Answer:
Q = 3440Kj
Explanation:
Given data:
Mass of gold = 2kg
Latent heat of vaporization = 1720 Kj/Kg
Energy required to vaporize 2kg gold = ?
Solution:
Equation
Q= mLvap
It is given that heat required to vaporize the one kilogram gold is 1720 Kj thus, for 2 kg
by putting values,
Q= 2kg × 1720 Kj/Kg
Q = 3440Kj
Answer:
0.311 mmol/L
Explanation:
109 μmol = 109*10^(-6) mol
109*10^(-6) mol = 109*10^(-6) mol*(10^3 mmol/1mol) = 109*10^(-3) mmol =
=0.109 mmol
350 mL = 0.350 L
0.109 mmol/0.350 L = 0.311 mmol/L
Answer:
It should be transferred to a Beaker to finish preparing the solution
Explanation:
NaNO₃ is sodium Nitrate. It is a salt that is soluble in water hence it's aqueous solution can be prepared by dissolving it in water.
A beaker is a laboratory glassware that is used for various functions; one of which includes the preparation of reagents or aqueous solutions of a compound. In the example narrated in the question, the measured NaNO₃ solid can be transferred to a 100 mL or 200 mL beaker and then added to it is the desired volume of water to make the solution 100 mL. This solution can then be stirred with a "glass rod"
Answer: 
Explanation:
Let consider that one mole of diatomic oxygen is used. So, the stoichometric can be modelled by using three variables:
Where 
are the required variables.
Now, three equations are constructed from the number of elements involved (Carbon, Hydrogen and Oxygen):
Carbon
Oxygen
Hydrogen
The coefficients can be found by solving the abovementioned 3 x 3 Linear System:
The whole-number coefficients are determined by multiplying every coefficient by 3, then:
