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2 years ago

1. What are the ways in which matter and energy can interact?

Chemistry

1 answer:

frozen [14]2 years ago

4 0

Answer:

Einstein tells us that the relationship between energy and matter is defined by the equation E = M C 2 , where E is energy, and M is mass.

Explanation:

By looking at Einstein's equation we can see that matter (or something that has mass) and energy are actually the same thing

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A 62.6-gram piece of heated limestone is placed into 75.0 grams of water at 23.1°C. The limestone and the water come to a final

VMariaS [17]

M₁ = mass of water = 75 g
T₁ = initial temperature of water = 23.1 °C
c₁ = specific heat of water = 4.186 J/g°C

m₂ = mass of limestone = 62.6 g
T₂ = initial temperature of limestone = ?
c₂ = specific heat of limestone = 0.921 J/g°C

T = equilibrium temperature = 51.9 °C
using conservation of heat
Heat lost by limestone = heat gained by water
m₂c₂(T₂ - T) = m₁c₁(T - T₁)
inserting the values
(62.6) (0.921) (T₂ - 51.9) = (75) (4.186) (51.9 - 23.1)
T₂ = 208.73 °C
in three significant figures
T₂ = 209 °C

7 0

3 years ago

Which of these processes are chemical reactions?

masya89 [10]

Rotting vegetables is a chemical change I think.

8 0

2 years ago

What is the partial pressure of radon if the total pressure is 780 torr and the

Brilliant_brown [7]

Answer:

0.03atm

Explanation:

Given parameters:

Total pressure = 780torr

Partial pressure of water vapor = 1.0atm

Unknown:

Partial pressure of radon = ?

Solution:

A sound knowledge of Dalton's law of partial pressure will help solve this problem.

The law states that "the total pressure of a mixture of gases is equal to the sum of the partial pressures of the constituent gases".

Mathematically;

P $_{t}$ = P $_{1}$ + P $_{2}$ + P $_{n}$

Since the total pressure is 780torr, convert this to atm;

760torr = 1 atm

780torr = $\frac{780}{760}$ atm = 1.03atm

For this problem;

Total pressure = Partial pressure of radon + Partial pressure of water vapor

1.03 = Partial pressure of radon + 1.0

Partial pressure of radon = 1.03 - 1.00 = 0.03atm

6 0

3 years ago

In a unimolecular reaction with twice as much starting material as product at equilibrium, what is the value of Keq? Is ΔG o pos

lubasha [3.4K]

Answer : The value of $K_{eq}$ is, 0.5 and $\Delta G$ = positive.

Explanation :

The unimolecular reaction is:

$A\rightarrow B$

In unimolecular reaction, the starting material is 2 times to the product.

$A=2B$ .........(1)

As we know that:

$K_{eq}=\frac{B}{A}$ ...........(2)

Now substitute equation 1 in 2, we get:

$K_{eq}=\frac{\frac{A}{2}}{A}$

$K_{eq}=0.5$

Now we have to calculate the value of $\Delta G^o$ at 298 K.

$\Delta G^o=-RT\ln K_{eq}$

where,

$\Delta G^o$ = standard Gibbs free energy = ?

R = gas constant = 8.314 J/mol.K

T = temperature = 298 K

$K_{eq}$ = equilibrium constant = 0.5

Now put all the given values on the above formula, we get:

$\Delta G^o=-(8.314J/mol.K)\times (298K)\ln (0.5)$

$\Delta G^o=1717.32J/mol$

Thus, the value of $\Delta G^o$ at 298 K is, 1717.32 J/mol

As we know that:

$\Delta G$ = +ve, reaction is non spontaneous

$\Delta G$ = -ve, reaction is spontaneous

$\Delta G$ = 0, reaction is in equilibrium

Thus, the $\Delta G$ = +ve. So, the reaction is non spontaneous.

8 0

2 years ago

For which of the following reactions is ΔH∘rxn equal to ΔH∘f of the product(s)?

nexus9112 [7]

<span>The correct answers are A, B, and D. The ΔHrxn is equal to the ΔHf when the ΔHf of the reactants is zero, and this can only occur when the reactants are pure elements (not compounds) in their natural phase at ambient temperature.

c) Chlorine's natural phase is gaseous, but since it is a liquid in this reaction, ΔHf ≠ 0.

e) CO is a compound, not a pure element.

f) BaCO3 is a compound, not a pure element.</span>

6 0

2 years ago

Read 2 more answers