Answer:
Empirical formula is Cr₂O₃.
Explanation:
Given data:
Percentage of Cr = 68.4%
Percentage of O = 31.6%
Empirical formula = ?
Solution:
Number of gram atoms of Cr = 68.4 / 52 = 1.3
2
Number of gram atoms of O = 31.6 / 16 = 1.98
Atomic ratio:
Cr : O
1.32/1.32 : 1.98/1.32
1 : 1.5
Cr : O = 1 : 1.5
Cr : O = 2(1 : 1.5)
Empirical formula is Cr₂O₃.
A.) Phosphate ion or Orthophosphate
d.) Hydroxide
D.) Ammonium
e.) Iron
C.) Nitrate
f.) Sulfur dioxide
Answer:
A. is insufficient to overcome intermolecular forces
Explanation:
Just took the review
Answer:
HClO (l) → H⁺ (aq) + ClO⁻ (aq)
proton and hypochlorite.
Explanation:
HClO (Hypochlorous acid)
This is a weak acid that can be dissociated as this:
HClO (l) → H⁺ (aq) + ClO⁻ (aq)
proton and hypochlorite.
It is a weak acid, so it can adopt a Ka for its equilibrium
HClO + H₂O ⇄ H₃O⁺ + ClO⁻ Ka
Molar mass = 52.46 g/m
Answer:
ΔG° of reaction = -47.3 x
J/mol
Explanation:
As we can see, we have been a particular reaction and Energy values as well.
ΔG° of reaction = -30.5 kJ/mol
Temperature = 37°C.
And we have to calculat the ΔG° of reaction in the biological cell which contains ATP, ADP and HPO4-2:
The first step is to calculate the equilibrium constant for the reaction:
Equilibrium Constant K = ![\frac{[HPO4-2] x [ADP]}{ATP}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BHPO4-2%5D%20x%20%5BADP%5D%7D%7BATP%7D)
And we have values given for these quantities in the biological cell:
[HP04-2] = 2.1 x
M
[ATP] = 1.2 x
M
[ADP] = 8.4 x
M
Let's plug in these values in the above equation for equilibrium constant:
K = ![\frac{[2.1x10^{-3}] x [8.4x10^{-3}] }{[1.2 x 10^{-2}] }](https://tex.z-dn.net/?f=%5Cfrac%7B%5B2.1x10%5E%7B-3%7D%5D%20x%20%5B8.4x10%5E%7B-3%7D%5D%20%7D%7B%5B1.2%20x%2010%5E%7B-2%7D%5D%20%7D)
K = 1.47 x
M
Now, we have to calculate the ΔG° of reaction for the biological cell:
But first we have to convert the temperature in Kelvin scale.
Temp = 37°C
Temp = 37 + 273
Temp = 310 K
ΔG° of reaction = (-30.5
) + (8.314)x (310K)xln(0.00147)
Where 8.314 = value of Gas Constant
ΔG° of reaction = (-30.5 x
) + (-16810.68)
ΔG° of reaction = -47.3 x
J/mol