I would say the answer is C
hope this helps :)
Thermoplastics and thermosetting polymers Examples include: polyethylene (PS) and polyvinyl choline (PVC). Common thermoplastics range from 20,000 to 50,000 amu, while thermosets are assumed to have infinite molecular weight.
Answer: E = 2.455 x 10^5 N/C
Explanation:
q1 = 1.2x10^-7C
q2 = 6.2x10^-8C
Electric field, E = kQ/r²
where k = 9.0x10^9
since the location is (27 - 5)cm from q1
hence electric field, E1 = k*q1/r²
E1= (9x10^9 x 1.2x10^-7)/(0.22)² = 22314.05 N/C
for q2:
E1 = k*q2/r²
E2 at 5cm
E2 = (9x10^9 x 6.2x10^-8)/(0.05)² = 223200 N/C
Hence, the total electric field at 5cm position is
E = E1 + E2
E = 22314.05 + 223200 = 245514.05 N/C
E = 2.455 x 10^5 N/C
Answer:
Explanation:
Hello,
In this case, we can consider the following chemical reaction for the oxidation of copper which only occurs at high temperatures:
In such a way, for 0.032 grams of copper, the following grams of copper (II) oxide (black product) are yielded:
Therefore, the percent yield is:
Best regards.
Your protons are correct but it’s 28 neutrons not 27!
