Water is a pure substance, however it is made up of oxygen and hydrogen atoms, hence the nickname H2O (Hydrogen2, Oxygen)
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>
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<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).
Given that you only know the mass in grams of the solute and the mass in grams of the solution, you can only calculate the concentration as percent mass/mass.
The formula is:
Percent m/m, % = [ (mass in grams of solute) / (mass in grams of solution) ] * 100
% = [ (1.2 * 10^ -3 g) / (800 g) ] * 100 = 0.0015%
Answer: 0.0015%
Answer:
B
Explanation:
The methyl prefix means theres a CH3 group bonded to a carbon in a hydrocarbon. This is called branching because it looks like a branched tree. This is literally cheating but you do you.
Answer:
1.634 molL-1
Explanation:
The mol ration between NH3 and HCl is 1 : 1
Using Ca Va / Cb Vb = Na / Nb where a = acid and b = base
Na = 1
Nb = 1
Ca = 0.208 molL-1
Cb = ?
Va = 19.64 mL
Vb = 25.00mL
Solving for Cb
Cb = Ca Va / Vb
Cb = 0.208 * 19.64 / 25.0
Cb = 0.1634 molL-1 (Concentration of diluted ammonia solution)
Using the dilution equation;
C1V1 = C2V2
Initial Concentration, C1 = ?
Initial Volume, V1 = 25.00 mL
Final Volume, V2 = 250 mL
Final Concentration, C2 = 0.1634 molL-1
Solving for C1;
C1 = C2 * V2 / V1
C1 = 0.1634 * 250 / 25.00
C1 = 1.634 molL-1