F=kq'q"/d²
charge of electron(q')= 1.6022*10^-19
charge of proton (q")= -1.6022*10^-19
Coulomb's constant (k)= 9*10^9
distance (d)=10^-10
substituting values into equation
F= (9*10^9)*(1.6022*10^-19)(-1.6022*10^-19)/(10^10)²
F= -2.31034036 × 10^-48N
Answer:
318 kPa
Explanation:
Step 1: Given data
- Initial volume (V₁): 0.375 L
- Final pressure (P₂): 95.5 kPa
- Final volume (V₂): 1.25 L
Step 2: Calculate the initial pressure of the gas
Assuming constant temperature and ideal behavior, we can calculate the initial pressure of the gas using Boyle's law.
P₁ × V₁ = P₂ × V₂
P₁ = P₂ × V₂ / V₁
P₁ = 95.5 kPa × 1.25 L / 0.375 L = 318 kPa
Answer:
= 1.5 eq
Explanation:
One definition of an equivalent weight is that it is mass of a substance that gains or loses 1 mole of electrons.
Al3+ has lost 3 e-, so there are 3 equivalent weights in 1 mol Al3+.
1 mol Al3+ =3 eq. wts.
1 mol Al x(27 g / 1 mol)x(1 mol / 3 eq. wts.) = 9.0 g = 1 eq. wts.
13.5 g Al3 + x (1 eq.wt. / 9.0 g) = 1.5 eq
Answer:
73.0g of HCl
Explanation:
Check the attachment below for explanation.
Answer:
that the second one surely represent the beta particle