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sdas [7]
2 years ago
12

Select the correct answer.

Chemistry
1 answer:
marissa [1.9K]2 years ago
3 0

Answer:

How does the equilibrium change with the removal of hydrogen (H2) gas from this equation? 2H2S ⇌ 2H2(g) + S2(g) A. ... Equilibrium shifts left to produce less reactant.

Explanation:

option A is the correct answer

Equilibrium shifts right to produce more product.

I hope it will help you.

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Someone help answer this
alexandr402 [8]

it is water energy :) this is easy

6 0
3 years ago
Read 2 more answers
How many protons are there in 20.02 mol of neon (ne)?
hodyreva [135]
In order to determine the number of protons in 20.02 mol of Ne, we use Avogadro's number to convert the number of moles to number of atoms, 1 mol = 6.022 x 10^23 atoms. From there, we must know the number of protons in a Neon atom, which is 10. Thus, the formula will be:

(20.02 mol Ne)x(6.022 x 10^23 atoms/mol)x(10 protons/1 atom Ne) =
1.2056 x 10^26 protons 
7 0
3 years ago
Read 2 more answers
An aqueous solution is found to be 3.64% potassium nitrate by mass, how many grams of solution should I evaporate the water from
Sidana [21]

Answer:

513.74 g of solution

Explanation:

% Mass grams are defined as the <em>grams that are dissolved in salt</em> (in this case, it would be <em>potassium nitrate</em>) <em>dissolved every 100 g of the solution</em>. Having this information, you can calculate the amount of solution that has dissolved 18.7 g of potassium nitrate, which is what we want to obtain.

The relationship is:

3.64 g of potassium nitrate _____ 100 g solution

18.7 g of potassium nitrate _____ X = 513.74 g of solution

Calculation: 18.7g x 100g / 3.64g = 513.74 g of solution

So, <em>I need 513.74 g of solution to get 18.7g of potassium nitrate by evaporating it</em>.

8 0
3 years ago
1 mol super cooled liquid water transformed to solid ice at -10 oC under 1 atm pressure.
Arada [10]

Answer:

Explanation:

Given that:

number of moles of super cooled liquid water = 1

Melting enthalpy of ice = 6020 J/mol

Freezing point =0 °C = (0 + 273 K)= 273 K

The decrease in entropy of the system during freezing for 1 mol (i.e during transformation from liquid water to solid ice )  = - 6020 J/mol × 1 mol /273 K = -22.051 J/K

Entropy change during further cooling from 0 °C (273 K) to -10 °C (263 K)

\Delta \ S = \int\limits^{T_2}_{T_1}\dfrac{nC_p(s)dT}{T}

\Delta \ S = {nC_p(s)In \dfrac{T_2}{T_1}

\Delta \ S = {(1*37.7)In \dfrac{263}{273}

Δ S = -1.4 J/K

Total entropy change of the system = - 22.05 J/K - 1.4 J/K = - 23.45 J/K

Entropy change of universe = entropy change of the system+ entropy change of the surrounding

According to the second law of thermodynamics

Entropy change of universe  >0

SO,

Entropy change of the system + entropy change in the surrounding > 0

Entropy change in the surrounding > - entropy change of the system

Entropy change in the surrounding > - (- 23.53 J/K)

Entropy change in the surrounding > 23.53 J/K

b) Make some comments on entropy changes from the obtained data.

From the data obtained; we will realize that the entropy of the system decreases as cooling takes place when water is be convert to ice , randomness of these molecules reduces and as cooling proceeds , hence, entropy reduces more as well and the liberated heat will go into the surrounding due to this entropy of the surrounding increasing.

4 0
3 years ago
In which location of the digestive system will you find both physical AND chemical digestion?
d1i1m1o1n [39]

Answer:

The stomach

Explanation:

In the stomach, physical digestion occurs because of peristaltic contractions (wave like contraction in the lean muscle) and chemical digestion occurs because of stomach acid, hydrochloric acid, breaking down your food.

5 0
3 years ago
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