Answer:
C-12 or C with a 12 superscripted on Upper left and 6 Subscripted on bottom left
Explanation:
Isotopic notation
Answer:
The answer to your question is 1.11 M
Explanation:
Data
volume 1 = 287 ml
concentration 1 = 1.6 M
volume 2= 412 ml
concentration 2 = ?
Formula
Volume 1 x concentration 1 = Volume 2 x concentration 2
Solve for concentration 2
concentration 2 = (volume 1 x concentration 1) / volume 2
Substitution
concentration 2 = (287 x 1.6) / 412
Simplification
concentration 2 = 459.2 / 412
Result
concentration 2 = 1.11 M
Answer:
The answers are explained below
Explanation:
a)
Given: concentration of salt/base = 0.031
concentration of acid = 0.050
we have
PH = PK a + log[salt]/[acid] = 1.8 + log(0.031/0.050) = 1.59
b)
we have HSO₃⁻ + OH⁻ ------> SO₃²⁻ + H₂O
Moles i............0.05...................0.01.................0.031.....................0
Moles r...........-0.01.................-0.01................0.01........................0.01
moles f...........0.04....................0....................0.041.....................0.01
c)
we will use the first equation but substituting concentration of base as 0.031 + 10ml = 0.031 + 0.010 = 0.041
Hence, we have
PH = PK a + log[salt]/[acid] = 1.8 + log(0.041/0.050) = 1.71
d)
pOH = -log (0.01/0.510) = 1.71
pH = 14 - 1.71 = 12.29
e)
Because the buffer solution (NaHSO3-Na2SO3) can regulate pH changes. when a buffer is added to water, the first change that occurs is that the water pH becomes constant. Thus, acids or bases (alkali = bases) Additional may not have any effect on the water, as this always will stabilize immediately.
Answer:
Si is the answer I hope this help
Answer: the electrostatic potential energy is -3.26 × 10⁻¹⁸ J
Explanation:
Given the data in the question;
I think the value of electrostatic potential energy of calcium sulfide CaS formed will be less(more negative) than KCl because the Ca2+ and S2- ions has more charge that is +2 and -2 respectively when compared to Kcl which has +1 and -1 charge.so it will be more negative.
radius = r
+ r
= 100 + 184 = 284 pm = 2.84 × 10⁻¹⁰ m
we know that; k = 2.31 × 10⁻²⁸ J
q1 = 2 ( charge on Ca⁺² )
q2 = -2( charge on S⁻²)
so
Ep = k × q1q2/r
so we substitute
Ep = 2.31 × 10⁻²⁸ × (+2×-2) / 2.84 × 10⁻¹⁰
Ep = (2.31 × 10⁻²⁸ × -4) / 2.84 × 10⁻¹⁰
Ep = -9.24 × 10⁻²⁸ / 2.84 × 10⁻¹⁰
Ep = -3.26 × 10⁻¹⁸ J
Therefore, the electrostatic potential energy is -3.26 × 10⁻¹⁸ J
