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Olin [163]
3 years ago
7

For acidic solutions, which element is added to balance half-reactions

Chemistry
2 answers:
mina [271]3 years ago
6 0
For balancing acidic solutions, we would need to add H+ ions to the correct side of the equation to balance the total number of atoms and the overall charge.
My name is Ann [436]3 years ago
5 0
The answer is Hydrogen
PLATOFAM#
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Q#5 Balance and list the coefficients from reactants to products.
skad [1K]

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

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8 0
2 years ago
Read 2 more answers
How does the type of chemical bonds present in a substance affect the substances properties?
SCORPION-xisa [38]
Once a chemical bond is formed, the atoms are rearranged to form a stronger bond, affecting the hardness, malleability, etc. The stronger the bond, the easier a substance will break, or, if its a liquid, it will resist seperation.
8 0
2 years ago
The specific heat of aluminum is 0.897j/(g•°C). If a 22.6 g sample of aluminum is heated from 25°C to 250°C, then how much heat
Oxana [17]

Q=mcat

=(22.6)(.897)(250-25)

4,561.245 or 4.5 * 10^-3

3 0
3 years ago
How many electrons can occupy the 4p sublevel of a neutral atom of bromine
VMariaS [17]

Answer:

6

Explanation

every p sublevel holds up to 6 electrons

so the 4p sublevel  must hold up to 6 electonsr

4 0
2 years ago
How many grams are in 3.14 x 1015 molecules of CO?
Vesnalui [34]

Answer:

Explanation:

Not Many

1 mol of CO has a mass of

C = 12

O = 16

1 mol = 28 grams.

1 mol of molecules = 6.02 * 10^23

x mol of molecules = 3.14 * 10^15        Cross multiply

6.02*10^23 x = 1 * 3.14 * 10^15             Divide by 6.02*10^23

x = 3.14*10^15 / 6.02*10^23

x = 0.000000005 mols

x = 5*10^-9

1 mol of CO has a mass of 28

5*10^-9 mol of CO has a mass of x                        Cross Multiply

x = 5 * 10^-9 * 28

x = 1.46 * 10^-7 grams

Answer: there are 1.46 * 10-7 grams of CO if only 3.14 * 10^15 molecules are in the sample

5 0
2 years ago
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