Answer:
C: object does not slide off the pan
Explanation:
Answer:
I think it's A or D im not sure which one..
<u>Solu</u><u>tion</u><u> </u><u>:</u><u>-</u>
First of all for comparing any two quantities , they must be in same units .
Here one unit is in grams and other in milligrams .
So let's convert gram to miligram .
We know that 1 gram = 1000 mg.
So , 0.7g = 0.7 × 1000 mg = 700mg .
Clearly 711mg > 700mg .
Hence 711mg is Greater than 0.7 g .
Answer:
The order of the reaction with respect to the gas = 2
Explanation:
Let the original gas pressure be [G₀]
Initial rate of reaction is given as
r = k [G₀]ⁿ
When 10% had reacted, amount of gas left = [0.9G₀], r = 9.71 Pa/s
r = k [0.9G₀]ⁿ = 9.71 (eqn 1)
when 20% had reacted, amount of gas left = [0.8G₀], r = 7.67 Pa/s
r = k [0.8G₀]ⁿ = 7.67 (eqn 2)
Dividing (eqn 1) by (eqn 2)
(9.71/7.67) = [0.9/0.8]ⁿ
1.266 = 1.125ⁿ
1.125ⁿ = 1.266
Take natural logarithms of both sides
n (In 1.125) = In 1.266
n = 0.236/0.118
n = 2.
Answer:
The bond order for C2 molecule is 2.
Explanation:
Bond order can be defined as the half of the difference between the number of electrons in the bonding orbital and the number of electrons in the antibonding orbitals. It can be represented mathematically by; .
Bond order,n= [number of electrons in the bonding molecular orbitals(BMO) - the number or electrons in the anti-bonding molecular orbitals(AMO) ] / 2.
The electronic configuration of the C2 molecule is given below;
C2 = (1sσ)^2 (1s^*σ)^2 (2sσ)^2 (2s^*σ)^2 (2pπ)^4.
The ones with the (*) are known as the Anti-bonding molecular orbitals while the ones without (*) are known as the bonding molecular orbitals. Hence, we have 8 Electrons from the bonding molecular orbitals and 4 Electrons from the anti-bonding molecular orbitals.
So, from the formula given above, the bond order of C2 molecule is;
===> 8-4/2= 4/2.
===> 2.
