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pochemuha
2 years ago
9

Q#5 Balance and list the coefficients from reactants to products.

Chemistry
2 answers:
skad [1K]2 years ago
8 0

A. 2Fe_2O_3(s) +3C(s) → 4Fe(s) + 3CO_2(g)

B. 2Al(s) + 3FeO(s) → Fe(s) + 3Al_2O_3(s)

C. 2Al(s) +3H_2SO_4(aq) → Al_2(SO_4)_3(aq) + 3H_2(g)

What is a balanced chemical equation?

An equation that has an equal number of atoms of each element on both sides of the equation is called a balanced chemical equation.

A. 2Fe_2O_3(s) +3C(s) → 4Fe(s) + 3CO_2(g)

B. 2Al(s) + 3FeO(s) → Fe(s) + 3Al_2O_3(s)

C. 2Al(s) +3H_2SO_4(aq) → Al_2(SO_4)_3(aq) + 3H_2(g)

Learn more about the balanced chemical equation here:

brainly.com/question/15052184

#SPJ1

nika2105 [10]2 years ago
4 0

A. 2\text{Fe}_{2}\text{O}_{3}+3\text{C} \longrightarrow 4\text{Fe} +3\text{CO}_{2}

B. 2\text{Al}+3\text{FeO} \longrightarrow \text{Al}_{2}\text{O}_{3}+3\text{Fe}

C. 2\text{Al}+3\text{H}_{2}\text{SO}_{4} \longrightarrow \text{Al}_{2}(\text{SO}_{4})_{3}+3\text{H}_{2}

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rite stepwise equations for protonation or deprotonation of this polyprotic species in water. What are the products of the first
adoni [48]

<u>Answer:</u> The products of the given chemical equation are HCO_3^-\text{ and }OH^-

<u>Explanation:</u>

Protonation equation is defined as the equation in which protons get added in the substance.

The chemical equation for the protonation of carbonate ion in the presence of water follows:

CO_3^{2-}+H_2O\rightarrow HCO_3^-+OH^-

By Stoichiometry of the reaction:

1 mole of carbonate ion reacts with 1 mole of water to produce 1 mole of hydrogen carbonate ion and 1 mole of hydroxide ion

Hence, the products of the given chemical equation are HCO_3^-\text{ and }OH^-

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What is the electron structure of sodium
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nydimaria [60]

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B

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3 years ago
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What mass of Sodium Chloride is required to make 100.0 mL of 3.0 M solution?
Julli [10]

Answer:

17.55 g of NaCl

Explanation:

The following data were obtained from the question:

Molarity = 3 M

Volume = 100.0 mL

Mass of NaCl =..?

Next, we shall convert 100.0 mL to L. This can be obtained as follow:

1000 mL = 1 L

Therefore,

100 mL = 100/1000

100 mL = 0.1 L

Therefore, 100 mL is equivalent to 0.1 L.

Next, we shall determine the number of mole NaCl in the solution. This can be obtained as follow:

Molarity = 3 M

Volume = 0.1 L

Mole of NaCl =?

Molarity = mole /Volume

3 = mole of NaCl /0.1

Cross multiply

Mole of NaCl = 3 × 0.1

Mole of NaCl = 0.3 mole

Finally, we determine the mass of NaCl required to prepare the solution as follow:

Mole of NaCl = 0.3 mole

Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol

Mass of NaCl =?

Mole = mass /Molar mass

0.3 = mass of NaCl /58.5

Cross multiply

Mass of NaCl = 0.3 × 58.5

Mass of NaCl = 17.55 g

Therefore, 17.55 g of NaCl is needed to prepare the solution.

5 0
3 years ago
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marta [7]

Answer:

okay lol, ima use this for points if okay

Explanation:

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