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Vika [28.1K]
3 years ago
15

Old thermometers contained very small amounts of mercury. The mercury in the photo has a melting point of -38.8 degrees Celsius.

What can you conclude about the melting point of the mercury in old thermometers? *
It's melting point can only be determined when the mercury is burned.
It's melting point changes as the temperature of mercury changes.
It's melting point equals -38.8 degrees Celsius because it is mercury.
It's melting point is less than -38.8 degrees Celsius because its volume is smaller.
Chemistry
1 answer:
Nikolay [14]3 years ago
7 0

Answer:

••It's melting point equals -38.8 degrees Celsius because it is mercury. |••

Explanation:

This one makes the most sense out of the answers.

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E=mc^2 and mass defect
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3 years ago
At a festival, spherical balloons with a radius of 280. Cm are to be inflated with hot air and released. The air at the festival
evablogger [386]

Answer:

8608.18 balloons

Explanation:

Hello! Let's solve this!

Data needed:

Enthalpy of propane formation: 103.85kJ / mol

Specific heat capacity of air: 1.009J · g ° C

Density of air at 100 ° C: 0.946kg / m3

Density of propane at 100 ° C: 1.440kg / m3

First we will calculate the propane heat (C3H8)

3000g * (1mol / 44g) * (103.85kJ / mol) * (1000J / 1kJ) = 7.08068 * 10 ^ 6 J

Then we can calculate the mass of the air with the heat formula

Q = mc delta T

m = Q / c delta T = (7.08068 * 10 ^ 6 J) / (1.009J / kg ° C * (100-25) ° C) =

m = 93566.96kg

We now calculate the volume of a balloon.

V = 4/3 * pi * r ^ 3 = 4/3 * 3.14 * 1.4m ^ 3 = 11.49m ^ 3

Now we calculate the mass of the balloon

mg = 0.946kg / m3 * 11.49m ^ 3 = 10.87kg

The amount of balloons is

93566.96kg / 10.87kg = 8608.18 balloons

5 0
3 years ago
Read 2 more answers
A system is composed of 7.14 grams of Ne gas at 298 K and 1 atm. When 2025 joules of heat are added to the system at constant pr
dlinn [17]

Answer:

(a) 1 atm, 8.72 L and 298 K respectively.

(b) 1 atm, 16.72 L and 573.35 K respectively.

(c) \Delta U=1215J

Explanation:

Hello,

(a) In this case, considering that neon could be considered as an ideal gas, we can compute its volume as follows:

PV=nRT\\\\V=\frac{nRT}{P}=\frac{7.14g*\frac{1mol}{20g}0.082\frac{atm*L}{mol*K}*298K}{1atm}\\  \\V=8.72L

Thus, initial conditions of pressure volume and temperature are 1 atm, 8.72 L and 298 K respectively.

(b) Since the process was carried out at constant pressure, the work is defined as:

W=P(V_2-V_1)

Thus, the final volume is:

V_2=\frac{W}{P} +V_1=\frac{810Pa*m^3}{1atm*\frac{101325Pa}{1atm} } *\frac{1000L}{1m^3}+8.72L\\ \\V_2=16.72L

And the final temperature is computed by considering the pressure-constant specific heat of neon (1.03 J/g*K) and the added heat:

Q=mCp(T_2-T_1)\\\\T_2=\frac{Q}{mCp}+T_1 =\frac{2025J}{7.14g*1.03J/(g*K)}+298K\\ \\T_2=573.35K

Therefore, final volume is 16.72 L, final pressure is also 1 atm and final temperature is 573.35 K for this expansion process.

(c) Finally, the change in the internal energy is computed via the first law of thermodynamics:

Q-W=\Delta U\\\\\Delta U=2025J-810J\\\\\Delta U=1215J

Best regards.

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Answer: True

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Incomplete combustion<span> occurs when the supply of air or oxygen is poor. Water is still produced, but carbon monoxide and carbon are produced instead of carbon dioxide</span>
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3 years ago
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