Answer:
- <u><em>Yes, 200 ml of fluid can be transferred to a 1-quart container.</em></u>
Explanation:
You must compare the two volumes, 200 ml and 1 quart. If 200 ml is less than or equal to 1 quart, then 200 ml of fluid can be transferred to a 1-quart container, else it is not possible.
To compare, the two volumes must be on the same system of units.
Quarts is a measure of volume equivalent to 1/4 of gallon.
One gallon is approximately 3.785 liters.
3.785 liter = 3.785 liter × 1,000 ml/liter
Then, to convert 1 quart to ml use the unit cancellation method:
- (1/4)gallon × 3.785 liter/gallon × 1,000ml / liter = 946.25 ml
Thus, you get that a 1-quart container has volume of 946.25 ml, which allows that 200ml of fluid be transferred to it.
<u>Answer: </u>The equation which is wrong is 
<u>Explanation:</u>
For the given reaction:
The expression for 
is given by:
![K_c=\frac{1}{[O_2]^3}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B1%7D%7B%5BO_2%5D%5E3%7D)
![K_p=\frac{1}{[O_2]^3}](https://tex.z-dn.net/?f=K_p%3D%5Cfrac%7B1%7D%7B%5BO_2%5D%5E3%7D)
The concentration of solids are taken to be 1, only concentration of gases and liquid states are taken. The pressure of only gases are taken.
Relationship between 
is given by the expression:
where,
= number of moles of gaseous products - number of moles of gaseous reactants
R = gas constant
T= temperature
For the above reaction,
= number of moles of gaseous products - number of moles of gaseous reactants = 0 - 3 = -3
Hence, the expression for 
is:
Therefore, the equation which is wrong is
Answer:
they are transfer from the towers
Explanation:
Answer:
The addition of sulfate ions shifts equilibrium to the left.
Explanation:
Hello!
In this case, according to the following ionization of strontium sulfate:
It is evidenced that when sodium sulfate is added, sulfate, 
is actually added in to the solution, which causes the equilibrium to shift leftwards according to the Le Ch athelier's principle. Thus, the answer in this case would be:
The addition of sulfate ions shifts equilibrium to the left.
Best regards!
Answer: Li is the reducing agentg and O is the oxidizing agent.
Explanation:
1) The oxidizing agent is the one that is reduced and the reducing agent is the one that is oxidized.
2) The given reaction is:
4Li(s) + O₂ (g) → 2 Li₂O(s)
3) Determine the oxidation states of each atom:
Li(s): oxidation state = 0 (since it is alone)
O₂ (g): oxidation state = 0 (since it is alone)
Li in Li₂O (s) +1
O in Li₂O -2
That because 2× (+1) - 2 = 0.
4) Determine the changes:
Li went from 0 to + 1, therefore it got oxidized and it is the reducing agent.
O went from 0 to - 2, therefore it got reduced and it is the oxidizing agent.
