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AleksAgata [21]
3 years ago
14

Ag2O(s) → 2Ag(s) + ½ O2(g) ΔH° = 31.05 kJ Which statements concerning the reaction above are true? (1) heat is released (2) heat

is absorbed (3) reaction is exothermic (4) reaction is endothermic (5) products have higher enthalpy content than reactants (6) reactants have higher enthalpy content than products A) 1, 3, and 5 B) 1, 3, and 6 C) 2, 4, and 6 D) 2, 4, and 5
Chemistry
1 answer:
Sidana [21]3 years ago
5 0

Answer:

D) 2, 4, and 5

Explanation:

In order to fully comprehend the answer choices we must take a close look at the value of ΔH° = 31.05. The enthalpy change of the reaction is positive. A positive value of enthalpy of reaction implies that heat was absorbed in the course of the reaction.

If heat is absorbed in a reaction, that reaction is endothermic.

Since ∆Hreaction= ∆H products -∆H reactants, a positive value of ∆Hreaction implies that ∆Hproducts >∆Hreactants, hence the answer choice above.

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The gas, 2 mol of H2, occupies the highest volume at STP since at STP the volume of this gas is approximately 44.8 mol as compared to other options this has the greatest amount.

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Why is zinc not extracted from ZnO through reduction using CO?​
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The standard Gibbs free energy of formation of ZnO from Zn is lower than that of CO2 from CO. Therefore, CO cannot reduce ZnO to Zn. Hence, Zn is not extracted from ZnO through reduction using CO
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1 year ago
When a plant is entering the Calvin cycle of photosynthesis_____.
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B). light energy is not required to proceed

Explanation:

In the Calvin cycle of photosynthesis, light energy is not required. The Calvin cycle is light independent and it is made up of a series of redox reactions.

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  • During the Calvin cycle aspect, light energy is not required for chemical reactions to take place. The light energy helps to move electrons.
  • The cycle is also known as dark reactions.
  • It is at this stage that carbon dioxide combines with water to form glucose.
  • The reaction is initiated with light energy which produces NADPH and ATP.
  • The Calvin cycle follows by using the NADPH and ATP to produce glucose in the dark phase.

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4 0
3 years ago
Please show all work for full credit.
koban [17]

Answer:

No, it is not sufficient

Please find the workings below

Explanation:

Using E = hf

Where;

E = energy of a photon (J)

h = Planck's constant (6.626 × 10^-34 J/s)

f = frequency

However, λ = v/f

f = v/λ

Where; λ = wavelength of light = 325nm = 325 × 10^-9m

v = speed of light (3 × 10^8 m/s)

Hence, E = hv/λ

E = 6.626 × 10^-34 × 3 × 10^8 ÷ 325 × 10^-9

E = 19.878 × 10^-26 ÷ 325 × 10^-9

E = 19.878/325 × 10^ (-26+9)

E = 0.061 × 10^-17

E = 6.1 × 10^-19J

Next, we work out the energy required to dissociate 1 mole of N=N. Since the bond energy is 418 kJ/mol.

E = 418 × 10³ ÷ 6.022 × 10^23

E = 69.412 × 10^(3-23)

E = 69.412 × 10^-20

E = 6.9412 × 10^-19J

6.9412 × 10^-19J is required to break one mole of N=N bond.

Based on the workings above, the photon, which has an energy of 6.1 × 10^-19J is not sufficient to break a N=N bond that has an energy of 6.9412 × 10^-19J

8 0
3 years ago
The combination of potassium-sparing diuretics and salt substitutes can result in dangerously high blood levels of:
alisha [4.7K]

Answer:

b. potassium.  

Explanation:

Potassium-sparing diuretics and salt substitutes are diuretics that eliminate salt and water but save potassium. They act by inhibiting the conducting sodium channels in the collecting tubule, such as amiloride and triamterene, or by blocking aldosterone, such as spironolactone.

Concomitant use of potassium-sparing diuretics together with salt substitutes may result in dangerously high blood levels of serum potassium. For this reason, it is important to consult a physician before taking these substances at the same time to avoid potential problems with potassium accumulation.

4 0
3 years ago
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