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OLEGan [10]
3 years ago
9

1 What is the approximate mass of a proton? (1)1 u (3) l g (2) 0.0005 u (4) 0.0005 g

Chemistry
1 answer:
joja [24]3 years ago
8 0
Proton mass = 1.672 x 10^-27 kilo. So the approximate mass of a proton would be 1 g or kg.
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Calculate the percentage difference in the fundamental vibrational wavenumbers of 23Na35Cl and 23Na37Cl on the assumption that t
stealth61 [152]

Answer:

1.089%

Explanation:

From;

ν =1/2πc(k/meff)^1/2

Where;

ν = wave number

meff = reduced mass or effective mass

k = force constant

c= speed of light

Let

ν =1/2πc (k/meff)^1/2  vibrational wave number for 23Na35 Cl

ν' =1/2πc(k'/m'eff)^1/2 vibrational wave number for 23Na37 Cl

The between the two is obtained from;

ν' - ν /ν  = (k'/m'eff)^1/2 - (k/meff)^1/2 / (k/meff)^1/2

Therefore;

ν' - ν /ν = [meff/m'eff]^1/2 - 1

Substituting values, we have;

ν' - ν /ν = [(22.9898 * 34.9688/22.9898 + 34.9688) * (22.9898 + 36.9651/22.9898 * 36.9651)]^1/2  -1

ν' - ν /ν = -0.01089

percentage difference in the fundamental vibrational wavenumbers of 23Na35Cl and 23Na37Cl;

ν' - ν /ν * 100

|(-0.01089)|  × 100 = 1.089%

4 0
3 years ago
Why is subatomic particles not made up of atoms
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Determine the heat energy needed to raise the temperature of 120 grams of ice at -5 to steam at 115°
CaHeK987 [17]

Answer:

Q = 30355.2 J

Explanation:

Given data:

Mass of ice = 120 g

Initial temperature = -5°C

Final temperature = 115°C

Energy required = ?

Solution:

Specific heat capacity of ice is = 2.108 j/g.°C

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

Q = m.c. ΔT

ΔT = T2 -T1

ΔT = 115 - (-5°C)

ΔT = 120 °C

Q = 120 g × 2.108 j/g.°C × 120 °C

Q = 30355.2 J

5 0
3 years ago
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