D. Sodium hydroxide aka naOH
(J) unit joule is the standard unit of heat (energy)
Answer: 15.8 g of 
will be required to produce 1.60 grams of 
Explanation:
To calculate the moles :
According to stoichiometry :
As 1 mole of is given by = 2 moles of
Thus 0.05 moles of is given by =
of
Mass of 
Thus 15.8 g of will be required to produce 1.60 grams of
Answer:
HNO₃
Explanation:
Data given
Nitrogen = 9.8 g
Hydrogen = 0.70 g
Oxygen = 33.6 g
Empirical formula = ?
Solution:
Convert the masses to moles
For Nitrogen
Molar mass of N = 14 g/mol
no. of mole = mass in g / molar mass
Put value in above formula
no. of mole = 9.8 g/ 14 g/mol
no. of mole = 0.7
mole of N = 0.7 mol
For Hydrogen
Molar mass of H = 1 g/mol
no. of mole = mass in g / molar mass
Put value in above formula
no. of mole = 0.70 g/ 1 g/mol
no. of mole = 0.7
mole of H = 0.7 mol
For Oxygen
Molar mass of O = 16 g/mol
no. of mole = mass in g / molar mass
Put value in above formula
no. of mole = 33.6 g / 16 g/mol
no. of mole = 2.1
mole of O = 2.1 mol
Now we have values in moles as below
N = 0.7
H = 0.7
O = 2.1
Divide the all values on the smallest values to get whole number ratio
N = 0.7 / 0.7 = 1
H = 0.7 / 0.7 = 1
O = 2.1 / 0.7 = 3
So all have following values
N = 1
H = 1
O = 3
So the empirical formula will be HNO₃ i.e. all three atoms in simplest small ratio.
Answer:
The right statement is " the Pathway A-C-D involves a catalyst and is faster than A-B-D"
Explanation:
Catalyst is a substance that increases the rate of a chemical reaction by lowering the activation energy Eₐ (i.e. by changing the reaction pathway), which Catalyst is not consumed in the catalyzed reaction and can continue to act repeatedly.
From the attached figure one can see that the path way A-C-D has different pathway and lower activation energy (Eₐ) than the pathway A-B-D, this means that the pathway A-C-D is a catalyzed pathway and it is faster than A-B-D.
So, the right statement is " the Pathway A-C-D involves a catalyst and is faster than A-B-D"
