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3 years ago

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A syringe containing 12.0 mL of dry air at 25 C is placed in a sterilizer and heated to 100.0 C. The syringe is sealed, but th

e plunger can move and the volume can change. What is the volume of the air in the syringe at 100.0 C, assuming no change in pressure?

1 answer:

monitta3 years ago

5 0

Answer:

15.01 Liters

Explanation:

T₁ = Initial temperature = 25°C = 298.15 K

T₂ = Final temperature = 100°C = 373.15 K

V₁ = Initial volume = 12 mL

Here, pressure is constant so we apply Charles Law

$\frac{V_1}{T_1}=\frac{V_2}{T_2}\\\Rightarrow {V_2}=\frac{V_1}{T_1}\times T_2\\\Rightarrow {V_2}=\frac{12}{298.15}\times 373.15\\\Rightarrow {V_2}=15.01 L$

∴ Final volume at 100°C is 15.01 Liters.

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The intensity of light from a central source varies inversely as the square of the distance. If you lived on a planet only half

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Answer:

the intensity will be 4 times that of the earth.

Explanation:

let us assume the following:

intensity of light on earth =J

distance of earth from sun = d

intensity of light on other planet = K

distance of other planet from sun = $\frac{d}{2}$ (from the question, the planet is half as far from the sun as earth)

from the question the intensity is inversely proportional to the square of the distance, hence

intensity on earth : J = $\frac{1}{d^{2} }$

J $d^{2}$ = 1 ... equation 1

intensity on other planet : K = $\frac{1}{(\frac{d}{2}) ^{2} }$ (the planet is half as far from the sun as earth)

K $(\frac{d}{2}) ^{2}$ = 1 ....equation 2

equating both equation 1 and 2 we have

J $d^{2}$ = K $(\frac{d}{2}) ^{2}$

J $d^{2}$ = K $\frac{d^{2}}{4}$

J = $\frac{K}{4}$

K = 4J

intensity of light on other planet (K) = 4 times intensity of light on earth (J)

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The energy change in an endothermic reaction is: A. Internal B. External C. Negative D. Positive

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Answer:

Positive

Explanation:

In an endothermic reaction, the products are at a higher energy than the reactants. This means that the enthalpy change of the reaction (∆H) is positive

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The maximum distance from the Earth to the Sun (at aphelion) is 1.521 1011 m, and the distance of closest approach (at perihelio

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Answer:

29274.93096 m/s

$2.73966\times 10^{33}\ J$

$-5.39323\times 10^{33}\ J$

$2.56249\times 10^{33}\ J$

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Explanation:

$r_p$ = Distance at perihelion = $1.471\times 10^{11}\ m$

$r_a$ = Distance at aphelion = $1.521\times 10^{11}\ m$

$v_p$ = Velocity at perihelion = $3.027\times 10^{4}\ m/s$

$v_a$ = Velocity at aphelion

m = Mass of the Earth = 5.98 × 10²⁴ kg

M = Mass of Sun = $1.9889\times 10^{30}\ kg$

Here, the angular momentum is conserved

$L_p=L_a\\\Rightarrow r_pv_p=r_av_a\\\Rightarrow v_a=\frac{r_pv_p}{r_a}\\\Rightarrow v_a=\frac{1.471\times 10^{11}\times 3.027\times 10^{4}}{1.521\times 10^{11}}\\\Rightarrow v_a=29274.93096\ m/s$

Earth's orbital speed at aphelion is 29274.93096 m/s

Kinetic energy is given by

$K=\frac{1}{2}mv_p^2\\\Rightarrow K=\frac{1}{2}\times 5.98\times 10^{24}(3.027\times 10^{4})^2\\\Rightarrow K=2.73966\times 10^{33}\ J$

Kinetic energy at perihelion is $2.73966\times 10^{33}\ J$

Potential energy is given by

$P=-\frac{GMm}{r_p}\\\Rightarrow P=-\frac{6.67\times 10^{-11}\times 1.989\times 10^{30}\times 5.98\times 10^{24}}{1.471\times 10^{11}}\\\Rightarrow P=-5.39323\times 10^{33}$

Potential energy at perihelion is $-5.39323\times 10^{33}\ J$

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Kinetic energy at aphelion is $2.56249\times 10^{33}\ J$

Potential energy is given by

$P=-\frac{GMm}{r_a}\\\Rightarrow P=-\frac{6.67\times 10^{-11}\times 1.989\times 10^{30}\times 5.98\times 10^{24}}{1.521\times 10^{11}}\\\Rightarrow P=-5.21594\times 10^{33}$

Potential energy at aphelion is $-5.21594\times 10^{33}\ J$

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