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Ne4ueva [31]
3 years ago
9

According to this climate graph, which month is the average temperature the lowest in Santa Cruz?

Chemistry
1 answer:
xxMikexx [17]3 years ago
7 0

Answer:

january

Explanation:

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During beta-oxidation of fatty acids, ___________ is produced in peroxisomes but not in mitochondria. A) acetyl-CoA B) FADH 2 C)
stepan [7]

Answer:

The correct option is D

Explanation:

Normally, beta-oxidation of fatty acid occurs in the mitchondrial matrix, however, when the fatty acid chains are too long, the beta-oxidation occurs in the peroxisomes <u>where the oxidation is not attached to ATP synthesis but rather transferred (i.e high energy electrons are transferred) to O₂ to form hydrogen peroxide</u> (H₂O₂). This is the major difference between the beta-oxidation that occurs in the peroxisomes to that which occurs in the mitochondria.

4 0
3 years ago
Predict, using Boyle’s Law, what will happen to a balloon that an ocean diver takes to a pressure of 202 kPa (normal atmospheric
kaheart [24]

Explanation:

Balloon that an ocean diver takes to a pressure of 202 k Pa will get reduced in size that is the volume of the balloon will get reduced. This is because pressure and volume of the gas are inversely related to each other.

According to Boyle's law: The pressure of the gas  is inversely proportional to the volume occupied by the gas at constant temperature(in Kelvins).

pressure\propto \frac{1}{volume} (At constant temperature)

The pressure beneath the sea is 202 kPa and the atmospheric pressure  is 101.3 kPa . This increase in pressure will result in decrease in volume occupied by the gas inside the balloon with decrease in size of a balloon. Hence, the size of the balloon will get reduced at 202 kPa (under sea).

7 0
3 years ago
A compound is 2.00% H by mass, 32.7% S by mass, and 65.3% O by mass. What is its empirical
Katarina [22]

Answer:

empirical formula: H_2SO_4

2 g H

32.7 g S

65.3 g O

Explanation:

Like the problem said, the first thing we can do is calculate the mass of each of the 3 elements in a 100-gram sample:

- 2.00% * 100g = 2 g H

- 32.7% * 100g = 32.7 g S

- 65.3% * 100g = 65.3 g O

Now we need to find the empirical formula from these. To do so, convert all of those masses into moles by using the molar mass for each element:

- the molar mass of H is 1.01 g/mol

- the molar mass of S is 32.06 g/mol

- the molar mass of O is 16 g/mol

2 g H ÷ 1.01 g/mol = 1.98 mol H

32.7 g S ÷ 32.06 g/mol = 1.02 mol S

65.3 g O ÷ 16 g/mol = 4.08 mol O

Our ratio of H : S : O is now:

1.98 mol : 1.02 mol : 4.08 mol

Divide them all by the smallest number, which is 1.02:

1.98/1.02  :  1.02/1.02  :  4.08/1.02

1.94 : 1 : 4

1.94 ≈ 2

So:

2 : 1 : 4

Thus, the empirical formula is: H_2SO_4.

7 0
3 years ago
For a particular isomer of C 8 H 18 , C8H18, the combustion reaction produces 5104.1 kJ 5104.1 kJ of heat per mole of C 8 H 18 (
vlada-n [284]

Answer:

The standard enthalpy of formation of this isomer of octane is -220.1 kJ/mol

Explanation:

Step 1: Data given

The combustion reaction of octane produces 5104.1 kJ per mol octane

Step 2: The balanced equation

C8H18(g) + 12.5 O2 ⟶ 8CO2 (g) + 9 H2O (g)  ∆H°rxn = -5104.1 kJ/mol

Step 3:

∆H°rxn = ∆H°f of products minus the ∆H° of reactants

∆H°rxn = ∆H°f products - [∆H°f reactants]

-5104.1 kJ/mol = (8*∆H°fCO2 + 9*∆H°fH20) - (∆H°fC8H18 + 12.5∆H°fO2)

∆H°f C8H18 = ∆H°f 8CO2 + ∆H°f 9H2O+ 5104.1 kJ/mol

∆H°f C8H18 = 8 * (-393.5 kJ)/mol + 9 * (-241.8 kJ/mol)] + 5104.1 kJ /mol

∆H°f C8H18 = -220.1 kJ/mol

The standard enthalpy of formation of this isomer of octane is -220.1 kJ/mol

7 0
3 years ago
Naphthalene, commonly found in moth balls, is composed of 93.7% carbon and 6.3% hydrogen. The molar mass of naphthalene is 128 g
Vikki [24]

Answer:

Molecular formula naphthalene → C₁₀H₈

Empirical formula naphthalene → C₅H₄

Explanation:

Centesimal composition means that in 100 g of compound we have x g of the element. Therefore in 100 g of naphthalene we have:

93.7 g of C

6.3 g of H

Let's make a rule of three:

In 100 g of naphthalene we have 93.7 g of C  and 6.3 g of H

In 128 g of naphthalene we would have:

128 . 93.7 / 100 = 120 g of C

128.  6.3 / 100 = 8 g of H

We convert the mass to moles, by molar mass:

120 g . 1mol / 12 g = 10 moles C

8 g . 1mol/ 1g = 8 moles H

Molecular formula naphthalene → C₁₀H₈

Empirical formula naphthalene → C₅H₄

(The sub-index of each element is divided by the largest possible number)

6 0
4 years ago
Read 2 more answers
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