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4vir4ik [10]
3 years ago
8

29 Points!!

Chemistry
2 answers:
bija089 [108]3 years ago
7 0
A single nucleus gives rise to two identical daughter nuclei. In mitosis<span>, identical copies of the chromosomes are produced and placed in each of the daughter cells.</span>
atroni [7]3 years ago
3 0
The last one would be false
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A molecule of glucose is comprised of 6 atoms of carbon, 12 atoms of hydrogen, and 6 atoms of oxygen.
Svetradugi [14.3K]

Answer:

C6H12O6 the third choice

3 0
2 years ago
Compared with halogens, the alkali metals in the same period has
Rufina [12.5K]
Alkali metals have the lowest electronegativities, while halogens have the highest.
3 0
3 years ago
Sulfur dioxide and oxygen react to form sulfur trioxide during one of the key steps in sulfuric acid synthesis. An industrial ch
guajiro [1.7K]

Answer:

Kp = 0.049

Explanation:

The equilibrium in question is;

2 SO₂ (g)  +  O₂ (g)   ⇄ 2 SO₃ (g)  

Kp = p SO₃² / ( p SO₂² x p O₂ )

The initial pressures are given, so lets set up the ICE table for the equilibrium:

atm        SO₂         O₂          SO₃

I              3.3        0.79           0

C              -2x           -x          2x

E             3.3 - 2x    0.79 - x    2x

We are told 2x = partial pressure of SO₃ is 0.47 atm at equilibrium, so we can determine the partial pressures of  SO₂ and O₂ as follows:

p SO₂  = 3.3 -0.47 atm = 2.83 atm

p O₂ = 0.79 - (0.47/2) atm = .56 atm

Now we can calculate Kp:

Kp = 0.47² /[ ( 2.83 )² x 0.56 ] = 0.049 ( rounded to 2 significant figures )

Note that we have extra data in this problem we did not need since once we setup the ICE table for the equilibrium we realize we have all the information needed to solve the question.

7 0
3 years ago
If 2.68 g of benzaldehyde are involved with the mixed aldol condensation reaction, how many moles of benzaldehyde are present? R
Zielflug [23.3K]

Answer:

The number of moles of benzaldehyde = 0.0253 moles

Explanation:

The molecular formula of benzaldehyde is C₇H₆O

Its molecular mass is calculated from the atomic masses of the constituent atoms.

C = 12.0 g: H = 1.0 g; O = 16.0 g

Molecular mass = ( 12 * 7) + (1 * 6) + (16 * 1) = 106.0 g/mol

Number of moles of  substance = mass of substance/ molar mass of the substance

mass of benzaldehyde = 2.68; molar mass = 106.0 g/mol

Number of moles of benzaldehyde = 2.68 g/ 106 g/mol = 0.0253 moles

Therefore, the number of moles of benzaldehyde = 0.0253 moles

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
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