Mass = 5 g
volume = 20 cm³
density = mass / volume
therefore:
D = m / V
D = 5 / 20
D = 0.25 g/cm³
2 <span>KOH +1 H3AsO4 →1 K2HAsO4 + 2 H2O</span>
It would be the same amount. So, 45 ml of NaOH is required to be added to the 45 ml of HCI to neutralize the acid fully. Here is a brief calculation:
Firstly, here is your formula: M(HCI) x V(HCI) = M(NaOh) x V(NaOH)
With the values put in: 0.35 x 45 = 0.35 x V(NaOH)
= 45 ml.
There is 45 ml of V(NaOH)
Let me know if you need anything else. :)
- Dotz
Answer : The molecular weight of a gas is, 128.9 g/mole
Explanation : Given,
Density of a gas = 5.75 g/L
First we have to calculate the moles of gas.
At STP,
As, 22.4 liter volume of gas present in 1 mole of gas
So, 1 liter volume of gas present in 
mole of gas
Now we have to calculate the molecular weight of a gas.
Formula used :
Now put all the given values in this formula, we get the molecular weight of a gas.
Therefore, the molecular weight of a gas is, 128.9 g/mole
Answer:
The correct answer is pOH= 11
Explanation:
From the aqueous acid-base equilibrium we know that
pH + pOH = 14
If we know pH, we can calculate pOH as follows:
pOH = 14 - pH
In this problem, the solution has a pH of 3, so:
pOH = 14 - 3 = 11
