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3 years ago

What happens to the average kinetic energy of the particles in a sample of matter as the temperature of the sample is increased?

Chemistry

1 answer:

antiseptic1488 [7]3 years ago

7 0

Answer:

As the temperature of a sample of matter is increased, the average kinetic energy of the particles in the sample increase.

Explanation:

The temperature of a substance is the measure of the average kinetic energy of its partilces.

The temperature, i.e. how hot or cold is a substance, is the result of the collisions of the particles (atoms or molecules) of matter.

The kinetic theory of gases states that, if the temperature is the same, the average kinetic energy of any gas is the same, regardless the gas and other conditions.

This equation expresses it:

Avg KE = (3/2) (R / N) T

Where Avg KE is the average kinetic energy, R is the universal constant of gases, N is Avogadro's constnat, and T is the temperature measure in absolute scale (Kelvin).

As you see, in that equation Avg KE is propotional to T, which means that as the temperature is increased, the average kinetic energy increases.

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How many moles of oxygen atoms are in 7.9E-1 moles of CO_2

Ilya [14]

Answer:

The number of moles of O atom in $(7.9\times10^{-1})$ mol of $CO_{2}$ = 1.6

Explanation:

1 molecule of $CO_{2}$ contains 2 atoms of O

So, $(6.023\times 10^{23})$ molecules of $CO_{2}$ contains $(2\times6.023\times10^{23})$ atoms of O.

We know that 1 mol of an atom/molecule/ion represents $6.023\times10^{23}$ numbers of atoms/molecules/ions respectively.

So, $(6.023\times 10^{23})$ molecules of $CO_{2}$ is equal to 1 mol of $CO_{2}$ .

$(2\times6.023\times10^{23})$ atoms of O is equal to 2 moles of O atom.

Hence, 1 mol of $CO_{2}$ contains 2 moles of O atom.

Therefore, $(7.9\times10^{-1})$ mol of $CO_{2}$ contains $(2\times7.9\times10^{-1})$ moles of O atom or 1.6 moles of O atom.

3 0

3 years ago

6CO2 + 6 H2O —> C6H12O6 + 6O2

PilotLPTM [1.2K]

Answer:

The answer to your question is 0.5 moles

Explanation:

Data

moles of Glucose = ?

moles of carbon dioxide = 3

Balanced chemical reaction

6CO₂ + 6H₂O ⇒ C₆H₁₂O₆ + 6O₂

Process

To solve this problem, use proportions, and cross multiplication.

Use the coefficients of the balanced equation.

6 moles of CO₂ ----------------- 1 mol of C₆H₁₂O₆

3 moles of CO₂ ---------------- x

x = (3 x 1) / 6

-Simplification

x = 3/6

-Result

x = 0.5 moles of Glucose

3 0

3 years ago

Which of these chemicals is definitely INorganic?

igor_vitrenko [27]

C. inorganic does not contain carbon

7 0

3 years ago

Read 2 more answers

Which of the following elements is a transition metal?

GenaCL600 [577]

Answer:

Scandium

Titanium

Vanadium

Chromium

Manganese

Iron

Cobalt

Nickel

Copper

Zinc

Yttrium

Zirconium

Niobium

Molybdenum

Technetium

Ruthenium

Rhodium

Palladium

Silver

Cadmium

Lanthanum

Hafnium

Tantalum

Tungsten

Rhenium

Osmium

Iridium

Platinum

Gold

Mercury

Actinium

Rutherfordium

Dubnium

Seaborgium

Bohrium

Hassium

Meitnerium

Darmstadtium

Roentgenium

Copernicium

Explanation:

all of those are transition metals lol

5 0

3 years ago

Read 2 more answers

A sample of Xe gas is observed to effuse through a pourous barrier in 4.83 minutes. Under the same conditions, the same number o

Solnce55 [7]

Answer:

28.93 g/mol

Explanation:

This is an extension of Graham's Law of Effusion where $\frac{R1}{R2} = \sqrt{\frac{M2}{M1} } = \frac{t2}{t1}$

We're only talking about molar mass and time (t) here so we'll just concentrate on $\sqrt{\frac{M2}{M1} } = \frac{t2}{t1}$ . Notice how the molar mass and time are on the same position, recall effusion is when gas escapes from a container through a small hole. The time it takes it to leave depends on the molar mass. If the gas is heavy, like Xe, it would take a longer time (4.83 minutes). If it was light it would leave in less time, that gives us somewhat an idea what our element could be, we know that it's atleast an element before Xenon.

Let's plug everything in and solve for M2. I chose M2 to be the unknown here because it's easier to have it basically as a whole number already.

$\sqrt{\frac{M2}{131} } = \frac{2.29}{4.83}$

The square root is easier to deal with if you take it out in the first step, so let's remove it by squaring each side by 2, the opposite of square root essentially.

$(\sqrt{\frac{M2}{131} } )^2= (\frac{2.29}{4.83})^2$

${\frac{M2}{131} } = (0.47)^2$

${\frac{M2}{131} } = 0.22$

M2= 0.22 x 131

M2= 28.93 g/mol

8 0

2 years ago