Answer:
1.427x10^-3mol per L
Explanation:
I could use ⇌ in the math editor so I used ----
from the question each mole of Y(IO3)3 is dissolved and this is giving us a mole of Y3+ and a mole of IO3^3-
Ksp = [Y^3+][IO3-]^3
So that,
1.12x10^-10 = [S][3S]^3
such that
1.12x10^-10 = 27S^4
the value of s is 0.001427mol per L
= 1.427x10^-3mol per L
so in conclusion
the molar solubility is therefore 1.427x10^-3mol per L
Follow
these steps to solve the given equation:
Multiply
the two decimal figures together and find the sum of the exponents, that is,
(1.5
* 1.89) * 10 ^4+3
(2.835)
* 10^7
10^7
can also be written as e.70
'e'
stands for exponential.
Therefore,
we have 2. 835 e 7.0 = 2.8 e 7.0.
Based on the calculations above, the correct option is A.
Answer:
q1 = mCpΔT
= 18.016g × 1.84J/g.K × (418.15-373.15)
= 1491.72 J
q2 = n×ΔH vap
= 1mol ×44.0kJ/mol
= 44KJ
∴ qtotal = q1+ q2
= 1.498kJ + 44.0kJ
= 45.498KJ
Explanation: The heat flow can be separated into steps.all that is being observed at a constant pressure,the heat flow is equal to the enthalpy.
Answer:
NaOBr (or) Na⁺ ⁻OBr
Explanation:
The Oxo-Acids of Bromine are as follow,
Hypobromous Acid = HOBr
Bromous Acid = HOBrO
Bromic Acid = HBrO₃
Perbromic Acid = HBrO₄
When these acids are converted to their conjugate bases their names are as follow,
Hypobromite = ⁻OBr
Bromite = ⁻OBrO
Bromate = ⁻OBrO₂
Perbromate = ⁻OBrO₃
According to rules, the positive part of ionic compound is named first and the negative part is named second. So, Sodium Hypobromite has a chemical formula of Na⁺ ⁻OBr or NaOBr.
Hot liquid rock comes in contact with Earth´s cold crust
