Answer: B or C
Explanation: The question does not include the variable or steps Brian is using so either one could be correct. It has to be the one that he is controlling though. This is because a control group is used to rule out any alternate explanations. Therefore the answer should be the one that he is trying to test out.
The answer is 6.022• 10^23 atoms
Answer:
8.279
Explanation:
The pH can be determined by hydrolysis of a conjugate base of weak acid at the equivalence point.
At the equivalence point, we have
= 25.00 x 0.200
= 5.00 m-mol
= 0.005 mol
Volume of the base that is added to reach the equivalence point is
Number of moles of
= 0.005 mol
Volume at the equivalence point is 25 + 5 = 30.00 mL
Therefore, concentration of
= 0.167 M
Now the ICE table :
I (M) 0.167 0 0
C (M) -x +x +x
E (M) 0.167-x x x
Now, the value of the base dissociation constant is ,
=
Base ionization constant,
So,
pOH =-
=
=5.72
Now, since pH + pOH = 14
pH = 14.00 - 5.72
= 8.279
Therefore the ph is 8.279 at the end of the titration.
It is going to be 11% of 100
11/100*100=11gram
Hey there!:
moles of AgNO3 = 100 x 0.1 / 1000 = 0.01
moles of NaCl = 100 x 0.200 / 1000 = 0.02
volume of solution = 100 + 100 = 200 mL
mass of solution = 200 x 1 = 200 g
temperature rise = 25.30 - 24.60 = 0.70ºC
Therefore:
Q = m* Cp* ΔT + Cp* ΔT =
200* 4.184 * 0.70 + 15.5 * 0.70 =
Q = 596.61 J
Given the reaction:
Q = 596.61 J
NaCl(aq) + AgNO3(aq)--------> AgCl(s) + NaNO3(aq)
1 mole NaCl ------ 1 mole AgNO3 ------ 1 mole AgCl
0.02 moles NaCl ----- 0.01 moles AgNO3
So:
ΔH = - Q / T
ΔH = - 596.61 * 10⁻³ / 0.01
ΔH = - 59.66 kJ
Hope that helps!