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eduard
3 years ago
8

Which has greater kinetic energy: a bullet that has a mass of 0.05kg travelling at 2,500m/s, or a lorry that has a mass of 4,500

kg and is moving at 50m/s?
lorry
bullet
Chemistry
2 answers:
Vladimir [108]3 years ago
6 0

Answer:

the answer will be lorry

FromTheMoon [43]3 years ago
6 0

Answer:

I'm really sorry man but I'll answer yours when I see you again

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If the temperature of a gas increased and the pressure remains the same what happens to the volume of gas?​
podryga [215]

Answer:

the answer is B.

Explanation:

The pressure increases... plus i had this same question

pls give me brainliest

3 0
3 years ago
An ideal gas is a gas. <br> Perfect<br> Theoretical<br> Real
dusya [7]

Answer:

An ideal gas is a theoretical gas composed of many randomly moving point particles that are not subject to interparticle interactions. The ideal gas concept is useful because it obeys the ideal gas law, a simplified equation of state, and is amenable to analysis under statistical mechanics.

5 0
3 years ago
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Assume the following potential energies for a Newman projection: anti Me/H or anti Me/Me = 0 kcal/mole; eclipsed H/H = 4 kcal/mo
kap26 [50]

Answer:

0 kcal/mole

Explanation:

The most stable Newman projection corresponds to the minimum potential energy: Answer (e) 0 kcal/mole

4 0
3 years ago
Which statement is the best example of pseudoscience?
pentagon [3]
Mary thinking her astrological sign, Leo, is a perfect fit for her personality is a belief in pseudoscience because there is not scientific evidence confirming such beliefs.
The answer is the second option.
8 0
3 years ago
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The normal boiling point of bromine is 58.8°C, and its enthalpy of vaporization is 30.91 kJ/mol. What is the approximate vapor p
saul85 [17]

Answer : The vapor pressure of bromine at 10.0^oC is 0.1448 atm.

Explanation :

The Clausius- Clapeyron equation is :

\ln (\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}\times (\frac{1}{T_1}-\frac{1}{T_2})

where,

P_1 = vapor pressure of bromine at 10.0^oC = ?

P_2 = vapor pressure of propane at normal boiling point = 1 atm

T_1 = temperature of propane = 10.0^oC=273+10.0=283.0K

T_2 = normal boiling point of bromine = 58.8^oC=273+58.8=331.8K

\Delta H_{vap} = heat of vaporization = 30.91 kJ/mole = 30910 J/mole

R = universal constant = 8.314 J/K.mole

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

\ln (\frac{1atm}{P_1})=\frac{30910J/mole}{8.314J/K.mole}\times (\frac{1}{283.0K}-\frac{1}{331.8K})

P_1=0.1448atm

Hence, the vapor pressure of bromine at 10.0^oC is 0.1448 atm.

4 0
3 years ago
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