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andreev551 [17]
3 years ago
10

What happens in the second step of metabolism

Chemistry
1 answer:
Roman55 [17]3 years ago
8 0

Answer:

The second process produces energy and is referred to as catabolic.

Explanation:

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What is the molarity of a solution containing 0.5 mol of zinc sulfate (ZnSO4) per liter?
QveST [7]

molarity = G/L/RFM

RFM = 120+32+48= 200

0.5/200 = 0.0025

7 0
4 years ago
To completely neutralize a 0.325 g sample of pure aspirin, 15.50 mL of a sodium hydroxide solution is added. If 16.25 mL of the
Kruka [31]
<span>a. 0.325 g / 63.55 g/mol = 5.11 X 10^-3 moles Cu. SHould form 5.11 X 10^-3 mol Cu2+

b. Should form 5.11 X 10^-3 mol Cu(OH)2

c. 1 g Zn / 65.4 g/mol = 0.0153 mol Zn
Excess Zn = 0.0153 - 0.0051 = 0.0102 moles excess zinc

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3 0
3 years ago
What is the answer?
nevsk [136]

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8 0
3 years ago
Consider stoichiometric combustion of gasoline and air. Assume a full tank of gasoline holds 50 L (approximately 13.2 gallons),
Inessa05 [86]

Answer:

520 kg

Explanation:

Let's consider the combustion of isooctane.

C₈H₁₈(l) + 12.5 O₂(g) → 8 CO₂(g) +  9 H₂O(l)

We can establish the following relations:

  • 1 mL of C₈H₁₈ has a mass of 0.690 g (ρ = 0.690 g/mL).
  • The molar mass of C₈H₁₈ is 114.22 g/mol.
  • The molar ratio of C₈H₁₈ to O₂ is 1:12.5.
  • The mole fraction of O₂ in air is 0.21.
  • The molar mass of air is 28.96 g/mol.

50 L of isooctane require the following mass of air.

50 \times 10^{3}mLC_{8}H_{18}.\frac{0.690gC_{8}H_{18}}{1mLC_{8}H_{18}} .\frac{1molC_{8}H_{18}}{114.22gC_{8}H_{18}} .\frac{12.5molO_{2}}{1molC_{8}H_{18}} .\frac{1molAir}{0.21molO_{2}} .\frac{28.96 \times 10^{-3}kgAir}{1molAir} =520kgAir

5 0
4 years ago
2. Suppose that 21.37 mL of NaOH is needed to titrate 10.00 mL of 0.1450 M H2SO4 solution.
AysviL [449]

Answer:

0.1357 M

Explanation:

(a) The balanced reaction is shown below as:

2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O

(b) Moles of H_2SO_4 can be calculated as:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Or,

Moles =Molarity \times {Volume\ of\ the\ solution}

Given :

For H_2SO_4 :

Molarity = 0.1450 M

Volume = 10.00 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 10×10⁻³ L

Thus, moles of H_2SO_4 :

Moles=0.1450 \times {10\times 10^{-3}}\ moles

Moles of H_2SO_4  = 0.00145 moles

From the reaction,

1 mole of H_2SO_4 react with 2 moles of NaOH

0.00145 mole of H_2SO_4 react with 2*0.00145 mole of NaOH

Moles of NaOH = 0.0029 moles

Volume = 21.37 mL = 21.37×10⁻³ L

Molarity = Moles / Volume = 0.0029 /  21.37×10⁻³  M = 0.1357 M

7 0
3 years ago
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