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3 years ago

6. At a glance, it appears like the law of

Chemistry

1 answer:

Mamont248 [21]3 years ago

4 0

Energy and mass equivalence

Explanation:

The law of conservation of energy does not apply to nuclear reactions, the law of conservation of mass-energy makes more sense in this regard.

In nuclear reactions mass is transformed into energy. Therefore, it does not conform with the law of conservation of energy.

According to the law of conservation of energy "energy is neither created no destroyed but they are transformed from one form to another. "
In nuclear reactions, mass is usually lost. Mass is not conserved.
But, the mass is converted into energy and we say there is mass and energy equivalence for nuclear reactions.

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(C=12.01 amu, H=1.008 amu, O=16.00 amu) [?] g/mol CH20

tekilochka [14]

CH₂O = C + 2.H + O

= 12.01 + 2 x 1.008 + 16

= 30.026 amu = 30.026 g/mol

8 0

2 years ago

How many moles of a gas sample are in a 10.0 L container at 298 K and 203 kPa? (Use the ideal gas law PV = nRT with R = 8.31 L-k

weeeeeb [17]

Ideal gas law: PV = nRT

P-pressure
V-volume
n-number of moles(m/M)
R-constant
T- temperature

State given info:
V=10L
T=298k
P=203kPa
R=8.31(only for kPa, for pressure at atm use 0.08206)

Sub in:
(203kPa)(10L)=n(8.31)(298)

Rearrange:
n= (203)(10) / (8.31)(298)
n = 0.819745mol in gas sample

8 0

3 years ago

Gaseous indium dihydride is formed from the elements at elevated temperature:

3241004551 [841]

Answer:

For 1: The value of $Q_p$ for above reaction is 36.83

For 2: The value of $Q_p$ for above reaction is 36.83

For 3: The equilibrium partial pressure of Indium is 0.126 atm

For 4: The equilibrium partial pressure of hydrogen gas is 0.094 atm

For 5: The equilibrium partial pressure of Indium dihydrogen is 0.018 atm

Explanation:

We are given:

Partial pressure of Indium gas = 0.0650 atm

Partial pressure of hydrogen gas = 0.0330 atm

Partial pressure of Indium dihydride = 0.0790 atm

The given chemical equation follows:

$ln(g)+H_2(g)\rightleftharpoons InH_2(g)$

Initial: 0.065 0.033 0.079

At eqllm: 0.065-x 0.033-x 0.079+x

For 1:

The expression of $Q_p$ for above reaction follows:

$Q_p=\frac{p_{InH_2}}{p_{In}\times p_{H_2}}$

Putting values in above equation, we get:

$Q_p=\frac{0.079}{0.065\times 0.033}=36.83$

Hence, the value of $Q_p$ for above reaction is 36.83

For 2:

We are given:

$K_p$ of the reaction = 1.48

There are 3 conditions:

When $K_{p}>Q_p$ ; the reaction is product favored.
When $K_{p}$ ; the reaction is reactant favored.
When $K_{p}=Q_p$ ; the reaction is in equilibrium.

As, $Q_{p}>K_p$ for the given reaction, the reaction is reactant favored.

Hence, the reaction proceed in the backward direction to attain equilibrium

For 3:

The expression of $K_p$ for above reaction follows:

$K_p=\frac{p_{InH_2}}{p_{In}\times p_{H_2}}$

Putting values in above equation, we get:

$1.48=\frac{(0.079+x)}{(0.065-x)\times (0.033-x)}\\\\x=-0.061,0.835$

Neglecting the value of x = 0.835 because the reaction is going backwards. So, by taking this value, the pressure of the reactants will decrease

So, equilibrium partial pressure of Indium = (0.065 - x) = [0.065 - (-0.061)] = 0.126 atm

For 4:

The equilibrium partial pressure of hydrogen gas = (0.033 - x) = [0.033 - (-0.061)] = 0.094 atm

For 5:

The equilibrium partial pressure of Indium dihydrogen = (0.079 + x) = [0.079 + (-0.061)] = 0.018 atm

6 0

3 years ago

Analysis of a 10.15 g sample of a compound known to contain only phosphorus and oxygen indicates a phosphorus content of 4.433 g

Anna007 [38]

find mass of oxygen by subtracting mass of phosphorus from the total mass

divide the masses by the molar mass to get moles

divide moles by the smallest amount of moles

multiply by 2 to get a nice number

P4O5

5 0

3 years ago

Carbonyl fluoride, COF2, is an important intermediate used in the production of fluorine-containing compounds. For instance, it

Sauron [17]

Answer:

The concentration of COF₂ at equilibrium is 0.296 M.

Explanation:

To solve this equilibrium problem we use an ICE Table. In this table, we recognize 3 stages: Initial(I), Change(C) and Equilibrium(E). In each row we record the concentrations or changes in concentration in that stage. For this reaction:

2 COF₂(g) ⇌ CO₂(g) + CF₄(g)

I 2.00 0 0

C -2x +x +x

E 2.00 - 2x x x

Then, we replace these equilibrium concentrations in the Kc expression, and solve for "x".

$Kc=8.30=\frac{[CO_{2}] \times [CF_{4}] }{[COF_{2}]^{2} } =\frac{x^{2} }{(2.00-2x)^{2} } \\8.30=(\frac{x}{2.00-2x} )^{2} \\\sqrt{8.30} =\frac{x}{2.00-2x}\\5.76-5.76x=x\\x=0.852$

The concentration of COF₂ at equilibrium is 2.00 -2x = 2.00 - 2 × 0.852 = 0.296 M

6 0

3 years ago