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serg [7]
3 years ago
9

Two samples of carbon come into contact. A heat transfer will occur between sample A and sample B. What must be true for heat to

transfer from sample A to sample B?
The average kinetic energy of A is greater than that of B.
The average kinetic energy of B is greater than that of A.
The average kinetic energy of both samples is equal.
The average kinetic energy does not determine the direction of heat transfer.
Chemistry
2 answers:
jarptica [38.1K]3 years ago
8 0

Answer:

The average kinetic energy of A is greater than that of B.

Explanation:

The temperature of an object is directly proportional to the average kinetic energy of the particles in the object. For instance, for an ideal gas, we have

KE=\frac{3}{2}kT

where

KE is the kinetic energy

k is the Boltzmann constant

T the absolute temperature of the gas

Therefore, this means that in a hotter object the average kinetic energy of the particles is higher than the average kinetic energy of the particles in a colder object.

Moreover, the laws of thermodynamics tell us that heat is always transferred from a hotter object (higher temperature) to a colder object (lower temperature).

In this problem heat is transferred from sample A to sample B. Therefore, this means that object A has higher temperature, and therefore, higher average kinetic energy. So the correct answer is

The average kinetic energy of A is greater than that of B.

Alekssandra [29.7K]3 years ago
6 0

Answer:

A.The average kinetic energy of A is greater than that of B.

Explanation:

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A sample of gas contains 6.25 × 10-3 mol in a 500.0 mL flask at 265°C. What is the pressure of the gas in kilopascals? Which var
RideAnS [48]

55.9 kPa; Variables given = volume (V), moles (n), temperature (T)

We must calculate <em>p</em> from <em>V, n</em>, and <em>T</em>, so we use <em>the Ideal Gas Law</em>:

<em>pV = nRT</em>

Solve for <em>p</em>: <em>p = nRT/V</em>

R = 8.314 kPa.L.K^(-1).mol^(-1)

<em>T</em> = (265 + 273.15) K = 538.15 K

<em>V</em> = 500.0 mL = 0.5000 L

∴ <em>p</em> = [6.25 x 10^(-3) mol x 8.314 kPa·L·K^(-1)·mol^(-1) x 538.15 K]/(0.5000 L) = 55.9 kPa

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3 years ago
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aseous methane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 5.5 g of methane is
maksim [4K]

Answer:

There is 9.6 grams of CO2 produced

Explanation:

Step 1: Data given

Mass of methane = 5.50 grams

Molar mass of methane = 16.04 g/mol

Mass of oxygen = 13.9 grams

Molar mass of oxygen = 32.0 g/mol

Step 2: The reaction

CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)

Step 3: Calculate number of moles

Moles = mass / molar mass

Moles methane = 5.50 grams / 16.04 g/mol

Moles methane = 0.343 moles

Moles oxygen = 13.9 grams / 32.0 g/mol

Moles oxygen = 0.434 moles

For 1 mol CH4 we need 2 moles O2 to produce 1 mol CO2 and 2 moles H2O

O2 is the limiting reactant. It will completely react (0.434 moles).

There will react 0.434/2 = 0.217 moles CH4

There will remain 0.343-0.217 = 0.126 moles CH4

There will be produced 0.434 moles of H2O and

0.434/2 =0.217 moles of CO2

Step 4: Calculate mass of products

Mass = moles * molar mass

Mass CO2 = 0.217 moles ¨44.01 g/mol

Mass CO2 = 9.6 grams

Mass H2O = 0.434 moles * 18.02

Mass H2O = 7.8 grams

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3 years ago
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5 0
3 years ago
CaBr + KOH – Ca(OH), + KBr (balance first) What mass, in grams, of
neonofarm [45]

Answer:

129.73 g of CaBr₂

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

CaBr₂ + 2KOH –> Ca(OH)₂ + 2KBr

Next, we shall determine the mass of CaBr₂ that reacted and the mass of Ca(OH)₂ produced from the balanced equation. This can be obtained as follow:

Molar mass of CaBr₂ = 40 + (80×2)

= 40 + 160

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Mass of CaBr₂ from the balanced equation = 1 × 200 = 200 g

Molar mass of Ca(OH)₂ = 40 + 2(16 + 1)

= 40 + 2(17)

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Mass of Ca(OH)₂ from the balanced equation = 1 × 74 = 74 g

SUMMARY :

From the balanced equation above,

200 g of CaBr₂ reacted to produce 74 g of Ca(OH)₂.

Finally, we shall determine the mass of CaBr₂ that react when 48 g of Ca(OH)₂ were produced. This can be obtained as follow:

From the balanced equation above,

200 g of CaBr₂ reacted to produce 74 g of Ca(OH)₂.

Therefore, Xg of CaBr₂ will react to produce 48 g of Ca(OH)₂ i.e

Xg of CaBr₂ = (200 × 48)/74

Xg of CaBr₂ = 129.73 g

Thus, 129.73 g of CaBr₂ were consumed.

6 0
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