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Irina-Kira [14]
3 years ago
7

Which of these elements has two valence electrons? A. hydrogen (H) B. barium (Ba) C. nitrogen (N) D. krypton (Kr) E. bromine (Br

)

Chemistry
2 answers:
oksian1 [2.3K]3 years ago
8 0

Barium has two valence electrons

9966 [12]3 years ago
3 0
So whenever you need to find valence electron look at Group Number and you can find it, Example barium it’s on group 2 it’s means it have 2 valence electrons.
let’s look for nitrogen group 5A that means it have 5 valence eletrons.
hope it’s help.
Let me know if you have any chemistry related questions

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The majority of energy used in industrialized countries is produced using renewable resources.
vaieri [72.5K]

Answer:

The answer is FALSE

Explanation:

I took the test and it was false,

also if it was true then in industrialized countries would be more eco-friendly and there wouldn't be a huge hole in our ozone layer. and no more wars over oil.

7 0
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Porfavor alguien que me explique los Cambios bioquimicos en los musculos ,sangre e higado durante los ejercicios fisicos y como
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8 0
3 years ago
How many hydrogen atoms are in 89.5 g of<br> C6H6 ?<br> Answer in units of atoms.
Elodia [21]

Solution :

Molar mass of C_6H_6 is :

M = 6×12 + 6×1 g

M = 78 g

78 gram of C_6H_6 contains 6.022 \times 10^{23} molecules.

So, 89.5 gram of C_6H_6 contains :

n = 6.022 \times 10^{23} \times \dfrac{89.5}{78}\\\\n = 6.91 \times 10^{23}

Now, from the formula we can see that one molecule of C_6H_6 contains 2 hydrogen atom . So, number of hydrogen atom are :

h = 2\times 6.91 \times 10^{23}\\\\h = 1.38 \times 10^{22}\ atoms

Hence, this is the required solution.

8 0
2 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
What would be the products for the reaction Br2 + KI →? (Just identify the correct products for the reaction. You do not need to
Savatey [412]

<u>Answer:</u> The products of the reaction will be I_2\text{ and }KBr

<u>Explanation:</u>

Single displacement reaction is defined as the reaction in which more reactive element displaces a less reactive element from its chemical reaction.

The general chemical equation for the single displacement reaction follows:

A+BC\rightarrow AC+B

The given chemical equation follows:

Br_2+2KI\rightarrow I_2+2KBr

Bromine element is more reactive than iodine element. Thus, can easily replace iodine from its chemical reaction.

Hence, the products of the reaction will be I_2\text{ and }KBr

5 0
3 years ago
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