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mamaluj [8]
3 years ago
13

Which of the following is NOT a characteristic of stars? A) Size B) temperature C) texture D) color

Chemistry
1 answer:
Lynna [10]3 years ago
7 0

Answer:

C

Explanation:

Characteristics used to classify stars include color, temperature, size, composition, and brightness.

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Nitrogen gas in an expandable container is cooled from 45.0 C to 12.3 C with the pressure held constant at 2.58 ∗ 105 Pa. The to
Tresset [83]

Answer:

The number of moles of the gas is: -27.14 mole

the charge in internal energy of the gas is -1.84 × 10⁴ J.

The work done by the gas is 4.42 × 10⁴ J

The heat liberated by the gas for the same temperature change while the volume was constant and is same as the change in internal energy.

As such ; Q = -1.84 × 10⁴ J.

Explanation:

The expression for the number of moles of a gas at constant pressure is as follows:

\mathbf{n = \frac{Q}{Cp \Delta T}}

\mathbf{n = \frac{Q}{Cp (T_2-T_1)}}

where ;

C_p is the specific heat at constant pressure of Nitrogen gas which is = 29.07 J/mol/K

Since heat is liberated from the gas ; then:

n = \dfrac{-2.58*10^4 }{29.07(45-12.3)}

n = -27.14 mole

The number of moles of the gas is: -27.14 mole

b) The expression to be used in order to determine the change internal energy is:

dU = nCv \Delta T

where ;

n= 27.14 mole

Cv = specific heat at constant volume of Nitrogen gas = 20.76 J/mol/K

ΔT = (12.3-45)

So;

dU = (27.14)(20.76)(12.3-45)

dU = 563.426(-32.7)

dU = -18424.04328

dU = -1.84 × 10⁴ J

Thus; the charge in internal energy of the gas is -1.84 × 10⁴ J.

c)  The workdone by the gas can be calculated as;

W = Q - ΔU

W = 2.58 × 10⁴ J - (-1.84 × 10⁴ J )

W = 2.58  × 10⁴ J + 1.84  × 10⁴ J

W = 4.42 × 10⁴ J

The work done by the gas is 4.42 × 10⁴ J

d) The expression to calculated the work done is given as:

W = pdV

since the volume is given as constant ; then dV = 0

so;

W = p(0)

W = 0

Replacing 0 for W in the equation W = Q - Δ U

0 = Q - ΔU

-Q = - ΔU

Q = ΔU

Thus , the heat liberated by the gas for the same temperature change while the volume was constant and is same as the change in internal energy.

As such ; Q = -1.84 × 10⁴ J.

5 0
3 years ago
Read 2 more answers
How does equation below demonstrate the law of conservation of mass?
gavmur [86]
The answer is always found if you think

so i would pick b
4 0
3 years ago
Someone help me with this. Imagine you drank a big glass of water today. Ten years later,if you go to Antarctica and drink a gla
Artyom0805 [142]
Well when you would "get rid of" the water it would eventually be led to some ocean, pond, lake, etc. There it would eventually evaporate, rising up to the sky. This is where it would move around with clouds until it rained again. 
3 0
3 years ago
The molarity of a 4.200 L solution is 1.230 M Na2CO3. What is the mass of Na2CO3
Lerok [7]

Answer:

547.5g

Explanation:

To get the mass, you need moles.

moles = (molarity)(Liters)

moles = (1.230M)(4.200L) = 5.166 moles Na2CO3

Now, just use stoichiometry

molar mass of Na2CO3 = 2(mass of Na) + (mass of C) + 3(mass of O)

= 2(22.9) + 12.01 + 3(16) = 105.99g/mol

5.166moles(105.99g/mol)

= 547.544

But, the measurements given had 4 significant figures, so in chemistry we write:

547.5g

7 0
3 years ago
We usually think of wires as perfect (R=0) conductors, which is not quite accurate. Real wires have resistance, and an effort is
SSSSS [86.1K]

Answer:

Length of copper wire is 5,22m and the diameter of copper wire is 2,14x10⁻³m.

Explanation:

The resistance of a wire is determined by:

R = ρL/A

Where R is resistance (25x10⁻³Ω); ρ is resistivity (1,72x10⁻⁸Ωm); R is length and A is cross sectional area.

Replacing you will obtain:

1,45x10⁶ m⁻¹ = L/A <em>(1)</em>

$1 of copper are:

$1×\frac{1 lb}{2,69} = 0,37 lb. In kg:

0,37lb×\frac{0,453592kg}{1lb} = 0,169 kg of copper.

As density of copper is 8960 kg/m³. The volume of this amount of copper is:

0,169kg×\frac{1m^{3}}{8960kg} = 1,88x10⁻⁵ m³

The volume of a wire is L×A. Thus:

1,88x10⁻⁵ m³ = L×A<em> (2)</em>

Replacing (1) in (2)

1,88x10⁻⁵ m³ = 1,45x10⁶ m⁻¹A²

A = 3,6x10⁻⁶ m²

Thus, L = 5,22 m

Thus, <em>length of copper wire is 5,22 m</em>

As area = 3,6x10⁻⁶ m² = \frac{pi}{4}d^{2} Where d is diameter.

The <em>diameter of copper wire is 2,14x10⁻³m</em>

I hope it helps!

4 0
3 years ago
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