Answer:
150 years i guess so? im nkt sure
The combustion reaction is as expressed,
CxHy + O2 --> CO2 + H2O
The mass fraction of carbon in CO2 is 3/11. Hence,
mass of C in CO2 = (3.14 g)(3/11) = 0.86 g C.
Given that we have 1 g of the hydrocarbon, the mass of H is equal to 0.14 g.
moles of C = 0.86 g C / 12 g = 0.0713
moles of H = 0.14 g H / 1 g = 0.14
The empirical formula for the hydrocarbon is therefore, CH₂.
Answer:- 335 kcal of heat energy is produced.
Solution:- The balanced equation for the combustion of glucose in presence of oxygen to give carbon dioxide and water is:

From given info, 2803 kJ of heat is released bu the combustion of 1 mol of glucose. We need to calculate the energy produced when 3.00 moles of oxygen react with excess of glucose.
We could solve this using dimensional analysis as:

= 1401.5 kJ
Now, let's convert kJ to kcal.
We know that, 1kcal = 4.184kJ
So, 
= 335 kcal
Hence, 335 kcal of heat energy is produced by the use of 3.00 moles of oxygen gas.
Answer:
The final volume will be "70.08 mL".
Explanation:
The given values are:
Molar mass,
M1 = 548 nM
or,
= 
M2 = 484 nM
or,
=
Volume,
V1 = 61.9 mL
V1 = ?
By using the expression, we get
⇒ 
or,
⇒ 
By substituting the values, we get



Answer:
a) 
2 moles of Zinc sulphide in solid form reacts with 3 moles of Oxygen in gaseous form to give 2 moles of Zinc oxide in solid form and 2 moles of sulphur dioxide in gaseous form.
b) 
1 mole of calcium hydride in solid form reacts with 2 moles of liquid water to give 1 mole of calcium hydroxide dissolved in water and 2 moles of hydrogen in gaseous form.
The chemical reactions are written by writing the chemical formula of the reactants on left side of the arrow followed by chemical formula of the products. The number of atoms of each element must be balanced to follow the law of conservation of mass.