Answer:
removing the Cl₂ as it is formed
.
adding more ICl(s)
.
removing some of the I₂(s).
Explanation:
<em>Le Châtelier's principle </em><em>states that when there is an dynamic equilibrium, and this equilibrium is disturbed by an external factor, the equilibrium will be shifted in the direction that can cancel the effect of the external factor to reattain the equilibrium.</em>
<em />
<u>1) Decreasing the volume of the container:</u>
- Decreasing the volume of the container will increase the pressure.
- When there is an increase in pressure, the equilibrium will shift towards the side with fewer moles of gas of the reaction. And when there is a decrease in pressure, the equilibrium will shift towards the side with more moles of gas of the reaction.
- The reactants side (left) has no moles of gases and the products side (right) has 1.0 mole of gases.
- So, increasing the pressure will shift the reaction to the side with lower moles of gas (left side) and so the total amount of Cl₂ produced is decreased.
so, decreasing the volume of the container will decrease the total amount of Cl₂ produced.
<u>2) Removing the Cl₂ as it is formed:</u>
- Removing Cl₂ gas will decrease the concentration of the products side, so the reaction will be shifted to the right side to suppress the decrease in the concentration of Cl₂ gas by removing and so the total amount of Cl₂ produced is increased.
so, removing the Cl₂ as it is formed will increase the total amount of Cl₂ produced.
<u><em>3) Adding more ICl(s)
:</em></u>
- Adding ICl(s) will increase the concentration of the reactants side, so the reaction will be shifted to the right side to suppress the increase in the concentration of ICl(s) by addition and so the total amount of Cl₂ produced is increased.
so, adding more ICl(s) will increase the total amount of Cl₂ produced.
<u>2) Removing some of the I₂(s):</u>
- Removing I₂ gas will decrease the concentration of the products side, so the reaction will be shifted to the right side to suppress the decrease in the concentration of Cl₂ gas by removing and so the total amount of Cl₂ produced is increased.
so, removing some of the I₂(s) will increase the total amount of Cl₂ produced.
<em>the following changes will increase the total amount of of Cl2 that can be produced:</em>
- removing the Cl₂ as it is formed
.
- removing some of the I₂(s).
1
Explanation:
The value of all conversion factors that are used in dimensional analysis must be 1.
Dimensional analysis is a beautiful and systematic way of converting one unit to another using a simple scientific method in which a conversion factor of 1 is used to multiply the number to be converted.
For example, let us convert 1200kg to g:
We know that;
1000g = 1kg
The conversion factor is;

The above value is the same as 1 since 1kg and 1000g are the same.
Now multiply with 1200kg;
1200kg x
= 1200000g
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Conversion brainly.com/question/555814
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Answer:
21 mL of NaOH is required.
Explanation:
Balanced reaction: 
Number of moles of HBr in 11.0 mL of 0.30 M HBr solution
=
moles = 0.0033 moles
Let's say V mL of 0.16 M NaOH solution is required to reach equivalence point.
So, number of moles of NaOH in V mL of 0.16 M NaOH solution
=
moles = 0.00016V moles
According to balanced equation-
1 mol of HBr is neutralized by 1 mol of NaOH
So, 0.0033 moles of HBr are neutralized by 0.0033 moles of NaOH
Hence, 

So, 21 mL of NaOH is required.
Answer:
- About 18 g of NH₄Cl will precipitate.
Explanation:
The <em>table G</em> is the graph of the solubility curves for several solutes which is attached.
The second picture identifies the solubilities for the NH₄Cl at 50ºC and 10ºC.
The solubility of NH₄Cl at 50ºC is about 52 g/ 100 g of water.
The solubility of NH₄Cl at 10ºC is about 34 g / 100 g of water.
Then, at 50ºC 100 g of water saturated with NH₄Cl contains about 52 g of NH₄Cl and 100 g of water saturated with NH₄Cl contains 34 g of NH₄Cl.
The difference, 52g - 34 g of NH₄Cl shall precipitate:
52 g - 34 g = 18g ← answer
Answer:
Option B. +3 and +6
Explanation:
<em>Zeff</em> = <em>z - s</em>
where <em>z</em> is the atomic number, <em>s</em> is the number of shielding(non-valence) electrons
For Boron, electronic configuration is 1s²2s³.
z = 5, s = 2
Zeff = 5 - 2 = +3
For Oxygen, electronic configuration is 1s²2s²2p⁴
z = 8, s = 2
Zeff = 8 - 2 = +6