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WITCHER [35]
3 years ago
6

Which statement about the density of cold water is most likely correct?

Chemistry
2 answers:
Tema [17]3 years ago
5 0

the answer is A) Cold water is more dense than warm water

podryga [215]3 years ago
3 0

Answer: Option (A) is the correct answer.

Explanation:

Density is defined as the mass of a substance divided by the volume of the substance.

Mathematically,    Density = \frac{mass}{volume}

In cold water, molecules are more closer to each other due to which there occurs increase in mass of the substance. As density is proportional to mass of the substance. So, density of cold water will be more.

On the other hand, in warm water molecules have more kinetic energy so, there will be more number of collisions. As a result, molecules will move away from each other due to which there will be decrease in density.

Thus, we can conclude that the statement cold water is more dense than warm water, is true about the density of cold water.

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Explanation:

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Why do you think we use heat to dissolve sodium tetraborate?
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3 years ago
Write out the word equations for these reactions.<br> ) Hydrogen reacts with oxygen to form water.
ale4655 [162]

Answer:

H2O

Explanation:

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4 0
3 years ago
The activation energy of an uncatalyzed reaction is 95kJ/mol. The addition of a catalyst lowers the activation energy to 55kJ/mo
notka56 [123]

Answer:

a) at 25°C the rate of reaction increases by a factor of 1,027*10^7

b) at 25°C the rate of reaction increases by a factor of 1,777*10^5

Explanation:

using the Arrhenius equation

k= ko*e^(-Ea/RT)

where

k= reaction rate

ko= collision factor

Ea= activation energy

R= ideal gas constant= 8.314 J/mol*K

T= absolute temperature

for the uncatalysed reaction

k1= ko*e^(-Ea1/RT)

for the catalysed reaction

k2= ko*e^(-Ea2/RT)

dividing both equations

k2/k1= e^(-(Ea2-Ea1)/RT)

a) at 25°C

k2/k1 = e^(-(55kJ/mol-95kJ/mol)/(8.314J/mol*K*298K)* (1000J/kJ ) ) = 1,027*10^7

therefore at 25°C , k2/k1 = 1,027*10^6

b) at 125°C

k2/k1 = e^(-(55kJ/mol-95kJ/mol)/(8.314J/mol*K*298K)* (1000J/kJ ) ) = 1,777*10^5

therefore at 125°C , k2/k1 = 1,777*10^5

Note:

when the catalysts is incorporated, the catalysed reaction and the uncatalysed one run in parallel and therefore the real reaction rate is

k real = k1 + k2 = k2 (1+k1/k2)

since k2>>k1 → 1+k1/k2 ≈ 1 and thus k real ≈ k2

6 0
3 years ago
Sodium hydride racts with excess water to produce
romanna [79]

Answer:

0.96g of sodium hydride

Explanation:

Equation of reaction:

NaH + H20 = NaOH + H2

Mass of hydrogen gas produced (m) = PVM/RT

P = 765torr - 28torr = 737torr = 737/760 = 0.97atm, V = 982mL = 982cm^3, M = 2g/mol, R = 82.057cm^3.atm/gmol.K, T = 28°C = 28 + 273K = 301K

m = (0.97×982×2)/(82.057×301) = 0.08g of hydrogen gas

From the equation of reaction

1 mole (24g) of sodium hydride produced 1 mole (2g) of hydrogen gas

0.08g of hydrogen gas would be produced by (24×0.08)/2 = 0.96g of sodium hydride

8 0
3 years ago
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