Answer:
7 gram of Na2SO4 should be required to prepare 0.5L of a 0.100 M solution.
Explanation:
First of all the molecular weight of Na2SO4 is 142.08 gram.Now we all know that if the molecular weight of a compound is dissolved in 1000ml or 1 litee of water then the strength of that solution becomes 1 M.
According to the given question we have to prepare 0.100 M solution
1000 ml of solution contain 142.08×0.1= 14.208 gram Na2SO4
1 ml of solution contain 14.208÷1000= 0.014 gram
0.5L or 500ml of solution contain 0.014×500= 7gram Na2SO4.
So it can be stated that 7 gram of Na2SO4 should be required to prepare 0.5L of a 0.100M solution.
I believe it might be c I might wrong
<span>C. 11.2 L
There are several different ways to solve this problem. You can look up the density of CO2 at STP and work from there with the molar mass of CO2, but the easiest is to assume that CO2 is an ideal gas and use the ideal gas properties. The key property is that a mole of an idea gas occupies 22.413962 liters. And since you have 0.5 moles, the gas you have will occupy half the volume which is
22.413962 * 0.5 = 11.20698 liters. And of the available choices, option "C. 11.2 L" is the closest match.
Note: The figure of 22.413962 l/mole is using the pre 1982 definition of STP which is a temperature of 273.15 K and a pressure of 1 atmosphere (1.01325 x 10^5 pascals). Since 1982, the definition of STP has changed to a temperature of 273.15 K and a pressure of exactly 10^5 pascals. Because of this lower pressure, one mole of an ideal gas will have the higher volume of 22.710947 liters instead of the older value of 22.413962 liters.</span>
The equation that shows the formation of chromium (ii) ion from neutral chromium atom is as follow
Cr ---> cr^2+ + 2e-
Cr^2+ is the chromium ion with oxidation state of two which is one of the common ion of chromium. Other common ion of chromium include chromium of oxidation state 6 and 3
X=107 because the sides of a hexagon must add up to 720
