We are going to use this equation:
ΔT = - i m Kf
when m is the molality of a solution
i = 2
and ΔT is the change in melting point = T2- 0 °C
and Kf is cryoscopic constant = 1.86C/m
now we need to calculate the molality so we have to get the moles of NaCl first:
moles of NaCl = mass / molar mass
= 3.5 g / 58.44
= 0.0599 moles
when the density of water = 1 g / mL and the volume =230 L
∴ the mass of water = 1 g * 230 mL = 230 g = 0.23Kg
now we can get the molality = moles NaCl / Kg water
=0.0599moles/0.23Kg
= 0.26 m
∴T2-0 = - 2 * 0.26 *1.86
∴T2 = -0.967 °C
The given question is incomplete. The complete question is:
Suppose a current of 0.920 A is passed through an electroplating cell with an aqueous solution of agno3 in the cathode compartment for 47.0 seconds. Calculate the mass of pure silver deposited on a metal object made into the cathode of the cell.
Answer: 0.0484 g
Explanation:
where Q= quantity of electricity in coloumbs
I = current in amperes = 0.920 A
t= time in seconds = 47.0 sec
96500 Coloumb of electricity electrolyzes 1 mole of Ag
43.24 C of electricity deposits =
of Ag
Thus the mass of pure silver deposited on a metal object made into the cathode of the cell is 0.0484 g
Oxygen gains two electrons when it bonds to form a complete outer shell and magnesium loses two electrons when bonding to gain its full outer shell.
As electrons are negative, the oxygen (which gains electrons) will become negative and the magnesium (which loses electrons) will become positive.
The negative and positive ions will then attract to one another due to the magnetic pull of the positive and negative.
Answer:
1.78 × 10⁹ μg
Explanation:
We have to convert 1.78 kg to μg.
Step 1: Convert 1.78 kilograms to grams
We will use the conversion factor 1 kg = 10³ g.
1.78 kg × 10³ g/1 kg = 1.78 × 10³ g
Step 2: Convert 1.78 × 10³ grams to micrograms
We will use the conversion factor 1 g = 10⁶ μg.
1.78 × 10³ g × 10⁶ μg/1 g = 1.78 × 10⁹ μg
Answer:
There are 0.5 mole in 20g of argon.
Explanation:
40 g of argon = 1mole
Then 20g of argon is,
→ 1/40 × 20
→ 0.5 mole
