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aksik [14]
3 years ago
15

What determines the average kinetic energy of particles in a gas ?

Chemistry
2 answers:
Thepotemich [5.8K]3 years ago
6 0

This in turn determines whether the substance exists in the solid, liquid, or gaseous state. Molecules in the solid phase have the least amount of energy, while gas particles have the greatest amount of energy. The temperature of a substance is a measure of the average kinetic energy of the particles.

Advocard [28]3 years ago
6 0

<u>Answer:</u> Temperature of the substance determines the average kinetic energy of particles in a gas.

<u>Explanation:</u>

Average kinetic energy is defined as the average of all the kinetic energies of the particles in a system. It is a measure of temperature of the particles in a gas.

The equation used to calculate the average kinetic energy of the particles is given as:

E=\frac{3RT}{2N_A}

where,

R = Gas constant

T = Temperature of the gas

N_A = Avogadro's number

E = Average kinetic energy of the particles

As, temperature is directly related to the average kinetic energy of the gas. Thus, the temperature of the particles increases, the average kinetic energy of the particles also increases and vice-versa.

Thus, temperature of the substance determines the average kinetic energy of particles in a gas.

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Please help with Chemistry! Very urgent! I’ll give you 40 points
kvv77 [185]

Answer: 3. No displacement, zinc is most reactive.

4. Calcium Chloride, Calcium is most reactive.

5. No displacement, Copper is most reactive

6. No displacement, Calcium is most reactive

7. Hydrogen Oxide, Hydrogen is most reactive

8. Carbon oxide, Carbon is most reactive

9. No displacement, Aluminum is most reactive

10. Potassium Kryptide + Lead, no displacement, Potassium is most reactive.

5 0
3 years ago
if 1.386 g of mg ribbon combusts to form 2.309 g of oxide product, calculate the experimental mass percent of oxygen from this d
Pavlova-9 [17]

1.386 g of Mg ribbon combusts to form 2.309 g of oxide product. The mass percent of oxygen in the oxide is 40.0 %.

Let's consider the reaction for the combustion of Mg.

Mg + 1/2 O₂ ⇒ MgO

1.386 g of Mg combusts to form 2.309 g of MgO. We want to determine the mass of oxygen in MgO. According to Lavoisier's law of conservation of mass, matter is not created nor destroyed over the course of a chemical reaction. Then, the mass of Mg in the reactants is equal to the mass of Mg in MgO. The mass of the magnesium oxide is the sum of the masses of magnesium and oxygen. The <u>mass of oxygen in the oxide</u> is:

mMgO = mMg + mO\\mO = mMgO - mMg = 2.309 g - 1.386 g = 0.923 g

We can calculate the mass percent of O in MgO using the following expression.

\% O = \frac{mO}{mMgO} \times 100\% = \frac{0.923 g}{2.309g} \times 100\%  = 40.0 \%

You can learn more about mass percent here: brainly.com/question/14990953

3 0
2 years ago
What element is Column 1, period 3
Finger [1]

Answer:

Sodium (Na)

Explanation:

The element on the periodic table at Column (group) 1, period 3 is Sodium (Na)

3 0
3 years ago
onvert the value of Kc to a value of Kp for the following reaction: N2(g)+3H2(g)⇌2NH3(g) Kc = 0.50 at 400 °C.
ale4655 [162]

Answer:

K_p= 0.00016

Explanation:

The relation between Kp and Kc is given below:

K_p= K_c\times (RT)^{\Delta n}

Where,  

Kp is the pressure equilibrium constant

Kc is the molar equilibrium constant

R is gas constant

T is the temperature in Kelvins

Δn = (No. of moles of gaseous products)-(No. of moles of gaseous reactants)

For the first equilibrium reaction:

N_2_{(g)}+3H_2_{(g)}\rightleftharpoons2NH_3_{(g)}

Given: Kc = 0.50

Temperature = 400^oC=[400+273]K=673K

R = 0.082057 L atm.mol⁻¹K⁻¹

Δn = (2)-(3+1) = -2

Thus, Kp is:

K_p= 0.50\times (0.082057\times 673)^{-2}

K_p= 0.00016

6 0
3 years ago
Which parts of sedimentary rock formation include the breakdown and carrying away of existing rock? Check all that apply.
stiks02 [169]

Answer

Weathering and erosion are two processes of which sedimentary rocks are broken down and carry away existing rocks.

Explanation:

3 0
3 years ago
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