The balanced equation for the neutralisation reaction is as follows
Ca(OH)₂ + H₂SO₄ ---> CaSO₄ + 2H₂O
stoichiometry of Ca(OH)₂ to H₂SO₄ is 1:1
equivalent number of acid reacts with base
number of H₂SO₄ mol reacting - 2 mol
according to molar ratio of 1:1
number of Ca(OH)₂ mol = number of H₂SO₄ moles
therefore number of Ca(OH)₂ moles required - 2 mol
C burning
Because melting is incorrect lol
Ok so 40 percent of 3000 grams which is 3 kg is 1200 grams
So you would need 1.2kg or 1200 grams to make a 40% solution
Answer:
Explanation:
q= mc theta
where,
Q = heat gained
m = mass of the substance = 670g
c = heat capacity of water= 4.1 J/g°C
theta =Change in temperature=(
66-25.7)
Now put all the given values in the above formula, we get the amount of heat needed.
q= mctheta
q=670*4.1*(66-25.7)
=670*4.1*40.3
=110704.1
Answer:
37.8 L OF CARBON MONOXIDE IS REQUIRED TO PRODUCE 18.9 L OF NITROGEN.
Explanation:
Equation for the reaction:
2 CO + 2 NO ------> N2 + 2 CO2
2 moles of carbon monoxide reacts with 2 moles of NO to form 1 mole of nitrogen
At standard temperature and pressure, 1 mole of a gas contains 22.4 dm3 volume.
So therefore, we can say:
2 * 22.4 L of CO produces 22.4 L of N2
44.8 L of CO produces 22.4 L of N2
Since, 18.9 L of Nitrogen is produced, the volume of CO needed is:
44.8 L of CO = 22.4 L of N
x L = 18.9 L
x L = 18.9 * 44.8 / 22.4
x L = 18.9 * 2
x = 37.8 L
The volume of Carbon monoxide required to produce 18.9 L of N2 is 37.8 L
