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In the background information, it was stated that CaF2 has solubility, at room temperature, of 0.00160 g per 100 g of water. How

____ [38]

Answer:

2.05*10⁻⁵ moles of CF₂ can dissolve in 100 g of water.

12.82 moles of CaF₂ will dissolve in exactly 1.00 L of solution

Explanation:

First, by definition of solubility, in 100 g of water there are 0.0016 g of CaF₂. So, to know how many moles are 0.0016 g, you must know the molar mass of the compound. For that you know:

Ca: 40 g/mole
F: 19 g/mole

So the molar mass of CaF₂ is:

CaF₂= 40 g/mole + 2*19 g/mole= 78 g/mole

Now you can apply the following rule of three: if there are 78 grams of CaF₂ in 1 mole, in 0.0016 grams of the compound how many moles are there?

$moles=\frac{0.0016 grams*1 mole}{78 grams}$

moles=2.05*10⁻⁵

<u><em>2.05*10⁻⁵ moles of CF₂ can dissolve in 100 g of water.</em></u>

Now, to answer the following question, you can apply the following rule of three: if by definition of density in 1 mL there is 1 g of CaF₂, in 1000 mL (where 1L = 1000mL) how much mass of the compound is there?

$mass of CaF_{2}=\frac{1000 mL*1g}{1mL}$

mass of CaF₂= 1000 g

Now you can apply the following rule of three: if there are 78 grams of CaF₂ in 1 mole, in 1000 grams of the compound how many moles are there?

$moles=\frac{1000 grams*1 mole}{78 grams}$

moles=12.82

<u><em>12.82 moles of CaF₂ will dissolve in exactly 1.00 L of solution</em></u>

5 0

3 years ago

The following reactions have the indicated equilibrium constants at a particular temperature: N2(g) + O2(g) ⇌ 2NO(g) Kc = 4.3 ×

Anuta_ua [19.1K]

Answer:

$Kc=~1.49x10^3^4}$

Explanation:

We have the reactions:

A: $N_2_(_g_) + O_2_(_g_) 2NO_(_g_)~~~~~~Kc = 4.3x10^-^2^5$

B: $2NO_(_g_)+~O_2_(_g_)~2NO_2_(_g_)~~~Kc = 6.4x10^9$

Our <u>target reaction</u> is:

$4NO_(_g_) N_2_(_g_) + 2NO_2_(_g_)$

We have $NO_(_g_)$ as a reactive in the target reaction and $NO_(_g_)$ is present in A reaction but in the products side. So we have to<u> flip reaction A</u>.

A: $2NO_(_g_) N_2_(_g_) + O_2_(_g_) ~Kc =\frac{1}{4.3x10^-^2^5}$

Then if we add reactions A and B we can obtain the target reaction, so:

A: $2NO_(_g_) N_2_(_g_) + O_2_(_g_) ~Kc =\frac{1}{4.3x10^-^2^5}$

B: $2NO_(_g_)+~O_2_(_g_)~2NO_2_(_g_)~Kc=6.4x10^9$

For the <u>final Kc value</u>, we have to keep in mind that when we have to <u>add chemical reactions</u> the total Kc value would be the <u>multiplication</u> of the Kc values in the previous reactions.

$4NO_(_g_) N_2_(_g_) + 2NO_2_(_g_)~~~Kc=\frac{6.4x10^9}{4.3x10^-^2^5}$

$Kc=~1.49x10^+^3^4}$

3 0

3 years ago

An atom of oxygen (atomic number 8) has a mass number of 17. How many neutrons are<br> in that atom?

ludmilkaskok [199]

<h2>Answer: # of neutrons = 9</h2>

<h3>Explanation:</h3>

# of neutrons = mass number - atomic number

= 17 - 8

= 9

<h3>∴ number of neutrons in an atom of oxygen of mass number 17, is 9.</h3>

3 0

3 years ago

2. What is the molarity of 100 mL of a 3.0% H2O2 (mass/volume) solution? What is the molarity of 100 mL of a 2.25% H2O2 solution

KatRina [158]

The 3% mass/volume H₂O₂ means 3 g of H₂O₂ in 100 ml of water.

Now, Molarity (M) = No. of moles of H₂O₂ / Volume of solution in liter

No. of moles of H₂O₂ = Mass / Molar mass = 3 g / 34 g/mol = 0.088 mol

So, molarity = 0.088 × 1000 ml / 100 ml = 0.88 M

In case of 2.25 % H₂O₂,

No of moles = 2.25 g / 34 g/mol = 0.066 mol

Molarity = 0.066 mol / 0.100 L = 0.66 M.

7 0

3 years ago

Give the hybridization for the O in H3O+

Damm [24]

Answer: It would be Sp2, because H3O+ has planar structure.

4 0

3 years ago