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3 years ago

Witch two conditions can limit the usefulness of the kinetic molecular theory in describing gas behavior?

Chemistry

2 answers:

zhuklara [117]3 years ago

6 0

Which two conditions can limit the usefulness of the kinetic-molecular theory in describing gas behavior? ... Increasing pressure is causing a smaller change in gas volume. The closeness of particles of gas and their low speeds allow intermolecular forces to become important at certain pressured and temperature.

qwelly [4]3 years ago

6 0

I believe its C Low pressure and low temps

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An empty beaker weighs 25.91 g. When completely filled with water, the beaker and its contents have a total mass of 333.85 g. Wh

Finger [1]

Answer:

The beaker holds 307.94 mL

Explanation:

As we know that the volume that beaker hold is the volume of water that occupied by it.

For this first we have to find mass of the water in the beaker

This can be calculated by the subtraction of beaker's weight from the weight of beaker and water.

weight of water (m) = total weight - weight of beaker

Empty weight of beaker = 25.91 g

Weight of beaker with water = 333.85 g

Weight of water = 333.85 - 25.91 = 307.94 g

Density of water = 1 g/mL

We have

Mass = Volume x density

307.94 = Volume x 1

Volume = 307.94 mL

The beaker holds 307.94 mL

3 0

2 years ago

1. Marisa determined the melting point of a substance to be 24.5C. Find the percent error of her measurement if the actual melti

USPshnik [31]

Answer:

$\%\ Error = 21.5\%$

Explanation:

Given

$Measured = 24.5$

$Actual = 31.2$

Required

Determine the percentage error

First, we need to determine the difference in the measurement

$Difference = |Actual - Measured|$

$Difference = |31.2 - 24.5|$

$Difference = |6.7|$

$Difference = 6.7$

The percentage error is calculated as thus:

$\%\ Error = \frac{Difference * 100\%}{Actual}$

$\%\ Error = \frac{6.7 * 100\%}{31.2}$

$\%\ Error = \frac{670\%}{31.2}$

$\%\ Error = 21.4743589744\%$

$\%\ Error = 21.5\%$ <em>approximated</em>

6 0

3 years ago

One difference between molecular formula and molecular weight.

jek_recluse [69]

Answer: The formula mass (formula weight) of a molecule is the sum of the atomic weights of the atoms in its empirical formula. The molecular mass (molecular weight) of a molecule is its average mass as calculated by adding together the atomic weights of the atoms in the molecular formula.

Hope this helps.... Stay safe and have a great weekend!!!!! :D

3 0

2 years ago

What is the molality of a solution made by dissolving 137.9g of sucrose in 414.1g of water?

Daniel [21]

Answer: 2.71 moles of solute for every 1 kg of solvent.

Explanation: As you know, the molality of a solution tells you the number of moles of solute present for every 1 kg of the solvent.This means that the first thing that you need to do here is to figure out how many grams of water are present in your sample. To do that, use the density of water.500.mL⋅1.00 g1mL=500. g Next, use the molar mass of the solute to determine how many moles are present in the sample.115g⋅1 mole NanO385.0g=1.353 moles NaNO3So, you know that this solution will contain 1.353moles of sodium nitrate, the solute, for 500. g of water, the solvent.In order to find the molality of the solution, you must figure out how many moles of solute would be present for 1 kg=103g of water.103g water⋅1.353 moles NaNO3500.g water=2.706 moles NaNO3You can thus say that the molality of the solution is equal to molality=2.706 mol kg−1≈2.71 mol kg−1 The answer is rounded to three sig figs.

8 0

2 years ago

For a particular reaction at 235.8 °C, ΔG=−936.92 kJ/mol , and ΔS=513.79 J/(mol⋅K) . Calculate ΔG for this reaction at −9.9 °C.

Rudik [331]

Answer:

-138.9 kJ/mol

Explanation:

Step 1: Convert 235.8°C to the Kelvin scale

We will use the following expression.

K = °C + 273.15 = 235.8°C + 273.15 = 509.0 K

Step 2: Calculate the standard enthalpy of reaction (ΔH°)

We will use the following expression.

ΔG° = ΔH° - T.ΔS°

ΔH° = ΔG° / T.ΔS°

ΔH° = (-936.92kJ/mol) / 509.0K × 0.51379 kJ/mol.K

ΔH° = -3.583 kJ (for 1 mole of balanced reaction)

Step 3: Convert -9.9°C to the Kelvin scale

K = °C + 273.15 = -9.9°C + 273.15 = 263.3 K

Step 4: Calculate ΔG° at 263.3 K

ΔG° = ΔH° - T.ΔS°

ΔG° = -3.583 kJ/mol - 263.3 K × 0.51379 kJ/mol.K

ΔG° = -138.9 kJ/mol

8 0

2 years ago