Answer:
C
Explanation:
the enthalpy of reaction is independent of the reaction path
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:
At the top of Group 11 above silver and gold.
Period 4
Explanation:
2Ca + O2 = 2CaO
First, determine which is the excess reactant
72.5 g Ca (1 mol) =1.8089725036
(40.078 g)
65 g O2 (1 mol) =2.0313769611
(15.999g × 2)
Since the ratio of to O2 is 2:1 in the balanced reaction, divide Ca's molar mass by 2 to get 0.9044862518. this isn't necessary because Ca is already obviously the limiting reactant. therefore, O2 is the excess reactant.
Now do the stoichiometry
72.5 g Ca (1 mol Ca) (1 mol O2)
(40.078 g Ca)(2 mol Ca)(31.998g O2)
=0.0282669621 g of O2 left over
a. 381.27 m/s
b. the rate of effusion of sulfur dioxide = 2.5 faster than nitrogen triiodide
<h3>Further explanation</h3>
Given
T = 100 + 273 = 373 K
Required
a. the gas speedi
b. The rate of effusion comparison
Solution
a.
Average velocities of gases can be expressed as root-mean-square averages. (V rms)

R = gas constant, T = temperature, Mm = molar mass of the gas particles
From the question
R = 8,314 J / mol K
T = temperature
Mm = molar mass, kg / mol
Molar mass of Sulfur dioxide = 64 g/mol = 0.064 kg/mol

b. the effusion rates of two gases = the square root of the inverse of their molar masses:

M₁ = molar mass sulfur dioxide = 64
M₂ = molar mass nitrogen triodide = 395

the rate of effusion of sulfur dioxide = 2.5 faster than nitrogen triodide