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Flauer [41]
3 years ago
8

Use coulomb\'s law to calculate the energy of a magnesium ion and an oxide ion at their equilibrium ion-pair separation distance

.
Chemistry
2 answers:
Salsk061 [2.6K]3 years ago
6 0

We know that

F = k * q1 * q2 / d^2

Force of attraction F

K = 8.99 x 10^9

q1 = charge on Mg = +2  = 2 X 1.9 x 10^-19 C

q2 = charge on O = -2 = 2 X 1.9 x 10^-19 C

d = distance between two ions = 72 + 140 pm = 212 pm

Force = 8.99 x 10^9 X 2 X 1.9 x 10^-19 C X 2 X 1.9 x 10^-19 C / (212 X 10^-12 )^2

Force = 0.002889 X 10^-5 = 2.889 X 10^-8 N

sammy [17]3 years ago
5 0

The energy required to separate the ion pair = 4.297×10^(−19)J .

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A reaction releases 489.34 J of heat how many calories of heat is that equal to 1 cal equals 4.186 J
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The answer is equal to 115.47 calories. Just devide the amount of heat release by 4.186 J because it is equal to 1 cal
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The existence of discrete (quantized) energy levels in an atom may be inferred from
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<span>It can be inferred from atomic line spectra. These are certain wavelengths of light that atoms give off when they become excited. The energy levels in an atom can be proven to exist from this phenomenon, and furthermore, they can be measured.</span>
8 0
3 years ago
Name:_____________________________________________________ Date:___________ Period:_________ 3/23 - 3/27 Assignment 1: Gas Law P
crimeas [40]

Answer:

The answers are;

1. 8.2 liters

2. 1214.84 ml

3. 318.027 K

4. 4.00 l.

Explanation:

1. Boyle's law states that the volume of  given mass of gas is inversely proportional to its pressure at constant temperature

that is

P₁·V₁ = P₂·V₂

Where:

P₁ = Initial pressure = 40.0 mm Hg

V₁ = Initial volume = 12.3 liters

P₂ = Final pressure = 60.0 mm Hg

V₂ = Final volume = Required

From P₁·V₁ = P₂·V₂, V₂ is given by

V_2=\frac{P_1\cdot V_1}{P_2} = \frac{40.0 mm Hg\cdot 12.3 l}{60.0 mm Hg} =  8.2 l

The volume reduces to V₂ = 8.2 liters

2. Here Charles law states that

\frac{T_1}{V_1} =\frac{T_2}{V_2}

T₁ = Initial temperature = 27.0 °C = 300.15 K

V₁ = Initial volume = 900.0 mL

T₂ = Final temperature = 132.0 °C = 405.15 K

V₂ = Final volume = Required

Therefore  V_2 =\frac{T_2\cdot V_1}{T_1} = \frac{405.15 K\times 900.0 mL}{300.15 K} = 1214.84 ml

V₂ = 1214.84 ml

3.  Gay-Lussac's Law states that

\frac{T_1}{P_1} =\frac{T_2}{P_2}

Where:

P₁ = Initial pressure = 15.0 atmospheres

T₁ = Initial temperature = 25.0 °C = 298.15 K

P₂ = Final pressure = 16.0 atmospheres

T₂ = Final temperature = Required

∴ T_2 = \frac{T_1\times P_2}{P_1}

=  \frac{298.15 K\times 16.0atm}{15.0atm} = 318.027 K

T₂ = 318.027 K

4. Avogadro's law states that,

Equal volume of all gases at the same temperature and pressure contain equal number of molecules.

Therefore if 5.00 moles of gas occupies 2.00 l volume, then

1 moles will occupy 2.00/5 l volume and

10 moles will occupy 2.00/5 × 10 or 4.00 l volume.

6 0
3 years ago
for the reaction C+2H2= Ch4 how many molecules of hydrogen are required to produce 14 grams of methane Ch4?
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Answer:

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No of mole of CH4 = 14/16.04

No of mole of CH4 = 0.8728mol

Stoichiometrically, from the above

2 Molecules of Hydrogen will produce 1molecule of methane

Therefore, 2*0.8728 mole of Hydrogen will produce 0.8728mol of Methane.

The no of molecule of hydrogen is 1.7456

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3 years ago
How many neutrons are in cesium-135 (135/55Cs)
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 there are 78 neutrons.

3 0
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