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Mila [183]
3 years ago
14

If you add a neutron to an atom how does it change ?

Chemistry
1 answer:
Kobotan [32]3 years ago
7 0

Answer:

Its atomic mass increases by 1

An isotope of that element is fotmed with mass differences by 1

Explanation:

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Determine the vapor pressure (atm) of rubbing alcohol (isopropanol) at 20.0 °C. The normal boiling point of isopropanol is 82.3
jeyben [28]
Here we apply the Clausius-Clapeyron equation:
ln(P₁/P₂) = ΔH/R x (1/T₂ - 1/T₁)

The normal vapor pressure is 4.24 kPa (P₁)
The boiling point at this pressure is 293 K (P₂)
The heat of vaporization is 39.9 kJ/mol (ΔH)
We need to find the vapor pressure (P₂) at the given temperature 355.3 K (T₂)

ln(4.24/P₂) = 39.9/0.008314 x (1/355.3 - 1/293)
P₂ = 101.2 kPa
8 0
3 years ago
a certain country, pivate enterprises provide various goods and services But the govemment provides tree public transportation a
puteri [66]
Mixed economy. Wherever the private and public sector have an active role, they are determined to be a mixed economy.
8 0
3 years ago
Which statement describes an open circuit? The circuit is incomplete and broken, and the lights in the circuit shine. The circui
SVEN [57.7K]

Answer:

the third option

Explanation:

The circuit is incomplete and broken, and the lights in the circuit do not shine.

An open circuit means an incomplete circuit.

The first option is wrong since the lights do not shine in an open circuit.

The second option is wrong since an open circuit is incomplete.

The fourth option is also wrong since an open circuit is broken or incomplete.

∴ The third option is correct.

8 0
3 years ago
Read 2 more answers
131i has a half-life of 8.04 days. assuming you start with a 1.53 mg sample of 131i, how many mg will remain after 13.0 days ___
inn [45]
For this problem we can use half-life formula and radioactive decay formula.

Half-life formula,
t1/2 = ln 2 / λ

where, t1/2 is half-life and λ is radioactive decay constant.
t1/2 = 8.04 days

Hence,         
8.04 days    = ln 2 / λ                         
λ   = ln 2 / 8.04 days

Radioactive decay law,
Nt = No e∧(-λt)

where, Nt is amount of compound at t time, No is amount of compound at  t = 0 time, t is time taken to decay and λ is radioactive decay constant.

Nt = ?
No = 1.53 mg
λ   = ln 2 / 8.04 days = 0.693 / 8.04 days
t    = 13.0 days 

By substituting,
Nt = 1.53 mg e∧((-0.693/8.04 days) x 13.0 days))
Nt = 0.4989 mg = 0.0.499 mg

Hence, mass of remaining sample after 13.0 days = 0.499 mg

The answer is "e"

8 0
3 years ago
A 34.4 L sample of oxygen gas at 229°C and 752 torr is cooled to 34°C at 668 torr. The volume of the sample is now
kenny6666 [7]

Answer:

23.55 L

Explanation:

USe the following 'identity' of gs laws

P1 V 1 / T1 = P2 V2 / T2         ( T must be in Kelvin)

re arrange to

P1 V 1  T2  /  (T1 P2)   = V2       NOW SUB IN THE VALUES

752 * 34.2 * ( 34 + 273.15) / [( 229 + 273.15) * 668]   = V2 = 23.55 L

5 0
1 year ago
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