Proton:
Positive
Found in Nucleus
Mass of 1 AMU
Neutron:
Neutral
Found in Nucleus
Mass of 1 AMU
Electron:
Negative
Found in orbitals
Mass of 0 AMU
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Given:
P1 = 13.0 atm
T1 = 20 °C
T2 = 102 °C
Required:
P2 of oxygen
Solution:
At constant volume,
we can apply Gay-Lussac’s law of pressure and temperature relationship
P1/T1=P2/T2
(13.0 atm) / (20 °C)
= P2 / (102 °C)
P2 = 66.3 atm
The answer is not in the choices given.
A the collisions between water particles and polllen grains
<span> Au</span>₂(SeO₄)₃
O = -2 × 4 = -8
Se = + 6
So,
(+6 - 8) = -2
Means (SeO₄) contains -2 charge, Now multiply -2 by 3
-2 ₓ 3 = -6
Means,
Au₂ + (-6) = 0
Au₂ = +6
Or,
Au = 6 / 2
Au = +3
Result:
Au = +3
Se = +6
O = -2
Ni(CN)₂
Cyanide (CN⁻) contains -1 charge,
So,
N = -3
C = +2
Then,
Ni + (-1)₂ = 0
Ni - 2 = 0
Or,
Ni = +2
Result:
N = -3
C = +2
Ni = +2
Answer: 207.2
Explanation:
In imprecise terms, one AMU is the average of the proton rest mass and the neutron rest mass. This is approximately 1.67377 x 10 -27 kilogram (kg), or 1.67377 x 10 -24 gram (g). The mass of an atom in AMU is roughly equal to the sum of the number of protons and neutrons in the nucleus.
