Answer : The number of molecules present in nitrogen gas are, 
Explanation :
First we have to calculate the moles of nitrogen gas by using ideal gas equation.
where,
P = Pressure of 
gas = 
(1 atm = 760 mmHg)
V = Volume of gas = 985 mL = 0.982 L (1 L = 1000 mL)
n = number of moles = ?
R = Gas constant = 
T = Temperature of gas = 
Now put all the given values in above equation, we get:
Now we have to calculate the number of molecules present in nitrogen gas.
As we know that 1 mole of substance contains 
number of molecules.
As, 1 mole of gas contains number of molecules
So, 
mole of gas contains 
number of molecules
Therefore, the number of molecules present in nitrogen gas are,
Answer:
an increase in 1-butene was observed when t-butoxide was used
Explanation:
When a base reacts with an alkyl halide, an elimination product is formed. This reaction is an E2 reaction.
Here we are to compare the reaction of two different bases with one substrate; 2-bromobutane. Both reactions occur by the E2 mechanism but follow different transition states due to the size of the base.
The Saytzeff product, 2-butene, is obtained when the methoxide is used while the non Saytzeff product, 1-butene, is obtained when t-butoxide is used.
The Saytzeff rule is reliable in predicting the major products of simple elimination reactions of alkyl halides given the fact that a small/strong bases is used for the elimination reaction. Therefore hydroxide, methoxide and ethoxide bases give similar results for the same alkyl halide substrate. Bulky bases such as tert-butoxide tend to yield a higher percentage of the non Saytzeff product and this is usually attributed to steric hindrance.
8.4 grams. I think but I’m not 100% sure
The equation structure for the above mentioned reaction can be written as
<u>Explanation:</u>
Considering the above reaction, When Boron sulfide, reacts with water more violently to form boric acid and hydrogen sulfide gas.
In order to balance the equation, we can do as follows.There are 2 B - atoms on both sides of the equation, but only 2 H - atoms, and one O - atom on LHS, so we have to balance it by putting 6 in front of water and 2 in front of Boric acid and 3 in front of hydrogen sulphide gas, so that we have 2 B - atoms, 3 - S atoms, 12 H - atoms on both sides of the equation, and it is balanced. Balanced equation is given as,
Thus a Balanced equation of the above mentioned reaction is written.
Ca + 2HCl = CaCl₂ + H₂
c=4.50 mol/l
v=2.20 l
n(HCl)=cv
m(Ca)/M(Ca)=n(HCl)/2
m(Ca)=M(Ca)cv/2
m(Ca)=40g/mol·4.50mol/l·2.20l/2=198 g
198 grams of Ca are needed
