${ \boxed{ \bf{vapour \: density = 2 \times molecular \: mass}}} \\{ \tt{ PV= (\frac{m}{ m_{r}}) RT}} \\ { \tt{3 \times 5.6 = \frac{11}{m _{r}} \times 0.0831 \times 273}} \\ { \tt{m _{r} = 14.85 \: g}} \\ \\ { \bf{vapour \: density = 2 \times m _{r}}} \\ = 2 \times 14.85 \\ = 29.7 \: { \tt{g {dm}^{ - 3} }}$

7 0

2 years ago

Determine the molecular

jeyben [28]

Answer:

MgO- magnesium oxide

Cu(NO3)2- copper(11)nitrate

Li2CO3- lithium carbonate

7 0

2 years ago

A sample of metal has a mass of 18.20g, and a volume of 7.38mL. What is the density in g/cm^3?

AlexFokin [52]

Density = Mass ÷ Volume. You get 2.466

8 0

3 years ago

The enthalpy change for converting 1.00 mol of ice at -50.0°c to water at 70.0°c is __________ kj. the specific heats of ice, wa

kogti [31]

First, calculate for the amount of heat used up for increasing the temperature of ice.

H = mcpdT
H = (18 g)*(2.09 J/g-K)(50 K) = 1881 J

Then, solve for the heat needed to convert the phase of water.
H = (1 mol)(6.01 kJ/mol) = 6.01 kJ = 6010 J

Then, solve for the heat needed to increase again the temperature of water.
H = (18 g)(4.18 J/gK)(70 k)
H = 5266.8 J

The total value is equal to 13157.8 J

Answer: 13157.8 J

8 0

3 years ago