Answer:
The whole molecule is polar because Sulfur has lone pairs but Carbon doesn't. Lone pairs count more toward polarity, shifting dipole toward S.
Explanation:
Even though carbon and sulfur have identical values of electronegativities, the molecule, 
is polar because of the presence of the lone pairs on the sulfur atom.
The C-S bond is not polar because the both the atoms have electronegatiivty. <u>But S has lone pairs which can attract the bond pairs of the bond between the S and H and thus acquires slightly negative charge and H acquires slightly positive charge.</u>
Answer:
The percent isotopic abundance of Ir-193 is 60.85 %
The percent isotopic abundance of Ir-191 is 39.15 %
Explanation:
we know there are two naturally occurring isotopes of iridium, Ir-191 and Ir-193
First of all we will set the fraction for both isotopes
X for the isotopes having mass 193
1-x for isotopes having mass 191
The average atomic mass is 192.217
we will use the following equation,
193x + 191(1-x) = 192.217
193x + 191 - 191x = 192.217
193x- 191x = 192.217 - 191
2x = 1.217
x= 1.217/2
x= 0.6085
0.6085 × 100 = 60.85 %
60.85% is abundance of Ir-193 because we solve the fraction x.
now we will calculate the abundance of Ir-191.
(1-x)
1-0.6085 =0.3915
0.3915× 100= 39.15 %
The equation for this question could be 
→
.
so for 6.75 moles of 
*3 moles of O_{2}/2 moles of
= 10.125
Answer:
N=3.05*10^-3g
Explanation:
Using Ln(No/N)=0.693/t½*t
Ln(No/N)=0.693/3*45
No/N=exp(10.396)
100/exp(10.396)=N
100/32695.7=N
N=3.05*10^-3g
Answer:
B. CH3COOH pH > 4.7 (4.8)
Explanation:
- CH3COOH + NaOH ↔ CH3COONa + H2O
- CH3COONa + NaOH ↔ CH3COONa
∴ mol NaOH = (5 E-3 L)*(0.10 mol/L) = 5 E-4 mol
⇒ mol CH3COOH = (0.05 L)*(0.20 mol/L) = 0.01 mol
⇒ <em>C</em> CH3COOH = (0.01 mol - 5 E-4 mol) / (0.105 L)
⇒ <em>C</em> CH3COOH = 0.0905 M
∴ mol CH3COONa = (0.05 L )*(0.20 mol/L) = 0.01 mol
⇒ <em>C</em> CH3COONa = (0.01 mol + 5 E-4 mol) / (0.105 L )
⇒ <em>C</em> CH3COONa = 0.1 M
∴ Ka = ([H3O+]*(0.1 + [H3O+])) / (0.0905 - [H3O+]) = 1.75 E-5
⇒ 0.1[H3O+] + [H3O+]² = (1.75 E-5)*(0.0905 - [H3O+])
⇒ [H3O+]² 0.1[H3O+] = 1.584 E-6 - 1.75 E-5[H3O+]
⇒ [H3O+]² + 0.1000175[H3O+] - 1.584 E-6 = 0
⇒ [H3O+] = 1.5835 E-5 M
∴ pH = - Log [H3O+]
⇒ pH = - Log (1.5835 E-5)
⇒ pH = 4.8004 > 4.7
