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aleksandrvk [35]
3 years ago
15

If 45.6 g of Fe2O3 reacts with excess water, how much heat is required?

Chemistry
2 answers:
tangare [24]3 years ago
8 0

Explanation:

I

have not yet learnt chemistry so sorry

Elena-2011 [213]3 years ago
8 0

Answer:

350

Explanation:

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please help!!!!!! What is the mass of 1 mole of magnesium (Mg)? Express your answer to four significant figures. The mass of 1 m
Semmy [17]
I believe you just look at your periodic table for this value. I don't think there is any math involved.
Therefore one mole of Mg = 24.305g.
5 0
3 years ago
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If a class has 120 students and the professor wants to purchase two cookies per student (or 2 coo/stu) he or she will need to bu
klio [65]
60 just divide 120 by 2
7 0
3 years ago
In the reaction C7H16+_______O2→ 8H2O + 7CO2, what coefficient should be placed in front og O2 to balence the reaction
Ira Lisetskai [31]
The answer is C because there are 22 oxygen atoms on the product side so to balance the equation the coefficient needed is 11
4 0
3 years ago
Write the balanced chemical equation between H2SO4 and KOH in aqueous solution. This is called a neutralization reaction and wil
emmainna [20.7K]

Answer:

0.166M

Explanation:

In a neutralization, the acid, H₂SO₄, reacts with a base, KOH, to produce a salt, K₂SO₄ and water. The reaction is:

H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O

To solve this problem, we need to determine moles of H2SO4 and moles of KOH that reacts to find the moles of sulfuric acid that remains after the reaction:

<em>Moles H2SO4:</em>

0.650L * (0.430mol /L) = 0.2795moles H2SO4

<em>Moles KOH:</em>

0.600L * (0.240mol / L) = 0.144 moles KOH

Moles of sulfuric acid that reacts with 0.144 moles of KOH are:

0.144 moles KOH * (1mol H2SO4 / 2 mol KOH) = 0.072 moles of H2SO4 react.

And remain:

0.2795moles H2SO4 - 0.072moles H2SO4 = 0.2075 moles of H2SO4 reamains.

In 0.650L + 0.600L = 1.25L:

Molar concentration of sulfuric acid:

0.2075 moles of H2SO4 / 1.25L =

<h3>0.166M</h3>
7 0
3 years ago
How many grams of zinc would be required to produce 9.65g of zinc hydroxide
Effectus [21]

The question is incomplete, here is the complete question:

How many grams of zinc would be required to produce 9.65g of zinc hydroxide

Zn+2MnO₂+H₂O→Zn(OH)₂+Mn₂O₃

<u>Answer:</u> The mass of zinc required is 6.35 grams

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

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

Given mass of zinc hydroxide = 9.65 g

Molar mass of zinc hydroxide = 99.4 g/mol

Putting values in equation 1, we get:

\text{Moles of zinc hydroxide}=\frac{9.65g}{99.4g/mol}=0.0971mol

The given chemical equation follows:

Zn+2MnO_2+H_2O\rightarrow Zn(OH)_2+Mn_2O_3

By Stoichiometry of the reaction:

1 mole of zinc hydroxide is produced from 1 mole of zinc

So, 0.0971 moles of zinc hydroxide will be produced from = \frac{1}{1}\times 0.0971=0.0971mol of zinc

Now, calculating the mass of zinc from equation 1, we get:

Molar mass of zinc = 65.4 g/mol

Moles of zinc = 0.0971 moles

Putting values in equation 1, we get:

0.0971mol=\frac{\text{Mass of zinc}}{65.4g/mol}\\\\\text{Mass of zinc}=(0.0971mol\times 65.4g/mol)=6.35g

Hence, the mass of zinc required is 6.35 grams

6 0
3 years ago
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