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3 years ago

Which statements describe or apply the principal of conservation of energy? Select all that apply.

Chemistry

2 answers:

laila [671]3 years ago

7 0

Answer:

A, C, and D

Explanation:

The correct options that apply to the principal conservation of energy are A, C, and D.

A is correct because energy can neither be created nor destroyed. However, energy can be transfered from one location to another or be converted from one form to another. Whether transferred to converted, the magnitude remains the same.

C is correct because energy cannot be destroyed but can be transferred or converted. Hence, if a body or a location loses temprature, then the loss is being gained by another body or location.

D is also correct. A closed system is a system that does not exchange matter with its surroundings. Hence, the total energy remains the same within the system.

Harrizon [31]3 years ago

6 0

Answer:

The total energy in a closed system such as Earth stays the same.

Explanation:

The simplest statement of the law of conservation of energy is that energy can neither be created nor destroyed but can be converted from one form to another.

A typical illustration of this principle of conservation of energy is a closed system of which our universe is an important example.

The total amount of energy in the universe is always a constant. Energy lost in one part must be equal to the energy gained on the other part.

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babymother [125]

Answer:

16.56 g

Explanation:

Mass is the production of Volume and Density.

m = V. d = 6 × 2.76 = 16.56 g

6 0

3 years ago

Pressure gauge at the top of a vertical oil well registers 140 bars. The oil well is 6000 m deep and filled with natural gas dow

andreyandreev [35.5K]

Explanation:

(a) The given data is as follows.

Pressure on top ( $P_{o}$ ) = 140 bar = $1.4 \times 10^{7} Pa$ (as 1 bar = $10^{5}$ )

Temperature = $15^{o}C$ = (15 + 273) K = 288 K

Density of gas = $\frac{PM}{ZRT}$

$\frac{dP}{dZ} = \rho \times g$

$\frac{dP}{dZ} = \int \frac{PM}{ZRT}$

$\int_{P_{o}}^{P_{1}} \frac{dP}{dZ} = \frac{Mg}{ZRT} \int_{0}^{4700} dZ$

$ln (\frac{P_{1}}{P_{o}}) = \frac{18.9 \times 10^{-3} \times 9.81 \times 4700 m}{0.80 \times 8.314 J/mol K \times 288 K}$

= 0.4548

$P_{1} = P_{o} \times e^{0.4548}$

= $1.4 \times 10^{7} Pa \times 1.5797$

= $2.206 \times 10^{7} Pa$

Hence, pressure at the natural gas-oil interface is $2.206 \times 10^{7} Pa$ .

(b) At the bottom of the tank,

$P_{2} = P_{1} + \rho \times g \times h$

= 2.206 \times 10^{7} Pa + 700 \times 9.81 \times (6000 - 4700)[/tex]

= $309.8 \times 10^{5} Pa$

= 309.8 bar

Hence, at the bottom of the well at $15^{o}C$ pressure is 309.8 bar.

6 0

3 years ago

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Tems11 [23]

Answer: A different group of scientists using different methods.

5 0

2 years ago

Given the following equilibrium constants: Kb B(aq) + H2O(l) ⇌ HB+(aq) + OH−(aq) 1/Kw H+(aq) + OH−(aq) ⇌ H2O(l) What is the equi

bija089 [108]

Answer: The value of $K_c$ for the net reaction is $\frac{K_b}{K_w}$

Explanation:

The given chemical equations follows:

Equation 1: $B(aq.)+H_2O(l)\rightleftharpoons HB^+(aq.)+OH^-(aq.);K_b$

Equation 2: $H^+(aq.)+OH^-(aq.)\rightleftharpoons H_2O(l);\frac{1}{K_w}$

The net equation follows:

$B(aq.)+H^+(aq.)\rightleftharpoons HB^+(aq.);K_c$

As, the net reaction is the result of the addition of first equation and the second equation. So, the equilibrium constant for the net reaction will be the multiplication of first equilibrium constant and the second equilibrium constant.

The value of equilibrium constant for net reaction is:

$K_c=K_1\times K_2$