The dissociation of a/an ________ releases hydrogen ions and increases the concentration of hydrogen ions in a solution.

Explain how mixture melting points can be helpful in distinguishing between two compounds that have essentially the same melting

grandymaker [24]

Ice and Vaseline. Those are two melting compounds. When melting ice it turns into water and when melting Vaseline it melts into a oily liquid form. You can also freeze both of them and you can melt them again.

4 0

3 years ago

The diameter of the He He atom is approximately 0.10 nm nm . Calculate the density of the He atom in g/cm 3 g/cm3 (assuming that

Sladkaya [172]

Answer:

Density of the He atom = 12.69 g/cm³

Explanation:

From the information given:

Since 1 mole of an atom = 6.022x 10²³ atoms)

1 atom of He = $1 \ atom \times (\dfrac{1 \ mole}{ 6.022 \times 10^{23} \ atoms}) \times ( \dfrac{4.003 \ grams}{ 1 \ mole})$

$=6.647 \times 10^{-24} \ grams$

The volume can be determined as folows:

since the diameter of the He atom is approximately 0.10 nm

the radius of the He = $\dfrac{0.10}{2}$ = 0.05 nm

Converting it into cm, we have:

$0.05 nm \times \dfrac{10^{-9} \ meters}{ 1 nm} \times \dfrac{ 1 cm }{10^{-2} \ meters}$

$=5 \times 10^{-9} \ cm$

Assuming that it is a sphere, the volume of a sphere is

= $\dfrac{4}{3}\pi r^3$

= $\dfrac{4}{3}\pi \times (5\times 10^{-9})^3$

= $5.236 \times 10^{-25} \ cm^3$

Finally, the density can be calcuated by using the formula :

$Density = \dfrac{mass}{volume}$

$D = \dfrac{6.647 \times 10^{-24} \ grams }{ 5.236 \times 10^{-25} \ cm^3}$

D = 12.69 g/cm³

Density of the He atom = 12.69 g/cm³

4 0

3 years ago

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NEED HELP!<br> Please refer to the picture for the question!

Mashutka [201]

Answer: i think the best bet i can give you is Option C (2:3)

Explanation: i apologize i haven't done chem in 2 years

but however to put it in retrospect the finished equation is 2(AL)^+3 3(O)^-2

4 0

3 years ago

A method used by the U.S. Environmental Protection Agency (EPA) for determining the concentration of ozone in air is to pass the

baherus [9]

Answer: 1. $9.08\times 10^{-6}$ moles

2. 90 mg

Explanation:

$O_3(g)+2NaI(aq)+H_2O(l) \rightarrow O_2(g)+I_2(s)+2NaOH(aq)$

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus $4.54 \times 10^{-6}$ moles of ozone is removed by = $\frac{2}{1}\times 4.54 \times 10^{-6}=9.08\times 10^{-6}$ moles of sodium iodide.

Thus $9.08\times 10^{-6}$ moles of sodium iodide are needed to remove $4.54\times 10^{-6}$ moles of $O_3$

2. $\text{Number of moles of ozone}=\frac{0.01331g}{48g/mol}=0.0003moles$

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus 0.0003 moles of ozone is removed by = $\frac{2}{1}\times 0.0003=0.0006$ moles of sodium iodide.

Mass of sodium iodide= $moles\times {\text {molar mass}}=0.0006\times 150g/mol=0.09g=90mg$ (1g=1000mg)

Thus 90 mg of sodium iodide are needed to remove 13.31 mg of $O_3$ .

3 0

3 years ago

Many scientists believe that the current period of global warming is related to increased levels of

OlgaM077 [116]

Option B is your answer.

5 0

3 years ago

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