The number of lone pairs that are most likely found on the central atom is zero. There are no lone pairs found on the central atom.
Answer:
Q14: 17,140 g = 17.14 kg.
Q16: 504 J.
Explanation:
<u><em>Q14:</em></u>
- To solve this problem, we can use the relation:
<em>Q = m.c.ΔT,</em>
where, Q is the amount of heat absorbed by ice (Q = 3600 x 10³ J).
m is the mass of the ice (m = ??? g).
c is the specific heat of the ice (c of ice = 2.1 J/g.°C).
ΔT is the difference between the initial and final temperature (ΔT = final T - initial T = 100.0°C - 0.0°C = 100.0°C).
∵ Q = m.c.ΔT
∴ (3600 x 10³ J) = m.(2.1 J/g.°C).(100.0°C)
∴ m = (3600 x 10³ J)/(2.1 J/g.°C).(100.0°C) = 17,140 g = 17.14 kg.
<u><em>Q16:</em></u>
- To solve this problem, we can use the relation:
<em>Q = m.c.ΔT,</em>
where, Q is the amount of heat absorbed by ice (Q = ??? J).
m is the mass of the ice (m = 12.0 g).
c is the specific heat of the ice (c of ice = 2.1 J/g.°C).
ΔT is the difference between the initial and final temperature (ΔT = final T - initial T = 0.0°C - (-20.0°C) = 20.0°C).
∴ Q = m.c.ΔT = (12.0 g)(2.1 J/g.°C)(20.0°C) = 504 J.
Answer:
15.5 gm
Explanation:
What is the mass of phosphorus that contains twice the number of atoms found in 14 g of iron?
[Relative atomic mass : P = 31; Fe = 56]
14 gm Fe = 14gm/ 56 gm/mole = 14 mole gm/56gm = 14/56 mole
0.25 moles
2 X 0.25 = 0.5 moles
1 mole P = 31 gm
so
0.5 moles P =31/2 =15.5 gm
Answer:
However, various hydrogen isotopes, such as H-2, have one proton and one neutron; H-3 has one proton and two neutrons, etc. The sum of the protons and neutrons in an atom's nucleus is its atomic mass. Thus, the atomic mass of the H-2 isotope is two, the atomic mass of the H-3 isotope is three, and so forth.
Explanation:
