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Umnica [9.8K]
3 years ago
9

8. Which statement correctly describes the location and charge of the electrons in an atom?

Chemistry
2 answers:
olchik [2.2K]3 years ago
5 0

Answer:

The protons are located in the nucleus and have a positive charge

Explanation:

   

Anna71 [15]3 years ago
4 0
D the electrons revolve around the nucleus and they contain negative charge
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Which of the following is an
eimsori [14]

Answer:

2 and 4

Explanation:

The rest of the changes are chemical. 1 has a chemical reaction happen which makes light sticks glow. 3 is because browning the meat actually causes some new compounds to form and cause caramelization.

For 2, the oxygen is simply heating up and expanding which pops the balloon. 4 is just a phase change of water vapor to liquid water.

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
Identifying polar regions within molecules is a very useful skill for chemists that helps us to think about the strength of inte
Ilia_Sergeevich [38]
 bond in the acetic acid molecule is the most polar <span>c. c=o bond
</span>which bond in the methyl amine molecule is the most polar <span>c. n–h bond</span>
4 0
4 years ago
As you move down through the family of alkali metals, the first ionization energy ________ because
kompoz [17]
I’m pretty sure the answer is B
5 0
3 years ago
Read 2 more answers
One way of obtaining pure sodium carbonate is through the decomposition of the mineral trona, Na3(CO3)(HCO3)·2H2O. 2Na3(CO3)(HCO
zhenek [66]
Percentage yield = (actual yield / theoretical yield) x 100%

The balanced equation for the decomposition is,
 2Na₃(CO₃)(HCO₃)·2H₂O(s) → 3Na₂CO₃(s) + CO₂(g) + 5H₂<span>O(g)

The stoichiometric ratio between </span>Na₃(CO₃)(HCO₃)·2H₂O(s)  and Na₂CO₃(s) is 2 : 3

The decomposed mass of Na₃(CO₃)(HCO₃)·2H₂O(s) = 1000 kg
                                                                                     = 1000 x 10³ g

Molar mass of Na₃(CO₃)(HCO₃)·2H₂O(s) = 226 g mol⁻¹
moles of Na₃(CO₃)(HCO₃)·2H₂O(s) = mass / molar mass
                                                         = 1000 x 10³ g / 226 g mol⁻¹
                                                         = 4424.78 mol

Hence, moles of Na₂CO₃ formed = 4424.78 mol x \frac{3}{2}
                                                     = 6637.17 mol

Molar mass of Na₂CO₃ = 106 g mol⁻¹

Hence, mass of Na₂CO₃ = 6637.17 mol x 106 g mol⁻¹
                                        = 703540.02 g
                                        = 703.540 kg

Hence, the theoretical yield of Na₂CO₃ =  703.540 kg
Actual yield of Na₂CO₃ = 650 kg

Percentage yield = (650 kg / 703.540 kg) x 100%
                            = 92.34%
7 0
3 years ago
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