Mercury (ii) oxide is made up of mercury and oxygen. The total mass of mercury (ii) oxide is 14.2 g, after decomposition 13.2 g of mercury were formed, therefore the mass of oxygen 1 g (14.2 g -13.2 g).
Percentage of oxygen = (1/14.2)×100 = 7.04%
Percentage of mercury = (13.2/14.2) × 100 = 92.96%
Therefore, percentage composition of the compound, oxygen is 7.04% and mercury is 92.96%.
ANSWERS:
Group 2 metal carbonates, nitrates and hydroxides decompose to heat to give the corresponding metal oxide and release CO2, NO2 and O2, and H2O respectively. The thermal stability increases down the group as theionic character of the compounds increases down the group.
PV=nRT will give you the answer I think. I haven’t worked with a certain unit in that problem
There are several ways of expressing concentration of solution. Few of them are listed below
1) mass percentage
2) volume percentage
3) Molarity
4) Normality
5) Molality
In most of the drugs, concentration is expressed either in terms of mass percentage or volume percentage. For, solid in liquid type systems, mass percentage is convenient way of expressing concentration, while for liquid in liquid type solutions, expressing concentration in terms of volume percentage is preferred. Present system is an example of liquid in liquid type solution
Here, concentration of H2O2 is given antiseptic = 3.0 % v/v
It implies that, 3ml H2O2 is present in 100 ml of solution
Thus, 400 ml of solution would contain 4 X 3 = 12 ml H2O2
Answer:
Explanation:
Volume of silver cube = 2.42³ = 14.17 cm³
mass of silver cube = volume x density
= 14.17 x 10.49 = 148.64 gm
Volume of gold cube = 2.75³ = 20.8 cm³
mass of gold cube = 20.8 x 19.3 = 401.44 gm
specific heat of silver and gold are .24 and .129 J /g°C
mass of 112 mL water = 112 g
Heat absorbed = heat lost = mass x specific heat x temperature fall or rise
Heat lost by metals
= 148.64 x .24 x ( 85.4 -T) + 401.44 x .129 x ( 85.4 - T )
= (35.67 + 51.78 ) x ( 85.4 - T )
87.45 x ( 85.4 - T )
= 7468.23 - 87.45 T
Heat gained by water
= 112 x 1 x ( T - 20.5 )
= 112 T - 2296
Heat lost = heat gained
7468.23 - 87.45 T = 112 T - 2296
199.45 T = 9764.23
T = 48.95° C
