 g

Molar mass of Al₂O₃ = (2×27) + (3×16)

= 54 + 48 = 102 g/mol

Mass of Al₂O₃ from the balanced equation = 1 × 102 = 102 g

SUMMARY:

From the balanced equation above,

160 g of Fe₂O₃ reacted with 54 g of Al to produce 112 g of Fe and 102 g of Al₂O₃

A. Determination of the mass of iron(III) oxide, Fe₂O₃, needed to produce 76.73 g of iron, Fe.

From the balanced equation above,

160 g of Fe₂O₃ reacted to produce 112 g of Fe.

Therefore, Xg of Fe₂O₃ will react to produce 76.73 g of Fe i.e

Xg of Fe₂O₃ = (160 × 76.73)/112

Xg of Fe₂O₃ = 109.61 g

Thus, 109.61 g of Fe₂O₃ is needed to produce 76.73 g of Fe.

B. Determination of the mass of aluminum, Al, to produce 76.72 g of iron, Fe.

From the balanced equation above,

54 g of Al reacted to produce 112 g of Fe.

Therefore, Xg of Al will react to produce 76.72 g of Fe i.e

Xg of Al = (54 × 76.72)/112

Xg of Al = 36.99 g

Thus, 36.99 g of Al is needed to produce 76.72 g of Fe.

C. Determination of the maximum mass of aluminum oxide, Al₂O₃, produced along with 76.75 g Fe.

We'll begin by calculating the mass of Fe₂O₃ needed to produce 76.75 g of Fe. This is illustrated below:

From the balanced equation above,

160 g of Fe₂O₃ reacted to produce 112 g of Fe.

Therefore, Xg of Fe₂O₃ will react to produce 76.75 g of Fe i.e

Xg of Fe₂O₃ = (160 × 76.75)/112

Xg of Fe₂O₃ = 109.64 g

Thus, 109.64 g of Fe₂O₃ is needed to produce 76.75 g of Fe.

Finally, we shall determine the maximum mass of Al₂O₃ produced along with 76.75 g Fe. this can be obtained as follow:

From the balanced equation above,

160 g of Fe₂O₃ reacted 102 g of Al₂O₃.

Therefore, 109.64 g of Fe₂O₃ will react to produce = (109.64 × 102)/160 = 69.90 g of Al₂O₃.

Thus, the maximum mass of Al₂O₃ produced is 69.90 g

6 0
2 years ago
Help please!
umka2103 [35]

Answer: In factories they burn coal under a giant container(a really big gumbo pot pretty much). when the fire is hot enough it starts to boil. When it boils it creates steam and the steam blows on a fan that makes electricity.

4 0
2 years ago
How much mass do 2.0 moles of uranium contain
jasenka [17]
The answer is going to be 476.06.
8 0
3 years ago
If 45.0 mL of ethanol (density = 0.789 g/mL) initially at 8.0 ∘C is mixed with 45.0 mL of water (density = 1.0 g/mL) initially a
strojnjashka [21]

Answer : The final temperature of the mixture is, 22.14^oC

Explanation :

First we have to calculate the mass of ethanol and water.

\text{Mass of ethanol}=\text{Density of ethanol}\times \text{Volume of ethanol}=0.789g/mL\times 45.0mL=35.5g

and,

\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1.0g/mL\times 45.0mL=45.0g

Now we have to calculate the final temperature of the mixture.

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

c_1 = specific heat of ethanol = 2.42J/g^oC

c_2 = specific heat of water = 4.18J/g^oC

m_1 = mass of ethanol = 35.5 g

m_2 = mass of water = 45.0 g

T_f = final temperature of mixture = ?

T_1 = initial temperature of ethanol = 8.0^oC

T_2 = initial temperature of water = 28.6^oC

Now put all the given values in the above formula, we get:

35.5g\times 2.42J/g^oC\times (T_f-8.0)^oC=-45.0g\times 4.18J/g^oC\times (T_f-28.6)^oC

T_f=22.14^oC

Therefore, the final temperature of the mixture is, 22.14^oC

3 0
3 years ago
Hydroxide is an example of _______ and it's formula is OH-.
Ksju [112]

Answer:

polyatomic ion

Explanation:

It is polyatomic ion have a great day marry christmass

6 0
2 years ago
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