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Soloha48 [4]
3 years ago
11

What does it mean for an acid and base to neutralize each other?

Chemistry
1 answer:
ruslelena [56]3 years ago
6 0
This means that they mixed .
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We want to find how much charge is on the electrons in a nickel coin. follow this method. a nickel coin has a mass of about 5 g.
Lady bird [3.3K]
The mass of a nickel coin is 5 g.
1 mol of Ni weighs 58 g. 1 mol contains 6.022 x 10²³ atoms of Ni.
therefore in 58 g  there are 6.022 x 10²³ atoms of Ni
then in 5 g the number of Ni atoms are - 6.022 x 10²³ /58 x 5 = 5.2 x 10²² Ni atoms
Therefore number of Ni atoms are 5.2 x 10²² atoms in a nickel coin

3 0
3 years ago
GIVING BRAINLIEST FOR RIGHT ANSWERS PLS HELP
Anuta_ua [19.1K]

Answer:

Answer: Chemical change; Chemical property of heat of combustion

Explanation:

A physical change is defined as a change in which there is alteration in shape, size etc. No new substance gets formed in these reactions.

Explanation:

4 0
2 years ago
Read 2 more answers
Use the equation n2 + 3h2=2nh3 if 8.0g N2 react, how many grams of NH3 will be produced
Eduardwww [97]

the balanced equation for the formation of ammonia is

N₂ + 3H₂ ---> 2NH₃

molar ratio of N₂ to NH₃ is 1:2

mass of N₂ reacted is 8.0 g

therefore number of N₂ moles reacted is - 8.0 g / 28 g/mol = 0.286 mol

according to the molar ratio,

1 mol of N₂ will react to give 2 mol of NH₃, assuming nitrogen is the limiting reactant

therefore 0.286 mol of N₂ should give - 2 x 0.286 mol = 0.572 mol of NH₃

therefore mass of NH₃ formed is - 0.572 mol x 17 g/mol = 9.72 g

a mass of 9.72 mol of NH₃ is formed

6 0
3 years ago
there about 5000 elements, which combine to form the vast number of diffrent substances in the world around us
marissa [1.9K]

Answer: false 100

Explanation:

if this is the true or false test here

5 0
3 years ago
A 0.1510 gram sample of a hydrocarbon produces 0.5008 gram CO2 and 0.1282 gram H2O in combustion analysis. Its
Over [174]
In a combustion of a hydrocarbon compound, 2 reactions are happening per element:

C + O₂ → CO₂
2 H + 1/2 O₂ → H₂O

Thus, we can determine the amount of C and H from the masses of CO₂ and H₂O produced, respectively.

1.) Compute for the amount of C in the compound. The data you need to know are the following:
Molar mass of C = 12 g/mol
Molar mass of CO₂ = 44 g/mol
Solution:
0.5008 g CO₂*(1 mol CO₂/ 44 g)*(1 mol C/1 mol CO₂) = 0.01138 mol C
0.01138 mol C*(12 g/mol) = 0.13658 g C

Compute for the amount of H in the compound. The data you need to know are the following:
Molar mass of H = 1 g/mol
Molar mass of H₂O = 18 g/mol
Solution:
0.1282 g H₂O*(1 mol H₂O/ 18 g)*(2 mol H/1 mol H₂O) = 0.014244 mol H
0.014244 mol H*(1 g/mol) = 0.014244 g H

The percent composition of pure hydrocarbon would be:
Percent composition = (Mass of C + Mass of H)/(Mass of sample) * 100
Percent composition = (0.13658 g + 0.014244 g)/(<span>0.1510 g) * 100
</span>Percent composition = 99.88%

2. The empirical formula is determined by finding the ratio of the elements. From #1, the amounts of moles is:

Amount of C = 0.01138 mol
Amount of H = 0.014244 mol

Divide the least number between the two to each of their individual amounts:
C = 0.01138/0.01138 = 1
H = 0.014244/0.01138 = 1.25

The ratio should be a whole number. So, you multiple 4 to each of the ratios:
C = 1*4 = 4
H = 1.25*4 = 5

Thus, the empirical formula of the hydrocarbon is C₄H₅.

3. The molar mass of the empirical formula is

Molar mass = 4(12 g/mol) + 5(1 g/mol) = 53 g/mol
Divide this from the given molecular weight of 106 g/mol
106 g/mol / 53 g/mol = 2
Thus, you need to multiply 2 to the subscripts of the empirical formula.

Molecular Formula = C₈H₁₀

4 0
3 years ago
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