The hydrogen Ion concentration of solution B is
1.0 x 10^-5 or 0.000 010 M
You can see that this will be proportional to the amount of B's PH compared to A's
hope this helps
Answer:
On edge nuity, it is 79.
Explanation:
Please read the explanation.
Since u have no coefficients, this problem is super easy.
You just add your products, 115.5+69.91 = 185.41
Then add your reactants: 49.8+56.5= 106.3
Then subtract your reactants from your product and round to the nearest whole number.
106.3-185.41=79
It's really easy. Once u understand it, you can do any problem like this. If there is coefficients, just multiply the coefficient by the amount given to you in the problem.
A polar molecule<span> has a net dipole as a result of the opposing charges (i.e. having partial positive and partial negative charges) from </span>polar<span> bonds arranged asymmetrically. Water (H</span>2<span>O) is an example of a </span>polar molecule<span> since it has a slight positive charge on one side and a slight negative charge on the other.</span>
167 mL
P1V1 = P2V2
P1 = .8 atm
V1 = 250 mL
P2 = 1.2 atm
Solve for V2 —> V2 = P1V1/P2
V2 = (0.8 atm)(250 mL) / (1.2 atm) = 167 mL
Na₂S(aq) + Cd(NO₃)₂(aq) = CdS(s) + 2NaNO₃(aq)
v=25.00 mL
c=0.0100 mmol/mL
M(Na₂S)=78.046 mg/mmol
n(Na₂S)=n{Cd(NO₃)₂}=cv
m(Na₂S)=M(Na₂S)n(Na₂S)=M(Na₂S)cv
m(Na₂S)=78.046*0.0100*25.00≈19.5 mg
