Answer:
3.2 x 10²¹molecules
Explanation:
Given parameters:
Volume of nitrogen gas = 120cm³
Unknown:
Mass of nitrogen gas = ?
Number of molecules = ?
Solution:
To solve this problem, note that;
1 mole of gas occupies a volume of 22.4L at STP
Now;
convert 120cm³ to L ;
1000cm³ = 1L
120cm³ gives 0.12L
Since;
22.4L of gas has 1 mole at STP
0.12L of gas will have 
= 0.0054mole at STP
So;
Mass of N₂ = number of moles x molar mass
Molar mass of N₂ = 2(14) = 28g/mol
Mass of N₂ = 0.0054 x 28 = 0.15g
Now;
1 mole of a gas will have 6.02 x 10²³ molecules
0.0054 mole of N₂ will contain 0.0054 x 6.02 x 10²³ =
3.2 x 10²¹molecules
Answer:
your answer should be B ANDC
<u>Answer:</u> The volume of NaOH solution required to reach the half-equivalence point is 0.09 mL
<u>Explanation:</u>
The chemical equation for the dissociation of butanoic acid follows:
The expression of 
for above equation follows:
![K_a=\frac{[CH_3CH_2CH_2COO^-][H^+]}{[CH_3CH_2CH_2COOH]}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7B%5BCH_3CH_2CH_2COO%5E-%5D%5BH%5E%2B%5D%7D%7B%5BCH_3CH_2CH_2COOH%5D%7D)
We are given:
![[CH_3CH_2CH_2COOH]=0.888M\\K_a=1.54\times 10^{-5}](https://tex.z-dn.net/?f=%5BCH_3CH_2CH_2COOH%5D%3D0.888M%5C%5CK_a%3D1.54%5Ctimes%2010%5E%7B-5%7D)
![[CH_3CH_2CH_2COO^-]=[H^+]](https://tex.z-dn.net/?f=%5BCH_3CH_2CH_2COO%5E-%5D%3D%5BH%5E%2B%5D)
Putting values in above expression, we get:
![1.54\times 10^{-5}=\frac{[H^+]^2}{0.888}](https://tex.z-dn.net/?f=1.54%5Ctimes%2010%5E%7B-5%7D%3D%5Cfrac%7B%5BH%5E%2B%5D%5E2%7D%7B0.888%7D)
![[H^+]=-0.0037,0.0037](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D-0.0037%2C0.0037)
Neglecting the negative value because concentration cannot be negative
To calculate the volume of base, we use the equation given by neutralization reaction:
where,
are the n-factor, molarity and volume of acid which is butanoic acid
are the n-factor, molarity and volume of base which is NaOH.
We are given:
Putting values in above equation, we get:
Hence, the volume of NaOH solution required to reach the half-equivalence point is 0.09 mL
Answer:
P.E = 493920 j
Explanation:
Given data:
Mass = 1200 kg
height = 42 m
Potential energy = ?
Solution:
Formula:
<em>P.E = mgh</em>
Now we will put the values in formula.
g = 9.8 m/s²
P.E = 1200 Kg × 9.8 m/s²× 42 m
P.E = 493920 Kg.m²/s²
Kg.m²/s² = j
P.E = 493920 j
