You would expect Na to have the larger ionic radius
Answer:
Empirical formula is C₃H₃O
Explanation:
Given data:
Mass of ethyl butyrate = 1.95 mg
Mass of CO₂ = 4.42 mg
Mass of water = 1.81 mg
Empirical formula = ?
Solution:
Percentage of C = 4.42/1.95 × 12/44×100
= 2.27× 0.273 ×100
= 62
percentage of H = 1.81/1.95 × 2/18 × 100
= 0.93 × 0.11 ×100
= 10.23
percentage of oxygen = 100 - (72.23)
= 27.77
number of gram atoms of C = 62/12 =5.17
number of gram atoms of H = 10.23 / 2 =5.115
number of gram atoms of O = 27.77 / 16 =1.74
Atomic ratio:
C : H : O
5.17/1.74 : 5.115/1.74 : 1.74/1.74
3 : 3 : 1
Empirical formula is C₃H₃O
Answer:
a. The rate of the reaction is not proportional to the concentration of the reactant.
Explanation:
The rate expression for a zero order reaction is given as;
A → Product
Rate = k[A]⁰
[A]⁰ = 1
Rate = K
GGoing through the options;
a) This is correct because in the final form of the rate expression, the rate is independent of the concentration.
b) This option is wrong
c) This option is also wrong
d) Like options b and c this is also wrong becaus ethere is no relationship between either the concentration or t.
<span>You can work out the following:
Mass = Density x Volume
Density = Mass Ă· Volume
Volume = Mass Ă· Density
The density of gold is 0.670205 oz/cm^3
The volume is length x width x depth = 7*3*2 = 42
The answer is Mass = 0.670205 x 42 = 281 ounces</span>
Answer:
The pressure of N₂O₄ in the reaction vessel after the reaction is 290 mmHg
Explanation:
Nitrogen gas reacts with oxygen gas to form dinitrogen tetroxide.
N₂ (g) + 2O₂ (g) → N₂O₄ (g)
Therefore since by Avogadro's law equal volumes of all gases contain equal numbers of molecules, there fore as the gases are within the same vessel, thier partial pressure is equivalent to their concentration
from the reaction, 1 mole of N₂ react with 2 moles of O₂ to produce 1 mole of N₂O₄
Thus
1 mmHg of N₂ react with 2 mmHg of O₂ to produce 1 mmHg of N₂O₄
337 mmHg N₂ ×(1 mmHg of N₂O₄/ 1 mmHg of N₂) = 337 mmHg N₂O₄
580 mmHg O₂ ×(1 mmHg of N₂O₄/ 2 mmHg of O₂) = 290 mmHg N₂O₄
As seen from the above calculation, the limting reactant is oxygen and the partial pressure of N₂O₄ = 290 mmHg