What is the volume of 60 g of ether if the density of ether is 70 g/mL

Other substances that give a positive test with AgNO3 are other chlorides present, iodides and bromide. However iodides and bromides have different colours hence they will not give a false positive test for KCl. Other chlorides present may lead to a false positive test for KCl.

Explanation:

In the qualitative determination of halogen ions, silver nitrate solution is used. Various halide ions give various colours of precipitate with silver nitrate. Chlorides yield a white precipitate, bromides yield a cream precipitate while iodides yield a yellow precipitate. All these ions or some of them may be present in the system.

However, if other chlorides are present, they will also yield a white precipitate just as KCl leading to a false positive test for KCl. Since other halogen ions yield precipitates of different colours, they don't lead to a false test for KCl. We can exclude other halides from the tendency to lead us to a false positive test for KCl but not other chlorides.

3 0

2 years ago

It has been suggested that the surface melting of ice plays a role in enabling speed skaters to achieve peak performance. Carry

Fittoniya [83]

Answer:

a) Pf = 689.4 bar

b) P = 226.6 bar

c) T = 269.99 K

Explanation:

a)

The molar volume of ice is equal to:

Vi = m/p = (18.02x10^-3 kg H2O/1 mol H2O)*(1 m^3/920 kg) = 1.96x10^-5 m^3 mol^-1

The molar volume of liquid water is equal to:

Vl = (18.02x10^-3 kg/1 mol)*(1 m^3/997 kg) = 1.8x10^-5 m^3 mol^-1

The change in volume is equal to:

ΔVchange = Vi-Vl = 1.96x10^-5 - 1.8x10^-5 = 1.5x10^-6 m^3 mol^-1

using the Clapeyron equation:

Pf = Pi + ((ΔHf*ΔT)/(ΔVf*Ti)) = 1.013x10^5 Pa + ((6010 J mol^-1 * 4.7 K)/(1.5x10^-6 * 273.15 K)) = 6.89x10^7 Pa = 689.4 bar

b)

For the pressure we will use the equation:

P = (m*g)/A, where m is the mass, g is the acceleration of gravity and A is the area. Replacing values:

P = (79 kg * 9.81 m s^-2)/(1.9x10^-4 m * 0.18 m) = 2.26x10^7 Pa = 226.6 bar

c)

From Clapeyron´s expression we need to clear ΔT:

ΔT = ((Pf-Pi)*ΔV*Ti)/ΔHf = ((2.26x10^7 - 6.89x10^7)*1.5x10^-6*273.15)/6010 = -3.16 K

you can evaluate the new melting point of ice:

T = Ti + ΔT = 273.15 K - 3.16 K = 269.99 K

4 0

3 years ago