It is two or more objects and different things that can be removed from each other.
I think it’s the third option but I’m not entirely sure
Answer:
H2Br + 2KOH ----- K2Br + 2H2O
Answer:
0.032 L or 32 mL
Explanation:
Use the dilution equation M1V1 = M2V2
M1 = 9.0 M
V1 = This is what we're looking for.
M2 = 0.145 M
V2 = 2 L
Solve for V1 --> V1 = M2V2/M1
V1 = (0.145 M)(2 L) / (9.0 M) = 0.032 L
Answer:
The volume of NO₂ gas collected over water at 25.0 °C is 1.68 Liters.
Explanation:

Moles of copper = 
According to reaction, 1 mol of copper gives 2 moles of nitrogen dioxide gas.
Then 0.03613 moles of copper will give:
of nitrogen dioxide gas
Moles of nitrogen dioxide gas = n = 0.06326 mol
Pressure of the gas = P
P = Total pressure - vapor pressure of water
P = 726 mmHg - 23.8 mmHg = 702.2 mmHg
P = 0.924 atm (1 atm = 760 mmHg)
Temperature of the gas = T = 25.0°C =298.15 K
Volume of the gas = V


V = 1.68 L
The volume of NO₂ gas collected over water at 25.0 °C is 1.68 Liters.