Balance Chemical Equation,
2 CO + O₂ → 2CO₂
Acc. to this reaction,
88 g (2 mole) of CO₂ was produced when = 56 g (2 mole)of CO was reacted
So,
24.7 g of CO₂ will be produced by reacting = X g of CO
Solving for X,
X = (56 g × 24.7 g) ÷ 88 g
X = 2.26 g ÷ 88 g
X = 0.0257 g of CO
Result:
0.0257 g of CO is required to be reacted with excess of O₂ to produce 24.7 g of CO₂.
The group of unsaturated hydrocarbons which 2 carbons are double bonded together, with H bonded to the left, and C H 2 bonded below left, above right, and below right is derived from <u>Alkenes</u>
<h3>What are organic compounds?</h3>
Organic compounds are compounds which contains carbon and hydrogen
Some few classes or organic compounds or hydrocarbons are as follows:
- Alkanes
- Alkenes
- Alkynes
- Alkanols
- Alkanoic acid
- Ketones
- Esters
So therefore, the group of unsaturated hydrocarbons which 2 carbons are double bonded together, with H bonded to the left, and C H 2 bonded below left, above right, and below right is derived from <u>Alkenes</u>
Learn more about organic compounds:
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Answer:
Explanation:
rate of reaction
= -ve change in pressure of ethanol / time
= - (250 -237 )/100 = - 13 / 100 torr/s
= - 0.13 torr/s
next
- (237 - 224 )/100 = - 13 / 100 torr/s
= - .13 torr/s
next
- (224 - 211 )/100 = - 13 / 100 torr/s
= - .13 torr/s
so on
So rate of reaction is constant and it does not depend upon concentration or pressure of reactant .
So order of reaction is zero.
rate of reaction =K [C₂H₅OH]⁰
K is rate constant
K = .13 torr/s
In 900 s decrease in pressure
= 900 x .13 = 117
So after 900s , pressure of ethanol will be
250 - 117 = 133 torr
Answer:
Molarity = 0.4M
Explanation:
Molar mass of NaOH (M)= 40
m= 8g, V= 500ml=0.5L
n= m/M=[8/40]= 0.2mol
Applying
n= CV
0.2= C×0.5
C= 0.4M
Answer:
Average atomic mass = 85.557 amu.
Explanation:
Given data:
Percent abundance of Rb-85 = 72.15%
Percent abundance of Rb-87 = 27.85%
Average atomic mass = ?
Solution:
Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) / 100
Average atomic mass = (72.15×85)+(27.85×87) /100
Average atomic mass = 6132.75 + 2422.95 / 100
Average atomic mass = 8555.7 / 100
Average atomic mass = 85.557 amu.
