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2 years ago

What do you think is the coldest temperature something can get to? What limits how cold something can get?

Chemistry

1 answer:

damaskus [11]2 years ago

4 0

Answer:

The colder an object it, the slower the atoms are. This is the rule that affects how cold something can get. There is a limit of how cold something can get though. This is called absolute zero. And, actually, we've gotten very close to it. Scientists in Finland have cooled rhodium atoms to a 10th of a billionth of a degree above absolute zero.

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Boron (B) has two isotopes, boron-10 and boron-11. The atomic weight of boron is ?

suter [353]

Answer:

Approximately 10,5

Explanation:

The question is not really very specific, because it would need the percentages of those isotopes in the nature. As they are not shown, it should be the median of those two numbers.

atomic weight ≈ $\frac{10+11}{2}$ = 10,5

If you check a periodic table, you'll see it's actually 10,8, but that's because of the thing I told you at first (percentages missing).

Hope I could help.

8 0

3 years ago

The reaction C4H8(g)⟶2C2H4(g) C4H8(g)⟶2C2H4(g) has an activation energy of 262 kJ/mol.262 kJ/mol. At 600.0 K,600.0 K, the rate c

crimeas [40]

Answer : The rate constant at 785.0 K is, $1.45\times 10^{-2}s^{-1}$

Explanation :

According to the Arrhenius equation,

$K=A\times e^{\frac{-Ea}{RT}}$

or,

$\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$K_1$ = rate constant at $600.0K$ = $6.1\times 10^{-8}s^{-1}$

$K_2$ = rate constant at $785.0K$ = ?

$Ea$ = activation energy for the reaction = 262 kJ/mole = 262000 J/mole

R = gas constant = 8.314 J/mole.K

$T_1$ = initial temperature = $600.0K$

$T_2$ = final temperature = $785.0K$

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

$\log (\frac{K_2}{6.1\times 10^{-8}s^{-1}})=\frac{262000J/mole}{2.303\times 8.314J/mole.K}[\frac{1}{600.0K}-\frac{1}{785.0K}]$

$K_2=1.45\times 10^{-2}s^{-1}$

Therefore, the rate constant at 785.0 K is, $1.45\times 10^{-2}s^{-1}$

7 0

4 years ago

If you have 1.00 mole of F2 at 1.00 atm of pressure and 0°C, what is the volume of F2?

ss7ja [257]

Answer:

$V=22.4L$

Explanation:

Hello,

In this case, by using the ideal gas equation, er can compute the volume of fluorine gas as shown below:

$PV=nRT\\\\V=\frac{nRT}{P}=\frac{1mol*0.082\frac{atm*L}{mol*K}+(0+273)K}{1.00atm} \\\\V=22.4L$

Best regards.

4 0

3 years ago

If the CaCO3 weighed 983 g and the CaO weighed 551 g, how many grams of CO2 were formed in the reaction?

stira [4]

Answer:

The answer to your question is 432 g of CO₂

Explanation:

Data

CaCO₃ = 983 g

CaO = 551 g

CO₂ = ?

Balanced reaction

CaCO₃ (s) ⇒ CaO (s) + CO₂ (g)

This reaction is balanced, to solve this problem just remember the Lavoisier Law of conservation of mass that states that the mass of the reactants is equal to the mass of the products.

Mass of reactants = Mass of products

Mass of CaCO₃ = Mass of CaO + Mass of CO₂

Solve for CO₂

Mass of CO₂ = Mass of CaCO₃ - Mass of CaO

Mass of CO₂ = 983 g - 551 g

Simplification

Mass of CO₂ = 432 g

5 0

3 years ago

An -ate or -ite at the end of a compound name ususally indiates that the compound contains...

Zarrin [17]

Answer: Oxygen.

Explanation: The -ate is used for the ion that has the largest number of Oxygen atoms. The -ite would be used for the ion with the smaller amount of oxygen atoms.

4 0

3 years ago