The molarity of KBr solution is 1.556 M
molarity is defined as the number of moles of solute in volume of 1 L solution.
the number of KBr moles in 1 L - 1.556 mol
Therefore in 200.0 L - 1.556 mol/L x 200.0 L = 311.2 mol
Molar mass of KBr - 119 g/mol
mass of Kbr - 311.2 mol x 119 g/mol = 37 033 g
mass of solute therefore is 37.033 kg
The characteristics of an Arrhenius acid was the acid donates H+ when it is in the aqueous solution, whether the solution is acid or base. If the solution is a acid, then be H+ builds up and increase its thing. The pH level for acid solution is the scale from 1-6. Even though the ph scale is lower than the base, the color also varies. It also applies to the base, too.
<u>Answer:</u> The value of
for the given reaction is 1.435
<u>Explanation:</u>
To calculate the molarity of solution, we use the equation:

Given mass of
= 9.2 g
Molar mass of
= 92 g/mol
Volume of solution = 0.50 L
Putting values in above equation, we get:

For the given chemical equation:

<u>Initial:</u> 0.20
<u>At eqllm:</u> 0.20-x 2x
We are given:
Equilibrium concentration of
= 0.057
Evaluating the value of 'x'

The expression of
for above equation follows:
![K_c=\frac{[NO_2]^2}{[N_2O_4]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BNO_2%5D%5E2%7D%7B%5BN_2O_4%5D%7D)
![[NO_2]_{eq}=2x=(2\times 0.143)=0.286M](https://tex.z-dn.net/?f=%5BNO_2%5D_%7Beq%7D%3D2x%3D%282%5Ctimes%200.143%29%3D0.286M)
![[N_2O_4]_{eq}=0.057M](https://tex.z-dn.net/?f=%5BN_2O_4%5D_%7Beq%7D%3D0.057M)
Putting values in above expression, we get:

Hence, the value of
for the given reaction is 1.435
Answer:
D. The higher energy chair conformation contains two axial methyl groups.
Explanation:
The diaxial trans-1,4-dimethylcyclohexane has the highest energy (the least stable) and the diequatorial chair conformation has the lowest energy (the most stable).
The diaxial conformation has a higher energy due to the interaction between the methyl groups.
Answer:
0.1M solution of NaOH
Explanation:
1 mole of NaOH - 40g
? moles - 1 g = 1/40 = 0.025 moles.
Molarity of 1.00g of NaOH in 0.25L (250 mL) = no. of moles/volume
= 0.025/0.25
= 0.1M.