Answer:
The answer to your question is: The mass number will be 4 units lower.
Explanation:
Alpha particles are Helium atoms, which have a mass number of 4 and atomic number of 2.
When an alpha particle is released, the original atom loses 2 protons and and 2 neutrons an we can see in the example.
²²⁶ ₈₈ Ra ⇒ ²²² ₈₆ Rn + ⁴₂ He
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
Answer:
B = basic
Explanation:
Given data:
[OH⁻] = 5.35×10⁻⁴M
pH = ?
Solution:
pOH = -log[OH⁻]
pOH = - [5.35×10⁻⁴]
pOH = 3.272
it is known that,
pH + pOH = 14
pH = 14- pOH
pH = 14 - 3.272
pH = 10.728
The acidic pH is range from zero to less than 7 while 7 pH is neutral and above 7 the pH is basic. So, the given solution is basic.
Cl2(g) -------> Cl-(aq) + ClO-(aq)
2e- + Cl2(g) -------> 2Cl-(aq) [reduction]
4OH-(aq) + Cl2(g) -----------> 2ClO-(aq) + 2H2O(l) + 2e- [oxidation]
______________________________________...
2OH-(aq) + Cl2(g) --------> Cl-(aq) + ClO-(aq) + H2O(l)
<span>the atomic mass of nitrogen is 14. There is 1 nitrogen atom in the molecule so the percentage of N is :
14/35 x100% = 40%</span>
