Just need help with these two questions

To me it looks like gas.
Gas- CORRECT!
Solid- incorrect because if this were a depiction of a solid then the particles would be much much closer together.
Liquid - incorrect because the particles would be a little more crowded together.
Plasma- I do not think that this answer is correct because, even though the spacing of particles is similar to the spacing of particles in plasma, the particles in plasma are charged particles(nuclei (+) and electrons(-)). Therefore, because the diagram doesn't have the particles labeled with charges, I believe that Plasma cannot be the correct answer.

Any questions?

7 0

3 years ago

How much faster will ammonia vapor travel across a room then carbon dioxide

FromTheMoon [43]

Answer:

D) 1.61 times faster

Explanation:

$vrms_{NH3}$ = √(3)RTM

R constant= 0.08206

T=constant, so in this problem we dont need a value for it

M=17.031 g/mol

√(3)(0.08206)(17.031)= 2.047

$vrms_{CO2}$ = √(3)RTM

R constant= 0.08206

T=constant, so in this problem we dont need a value for it

M= 44.01 g/mol

√(3)(0.08206)(44.01)= 3.29

Since we are trying to measure how much faster NH3 will be, we have to realize that mass and speed have an inverse relationship.

So instead of doing (2.047)/(3.29) = 6.22

we have to flip the values to get (3.29)/(2.047)= 1.61

6 0

2 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