Which of the following technologies would produce the least energy in light

A rigid container holds 0.30g of hydrogen gas.

Strike441 [17]

Answer:

Part A: $\mathbf{Q =94 \ J}$ to two significant figures

Part B: $\mathbf{Q =160 \ J}$ to two significant figures

Part C: $\mathbf{Q =220 \ J}$ to two significant figures

Explanation:

Given that :

mass of the hydrogen = 0.30 g

the molar mass of hydrogen gas molecule = 2 g/mol

we all know that:

number of moles = mass/molar mass

number of moles = 0.30 g /2 g/mol

number of moles = 0.15 mol

For low temperature between the range of 50 K to 100 K, the specific heat at constant volume for a diatomic gas molecule = $C_v=\dfrac{3}{2}R$

For Part A:

$Q = mC_v\Delta T$

$Q= 0.15 \ mol (\dfrac{3}{2})(8.314 \ J/mole.K )(100-50)K$

$Q= 0.15 \times (\dfrac{3}{2}) \times (8.314 \ J )\times (50)$

$Q=93.5325 \ J$

$\mathbf{Q =94 \ J}$ to two significant figures

Part B. For hot temperature, $C_v=\dfrac{5}{2}R$

$Q = mC_v\Delta T$

$Q= 0.15 \ mol (\dfrac{5}{2})(8.314 \ J/mole.K )(300-250)K$

$Q= 0.15 \times (\dfrac{5}{2}) \times (8.314 \ J )\times (50)$

$Q=155.8875 \ J$

$\mathbf{Q =160 \ J}$ to two significant figures

Part C. For an extremely hot temperature, $C_v=\dfrac{7}{2}R$

$Q = mC_v\Delta T$

$Q= 0.15 \ mol (\dfrac{7}{2})(8.314 \ J/mole.K )(2300-2250)K$

$Q= 0.15 \times (\dfrac{7}{2}) \times (8.314 \ J )\times (50)$

$Q=218.2425 \ J$

$\mathbf{Q =220 \ J}$ to two significant figures

6 0

3 years ago

Two bulbs are connected in parallel across a source of EMF = 8.0V with a negligible internal resistance. One bulb has a resistan

Sonja [21]

<h2>Answer:</h2>

(a) 3.18Ω

(b) 3.18Ω

<h2>Explanation:</h2>

Let the two bulbs be A and B

Given;

$R_{A}$ = Resistance in bulb A = 3.0Ω

$R_{B}$ = Resistance in bulb B = 2.5Ω

Since the two bulbs are connected in parallel;

i. their effective resistance ( $R_{X}$ ) is given by

$\frac{1}{R_{X}}$ = $\frac{1}{R_{A} }$ + $\frac{1}{R_{B} }$ ---------------(i)

Substitute the values of $R_{A}$ and $R_{B}$ into equation (i)

=> $\frac{1}{R_{X}}$ = $\frac{1}{3.0}$ + $\frac{1}{2.5}$

Solve for $R_{X}$

$R_{X}$ = 1.36Ω

ii. voltage (potential difference), V, across them is the same;

Therefore we can get the total current (I) that will flow through them if the voltage to be supplied is 2.4V.

Use the Ohm's law;

V = I x R -----------------(ii)

Where;

V = voltage across them = 2.4V

I = total current flowing through them

R = their effective resistance = $R_{X}$ = 1.36Ω

Substitute these values into equation (ii) as follows;

2.4 = I x 1.36

I = 2.4 / 1.36

I = 1.76A

(a) Now get the value of R

Since the voltage across the two bulbs is 2.4V out of the 8.0V supplied by the source, then the remaining (8.0 - 2.4 = 5.6)V will pass across the resistor R.

Also, since the two bulbs make a series connection with the resistor R, the same total current (I = 1.76A) that flows through these bulbs will flow through the resistor R.

Therefore, to get the value of R, we use the relation

V = I x R ------------------------------(iii)

Where;

V = voltage across the resistor = 5.6V

I = current through the resistor = 1.76A

<em>Substitute these values into equation (iii)</em>

=> 5.6 = 1.76 x R

=> R = 5.6 / 1.76

=> R = 3.18Ω

Therefore, the value of R to be chosen in order to supply each bulb with a voltage of 2.4V is 3.18Ω

(b) The potential difference and voltage across refer to the same thing. Therefore, the value of R that would make the potential difference across each of the bulbs be 2.4V is the same as the one calculated in (a) which is 3.18Ω

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3 years ago

Question #4 Multiple Choice Which of the following is a similarity between an industrial metrologist and a legal metrologist? Bo

sdas [7]

Answer:

choice B

Explanation:i took the test

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Prove that g is inversely proportional to the radius

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Explanation:

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Explain how evidence of energy transformation in fireworks supports the law of conservation of energy.

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