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Answer:
13mL
Explanation:
Step 1:
The balanced equation for the reaction. This is given below:
HNO3 + KOH —> KNO3 + H2O
From the balanced equation above, we obtained the following data:
Mole ratio of the acid (nA) = 1
Mole ratio of the base (nB) = 1
Step 2:
Data obtained from the question.
This includes the following:
Molarity of the acid (Ma) = 6M
Volume of the acid (Va) =?
Volume of the base (Vb) = 39mL
Molarity of the base (Mb) = 2M
Step 3:
Determination of the volume of the acid.
Using the equation:
MaVa/MbVb = nA/nB, the volume of the acid can be obtained as follow:
MaVa/MbVb = nA/nB
6 x Va / 2 x 39 = 1/1
Cross multiply to express in linear form
6 x Va = 2 x 39
Divide both side by 6
Va = (2 x 39)/6
Va = 13mL
Therefore, the volume of the acid (HNO3) needed for the reaction is 13mL
Answer:
44.8 litres is the answer to the question
126 grams of H2O is formed.
Explanation:
Data given:
volume of the gas = 88 Liters
pressure = 720 mm Hg or 0.947 atm
temperature T = 22 Degrees or 295.15 K
R = 0.08021 atm L/mole K
n =?
The formula is used is of ideal gas law to know the number of moles of CH4 undergoing combustion.
PV = nRT
n = 
putting the values in the equation
= 0.947 X 88/ 0.08021 X 295.15
n = 3.5 moles
balanced reaction for combustion of methane
CH4 + O2 ⇒ CO2 + 2H20
1 mole of CH4 undergoes combustion to form 2 moles of water
3.5 moles will give x moles of water
2/1 = x/3.5
x = 7 moles of water (atomic mass of water = 18 gram/mole)
mass = atomic mass x number of moles
mass = 18 x 7
=126 grams of water is formed.