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Tamiku [17]
3 years ago
9

Which observation indicates that the kinetic-molecular theory has limited use for describing a certain gas? Gas particles are ac

ting like tiny, solid spheres. Gas particles are obeying Newton’s laws of motion. Increasing pressure is reducing the volume of the gas. Increasing collisions of gas molecules will increase energy between them.
Chemistry
2 answers:
klasskru [66]3 years ago
8 0

Answer:

D. Increasing collisions of gas molecules will increase energy between them.

Explanation:

soldier1979 [14.2K]3 years ago
7 0

My answer to the question is "Gas particles are acting like tiny,solid spheres".

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Explain the concept of "like dissolves like"
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Answer:

Like dissolves like" is an expression used by chemists to remember how some solvents work. It refers to "polar" and "nonpolar" solvents and solutes. Basic example: Water is polar. Oil is non polar. ... Like dissolves like, that means polar dissolves polar, so water dissolves salt.

Explanation:

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Describe several main ideas in chemistry
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Matter.

A force of attraction that holds atom together 
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3 years ago
Compare the life cycles of a flowering and a conifer non-flowering plant.
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The main difference between flowering and nonflowering plants is their method of reproduction. Flowering plants rely on pollination for reproduction, where as nonflowering plants rely on dispersion to continue their life cycle.

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2 years ago
Describe the function of a cilia.
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Answer:

hope it helps..

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Read 2 more answers
When the volume of a gas is
Montano1993 [528]

Answer:

\boxed {\boxed {\sf 232.9 \textdegree C}}

Explanation:

This question asks us to find the temperature change given a volume change. We will use Charles's Law, which states the volume of a gas is directly proportional to the temperature. The formula is:

\frac {V_1}{T_1}= \frac{V_2}{T_2}

The volume of the gas starts at 250 milliliters and the temperature is 137 °C.

\frac{250 \ mL}{137 \textdegree C}= \frac{V_2}{T_2}

The volume of the gas is increased to 425 milliliters, but the temperature is unknown.

\frac{250 \ mL}{137 \textdegree C}= \frac{425 \ mL}{T_2}

We are solving for the new temperature, so we must isolate the variable T₂. First, cross multiply. Multiply the first numerator and second denominator, then multiply the first denominator and second numerator.

250 \ mL * T_2 = 137 \textdegree C * 425 \ mL

Now the variable is being multiplied by 250 milliliters. The inverse of multiplication is division. Divide both sides of the equation by 250 mL.

\frac{250 \ mL * T_2}{250 \ mL}=\frac{ 137 \textdegree C * 425 \ mL}{250 \ mL}

T_2=\frac{ 137 \textdegree C * 425 \ mL}{250 \ mL}

The units of milliliters (mL) cancel.

T_2=\frac{ 137 \textdegree C * 425 }{250 }

T_2= \frac{58225}{250} \textdegree C

T_2=232.9 \textdegree C

The temperature changes to <u>232.9 degrees Celsius.</u>

3 0
2 years ago
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