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Lady bird [3.3K]
3 years ago
10

If an atom has 35 protons in the nucleus how many electrons will have orbiting the nucleus

Chemistry
1 answer:
Andru [333]3 years ago
4 0

35 electronics orbiting....

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If I initially have a gas at a pressure of 10.0 atm, a volume of 24.0 liters, and a temperature
Bezzdna [24]

Answer:

The new volume of gas is 25.7 L.

Explanation:

Given data:

Initial volume = 24.0 L

Initial pressure = 10.0 atm

Initial temperature = 200 K

Final temperature = 300 K

Final volume = ?

Final pressure = 14.0 atm

Formula:  

P₁V₁/T₁ = P₂V₂/T₂  

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

Solution:

V₂ = P₁V₁ T₂/ T₁ P₂  

V₂ = 10.0 atm × 24.0 L × 300K / 200 K × 14.0 atm

V₂ = 72000 atm .L. K / 2800 K.atm

V₂ = 25.7 L

The new volume of gas is 25.7 L.

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4 years ago
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3 years ago
A galvanic cell consists of a Cu(s)|Cu2+(aq) half-cell and a Cd(s)|Cd2+(aq) half-cell connected by a salt bridge. Oxidation occu
Kryger [21]

Answer:

Cd(s)|Cd^{2+}(aq) || Cu^{2+}(aq)|Cu(s)

Explanation:

A galvanic cell is composed of two electrodes immersed in a suitable electrolyte and connected via a salt bridge. One of the electrodes serves as a cathode where reduction or gain of electrons takes place. The other half cell functions as an anode where oxidation or loss of electrons occurs.

The representation is given by writing the anode on left hand side followed by its ion with its molar concentration. It is followed by a salt bridge. Then the cathodic ion with its molar concentration is written and then the cathode.

As it is given that cadmium acts as anode, it must be on the left hand side and copper must be on right hand side.

Cd(s)|Cd^{2+}(aq) || Cu^{2+}(aq)|Cu(s)

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3 years ago
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3 years ago
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I am having some trouble figuring out how to approach the following problem: A 75.0-mL volume of 0.200 M NH3 (Kb=1.8×10−5) is ti
Tanya [424]
<span>We look at the end of the day:

n(HNO3) added = 0.500*17.0/1000 = 0.00850 mol
n(NH3) = 0.200*75.0/1000 - 0.00850 = 0.00650 mol
[NH3] left = 0.00650*1000/(17.0+75.0) = 0.070652
M [OH-] = Kb * [NH3] = 0.070652*1.8*10^(-5) = 1.27174 x 10^(-6)
pOH = -log[OH-] ≈ 5.8956 pH = 14 - pOH ≈ 8.10</span>
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3 years ago
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