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kotykmax [81]
3 years ago
12

What is the lightest material ever created by the hands of men

Chemistry
1 answer:
lutik1710 [3]3 years ago
6 0
The world's lightest material is carbon aerogel which has a mass of only 0.16 milligram and it is even lighter than aerographite.
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To break the oxygen-oxygen bonds in a single O2 molecule, 1 × 10–18 J of energy is required. Which of the following wavelengths
Sonbull [250]

E = hc/(lamda)

The lamda symbol is wavelength, which this site does not have. I can represent it with an "x" instead.

Plancks constant, h = 6.626×10^-32 J·s

Speed of light, c = 3.00×10^8 m/s

The energy must be greater than or equal to 1×10^-18 J

1×10^-18 J ≤ (6.626×10^-32 J·s)*(3.0×10^8 m/s) / x

x ≤ (6.626×10^-32 J·s)*(3.0×10^8 m/s) / (1×10^-18 J)

x ≤ 1.99×10^-7 m or 199 nm

The wavelength of light must be greater than or equal to 199 nm

8 0
3 years ago
Mg(OH)2 + 2HNO3 → Mg(NO3)2 + 2H20
LenaWriter [7]

Answer:

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3 0
3 years ago
What is the entropy change for the freezing process of 1 mole of liquid methanol at its freezing temperature (–97.6˚C) and 1 atm
Rudiy27

Answer : The value of change in entropy for freezing process is, -18.07 J/mol.K

Explanation :

Formula used :

\Delta S=\frac{\Delta H_{freezing}}{T_f}

where,

\Delta S = change in entropy

\Delta H_{fus} = change in enthalpy of fusion = 3.17 kJ/mol

As we know that:

\Delta H_{fus}=-\Delta H_{freezing}=-3.17kJ/mol=-3170J/mol

T_f = freezing point temperature = -97.6^oC=273+(-97.6)=175.4K

Now put all the given values in the above formula, we get:

\Delta S=\frac{\Delta H_{freezing}}{T_m}

\Delta S=\frac{-3170J/mol}{175.4K}

\Delta S=-18.07J/mol.K

Therefore, the value of change in entropy for freezing process is, -18.07 J/mol.K

4 0
3 years ago
1)Why do clouds usually form high in the air instead of near Earth's surface?
djverab [1.8K]
The answer is letter A definitively .
6 0
2 years ago
Convert the following to scientific notation 24,212,000
UNO [17]
2.4212 X 10^ 7
How I at least figure this problem out is I take a pencil and start on the right side of the 0 and make a loop to the left for each number and count until I get to the first two numbers that are between 1-9 when reading from left to right. This is where you put the decimal point. Some teachers rather you keep the 0's there, while others prefer one to get rid of them. Anyways with that new decimal number, you multiply the decimal by ten to what ever number you counted, which was 7.
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3 years ago
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