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3 years ago

Can convection occur in both liquids and

Chemistry

1 answer:

Triss [41]3 years ago

6 0

Answer:

Yes, Convection can occur in both liquids and gases

Explanation:

The Particle Theory suggests that Particles are always moving. Convection occurs in the breeze you feel, when the warm particles quickly move upwards resulting in cold air quickly sinking and creating a breeze. Convection also occurs when you boil water, the molecules of warm water quickly move to the top and cold water quickly moves to the bottom resulting in a circle the constantly keeps the water circulating, making it boil.

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This is the energy that all moving objects have. 1.

Delvig [45]

Every moving object has kinetic energy.
Kinetic energy is an energy possessed by an object in motion

4 0

4 years ago

The following properties are either physical or chemical. Which one is different from the rest based on those two category? A.)

Oduvanchick [21]

Answer:

A.)physical B.)physical C.)physical D.)chemical

5 0

4 years ago

Read 2 more answers

Select all possible answers:

Dafna11 [192]

Answer:

c and d are correct

Explanation:

In A, false because in Valence Electrons, the more the valences, the more stable an atom is.

In B, false because atoms cannot readily gain or lose valence electrons as the number of valence electrons is determined by the column they are in.

In C, true because the more the valence electrons, the more the stability of an atom.

In D, true as electron placing is important and the reactivity of an atom is important.

So C and D are true!

7 0

3 years ago

Read 2 more answers

Given that the initial rate constant is 0.0110s−1 at an initial temperature of 21 ∘C , what would the rate constant be at a temp

gulaghasi [49]

The rate constant is mathematically given as

K2=2.67sec^{-1}

<h3>What is the Arrhenius equation?</h3>

The rate constant for a particular reaction may be calculated with the use of the Arrhenius equation. This constant can be stated in terms of two distinct temperatures, T1 and T2, as follows:

$ln(\frac{K2}{K1})= (\frac{Ea}{R})*(\frac{1}{T1}-\frac{1}{T2})$

Therefore

KT1= 0.0110^{-1}

T1= 21+273.15

T1= 294.15K

T2= 200

T2=200+273.15

T2= 473.15K

Ea= 35.5 Kj/Mol

Hence, in j/mol R Ea is

Ea=35.5*1000 j/mol R

$ln(\frac{K2}{0.0110})= (\frac{35.5*1000}{8.314})*(\frac{1}{294.15}-\frac{1}{473.15}\\\\ln(\frac{K2}{0.0110})=5.492$

K2/0.0110 =e^(5.492)

K2/0.0110 =242.74

K2= 242.74*0.0110

K2=2.67sec^{-1}

In conclusion, rate constant

K2=2.67sec^{-1}