Answer:
19.8 kg of C₂H₂ is needed
Explanation:
We solve this by a rule of three:
If 1251 kJ of heat are relased in the combustion of 1 mol of acetylene
95.5×10⁴ kJ of heat may be released by the combustion of
(95.5×10⁴ kJ . 1) /1251kJ = 763.4 moles of C₂H₂
Let's convert the moles to mass → 763.4 mol . 26 g/1 mol = 19848 g
If we convert the mass from g to kg → 19848 g . 1kg / 1000g = 19.8 kg
Answer:
1.2*10^24 molecules of CF4
Explanation:
the molar mass of cf4 is 88.0043 g/mol
176/88.0043 = 2 moles of CF4
Then multiply by avogadro's number (6.022*10^23) to get the number of molecules
2*6.022*10^23 = 1.2*10^24 molecules of CF4
4 total bonds if u have double,or triple then subtract from 4
The CaCO3 produced if 47.5 moles of NH3 produced is calculated as follows
CaCN2 +3H2O = CaCO3 + 2NH3
by use of mole ratio between CaCO3 to NH3 which is 1:2 the moles of CaCO3 is therefore = 47.5 /2= 23.75 moles
mass of CaCO3 is therefore = moles x molar mass
= 23.75 moles x 100g/mol= 2375 grams which is approximate 2380 grams(answer 6)
<span>The cell must exchange materials with the environment across the surface membrane. An increase in size will result in a relatively greater increase in volume and mass than in surface area, so that the cell will lose effective exchange capacity.
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