The only metal that is a liquid at room temperature is mercury!!
Answer:
1.327 g Ag₂CrO₄
Explanation:
The reaction that takes place is:
- 2AgNO₃(aq) + K₂CrO₄(aq) → Ag₂CrO₄(s) + 2KNO₃(aq)
First we need to <em>identify the limiting reactant</em>:
We have:
- 0.20 M * 50.0 mL = 10 mmol of AgNO₃
- 0.10 M * 40.0 mL = 4 mmol of K₂CrO₄
If 4 mmol of K₂CrO₄ were to react completely, it would require (4*2) 8 mmol of AgNO₃. There's more than 8 mmol of AgNO₃ so AgNO₃ is the excess reactant. <em><u>That makes K₂CrO₄ the limiting reactant</u></em>.
Now we <u>calculate the mass of Ag₂CrO₄ formed</u>, using the <em>limiting reactant</em>:
- 4 mmol K₂CrO₄ *
= 1326.92 mg Ag₂CrO₄
- 1326.92 mg / 1000 = 1.327 g Ag₂CrO₄
128 ml is the voume of the balloon if the temperature of the gas increases to 320.0k.
Explanation:
given that:
T1 (initial temperature) = 300K
V1 ( initial volume) = 120ml
T2 (final temperature) = 320 K
V2 (final volume) = ?
Pressure remained constant throughout the process.
From the equation
= ![\frac{P2V2}{T2}](https://tex.z-dn.net/?f=%5Cfrac%7BP2V2%7D%7BT2%7D)
Since pressure is constant the equation will be:
= ![\frac{V2}{T2}](https://tex.z-dn.net/?f=%5Cfrac%7BV2%7D%7BT2%7D)
V2 = ![\frac{V1 X T2}{T1}](https://tex.z-dn.net/?f=%5Cfrac%7BV1%20X%20T2%7D%7BT1%7D)
Putting the values in the above formula:
V2 = ![\frac{120 X 320}{300}](https://tex.z-dn.net/?f=%5Cfrac%7B120%20X%20320%7D%7B300%7D)
= 128 ml
128 ml is the volume of the gas if temperature increases from 3OO K to 320k
Answer:
B: Adding water, then adding solute
Explanation:
This is because, say you have a solution with a certain concentration.
If you add more water, it will become more diluted (less concentrated)
If you add more solute, it will become more concentrated.
Therefore if you add water and solute, it could cancel out, and the concentration would remain the same.
Hope this helps! Let me know if you have any questions/ would like anything further explained :)
There is no image , so you cannot answer , sorry