The ideal gas law tends to become inaccurate when:

Lighting a match is an example of what kind of energy conversion?

Alika [10]

Answer:

Mechanical Energy to Thermal Energy

When you strike a match, it moves through the air until it rubs against a surface. The rubbing produces the heat required to light the match. This is a transformation from mechanical energy to thermal (heat) energy.

Explanation:

4 0

2 years ago

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A researcher is measuring changes in the density of the upper atmosphere over time. Which units should be used for the density m

valina [46]

Answer:

lbm/ft³ or psi/ft

Explanation:

The description in the question shows the density to measured is that of gases (upper atmosphere). The density of a gas can be described as the mass of the gas divided by the volume of the gas at a certain temperature and pressure. It's unit can be represented as lbm/ft³ or psi/ft.

NOTE:

"lbm" means "lean body mass" which is a unit for mass or weight and can be described as the pound mass

"psi" means "pound per square inch"; is also a unit for mass or weight

5 0

3 years ago

quien y en que cantidad sera el reactivo limite, si utilizamos 125 g de ácido, H (CH3COO) y 275 g de hidróxido , Al(OH)3

Deffense [45]

Answer:

The limiting reactant is acetic acid. All 125 g will react.

Explanation:

1. Assemble the information

We will need a balanced equation with masses and molar masses, so let’s gather all the information in one place.

Mᵣ: 60.05 78.00

3CH₃COO-H + Al(OH)₃ ⟶ (CH₃COO)₃Al + 3H₂O

Mass/g: 125 275

2. Calculate the moles of each reactant

$\text{Moles of CH$_{3}$COOH} = \text{125 g CH$_{3}$COOH} \times \dfrac{\text{1 mol CH$_{3}$COOH}}{\text{60.05 g CH$_{3}$COOH}} = \text{2.082 mol CH$_{3}$COOH }\\\\\text{Moles of Al(OH)}_{3} = \text{275 g Al(OH)}_{3} \times \dfrac{\text{1 mol Al(OH)}_{3}}{\text{78.00 g Al(OH)}_{3}} = \text{3.526 mol Al(OH)}_{3}$

3. Calculate the moles of (CH₃COO)₃Al from each reactant

$\textbf{From CH$_{3}$COOH:}\\\text{Moles of (CH$_{3}$COO)$_{3}$Al} = \text{2.082 mol CH$_{3}$COOH} \times \dfrac{\text{1 mol (CH$_{3}$COO)$_{3}$Al}}{\text{3 mol CH$_{3}$COOH}}\\\\= \text{0.6939 mol (CH$_{3}$COO)$_{3}$)Al}\\\textbf{From Al(OH)}_{3}:\\\text{Moles of (CH$_{3}$COO)$_{3}$Al } =\text{3.526 Al(OH)}_{3} \times \dfrac{\text{1 mol (CH$_{3}$COO)$_{3}$Al }}{\text{1 mol Al(OH)}_{3}}\\\\= \text{3.526 mol (CH$_{3}$COO)$_{3}$Al}$

$\text{Acetic acid is the $\textbf{limiting reactant}$ because it gives fewer moles of} \\\text{(CH$_{3}$COO)$_{3}$Al. All $\textbf{125 g}$ will react.}$

5 0

3 years ago

What are the two main reasons why most metals are more dense than most plastic and wood?

Viktor [21]

Answer:

Most common metals like aluminum, copper, and iron are denser than wood. The atoms that makeup metals are generally heavier than the atoms in plastic and wood and they are packed closer together.

3 0

3 years ago

in a plant 1.500kg of nitrogen oxide is consumed per day to produce 1.500 kg of nitrogen dioxide per day what is the percent yie

mafiozo [28]

Answer: The answer is 100%. Hope this helps :)

Explanation:

8 0

3 years ago