The given mass of cobalt chloride hydrate = 2.055 g
A sample of cobalt chloride hydrate was heated to drive off waters of hydration and the anhydrate was weighed.
The mass of anhydrous cobalt chloride = 1.121 g anhydrate.
The mass of water lost during heating = 2.055 g - 1.121 g = 0.934 g
Converting mass of water of hydration present in the hydrate to moles using molar mass:
Mass of water = 0.934 g
Molar mass of water = 18.0 g/mol
Moles of water = 
Answer:
The volume of the gas will be 78.31 L at 1.7 °C.
Explanation:
We can find the temperature of the gas by the ideal gas law equation:

Where:
n: is the number of moles
V: is the volume
T: is the temperature
R: is the gas constant = 0.082 L*atm/(K*mol)
From the initial we can find the number of moles:

Now, we can find the temperature with the final conditions:

The temperature in Celsius is:

Therefore, the volume of the gas will be 78.31 L at 1.7 °C.
I hope it helps you!
The balanced chemical reaction is given as follows:
<span>2 KClO3(s) → 2 KCl(s) + 3 O2(g)
The starting amount of the reactant are given above. These values would be used for the calculations. We do as follows:
</span>2.72 g KClO3 (1 mol / 122.50g )( 3 mol O2 / 2 mol KClO3 ) ( 32 g O2 / 1 mol O2 ) = 1.06 g O2
<span>
0.361 g KClO3 </span>(1 mol / 122.50g )( 3 mol O2 / 2 mol KClO3 ) ( 32 g O2 / 1 mol O2 ) = 0.14 g O2
<span>
83.6 kg KClO3 (1000g / 1kg) </span>(1 mol / 122.50g )( 3 mol O2 / 2 mol KClO3 ) ( 32 g O2 / 1 mol O2 ) = 3275.76 g O2
<span>
22.5 mg KClO3</span> (1 g / 1000 mg) (1 mol / 122.50g )( 3 mol O2 / 2 mol KClO3 ) ( 32 g O2 / 1 mol O2 ) = 0.009 g O2
Given the molar mass of Nitrogen is 14.01g/mol you can use that to solve for the moles of nitrogen.
0.235g(1mol/14.01g) = .0168 moles.
Answer:
Explanation:
When potassium chloride reacts with lead acetate in aqueous medium, it gives white precipitate of lead chloride and potassium acetate in aqueous medium as a product.
As shown in chemical reaction given below: