Calculate the partial pressure of each gas and the total pressure if the temperature of the gas is 21 ∘C∘C. Express the pressures in atmospheres to three significant digits separated by commas.

Answers

Answer 1

Answer:

The question is incomplete, complete question is ;

A deep-sea diver uses a gas cylinder with a volume of 10.0 L and a content of 51.8 g of $O_2$ and 33.1 g of He. Calculate the partial pressure of each gas and the total pressure if the temperature of the gas is 21°C.Express the pressures in atmospheres to three significant digits separated by commas.

Answer:

Partial pressure of the oxygen gas is 3.91 atm.

Partial pressure of the helium gas is 20.0 atm

Total pressure of the gases is 24.0 atm

Explanation:

Moles of oxygen gas = $n_1=(51.8)/(32 g/mol)=1.619 mol$

Moles of helium gas = $n_2=(33.1 g)/(4 g/mol)=8.275 mol$

Total moles of gas = $n_1+n_2=(1.619 +8.275 ) mole=9.894 mol$

Volume of the cylinder = V = 10.0 L

Total pressure in the cylinder = P = ?

Temperature of the gas in cylinder = T = 21°C = 21 + 273 K = 294 K

PV = nRT ( ideal gas equation )

$P=(nRT)/(V)$

$=(9.894 mol* 0.0821 atm L/mol K* 294 K)/(10.0 L)$

P = 23.88 atm ≈ 23.9

Partial pressure of the individual gas will be determined by the help of Dalton's law:

partial pressure = Total pressure × mole fraction of gas

Partial pressure of the oxygen gas

$p_(1)=P* \chi_(1)=P* (n_1)/(n_1+n_2)$

$p_1=23.88 atm* (1.619 mol)/(9.894 mol)=3.91 atm$

Partial pressure of the helium gas

$p_(2)=P* \chi_(2)=P* (n_2)/(n_1+n_2)$

$p_2=23.88 atm* (8.275 mol)/(9.894 mol)=19.97 atm\approx 20.0 atm$

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Refer to the following balance equation : 2c2h6 + 7o2 = 4co2 + 6 h20. How many moles of c2h6 will combust completely with 2.0 moles of oxygen gas. How many grams of water will be produced when 30.0 g of c2h6 reacts completely with oxygen?

Consider two aqueous solutions of NaCl. Solution 1 is 4.00 M and solution 2 is 0.10 M. In what ratio (solution 1 to solution 2) must these solutions be mixed in order to produce a 0.86 M solution of NaCl

Answers

Answer:

The ratio of solution 1 to solution 2 is 24.20 to 100.00.

Explanation:

We will mix V₁ (L) of solution 1 with V₂ (L) of solution 2 to get the final solution.

So the mole concentration in the final solution is calculated as below, note that C₁ is the concentration of solution 1, and C₂ is the concentration of solution 2

$[M] = (V_(1) C_(1) +V_(2)C_(2))/(V_(1)+V_(2)) = \frac{4 V_(1) + 0.1 V_(2)}_{V_(1)+V_(2)}}=0.86$

Then we can calculate for the ratio

$(V_(1))/(V_(2))=(0.86-0.10)/(4.00-0.86) =(0.76)/(3.14)$ or $(24.20)/(100.00)$

4-methyl-3-hexanol was prepared by reacting an alkene with either hydroboration-oxidation or oxymercuration-reduction. Draw the structure of the alkene that was used to prepare the alcohol in highest yield.

Answers

Answer:

Structure in attachment.

Explanation:

The oxymercuration-demercuration of an asymmetric alkene usually produces the Markovnikov orientation of an addition. The electrophile ⁺Hg(OAc), formed by the electrophile attack of the mercury ion, remains attached to least substituted group at the end of the double bond. This electrophile has a considerable amount of positive charge on its two carbon atoms, but there is more positive charge on the more substituted carbon atom, where it is more stable. The water attack occurs on this more electrophilic carbon, and the Markovnikov orientation occurs.

In hydroboration, borane adds to the double bond in one step. Boron is added to the less hindered and less substituted carbon, and hydrogen is added to the more substituted carbon. The electrophilic boron atom adds to the less substituted end of the double bond, positioning the positive charge (and the hydrogen atom) at the more substituted end. The result is a product with the anti-Markovnikov orientation.

The element oxygen has valence electrons

Answers

Answer:

it’s electron configuration is 1s^2 2s^2 2p^4. To determine valence electrons, add the outermost s and p orbitals. In an oxygen atom, 8 electrons are present. Electron present in the first shell (n=1) 2n^2=2 (1)^2=2 (1)=2.

What term is used to describe the formation of ions in an aqueous solution from a molecular compound

Answers

Answer:

Ionization

Explanation:

Molecular compounds are chemical compounds composed of discrete molecules. A molecular compound undergoes ionization when being dissolved in water and the formation of ions are being produced. For example, hydrogen chloride is a molecular compound, when it dissolves in water, ionization is being carried out, and ions are being formed.

$\mathbf{HCl \to H^+_((aq)) + Cl^-_((aq))}$

A potassium ion (K+) would most likely bond with _____.

Mg+
O
Cl-
Na+

Answers

Answer:

Chlorine

Explanation:

Answer:

That would be the negative ion: Cl-.

If 35.2 g of magnesium react with excess hydrochloric acid, how many grams of magnesium chloride are produced? Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(aq)

Answers

Answer:

139.33 g of magnesium chloride, MgCl2.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(aq)

Next, we shall determine the mass of Mg that reacted and the mass of MgCl2 from the balanced equation.

This is illustrated below:

Molar mass of Mg = 24 g/mol

Mass of Mg from the balanced equation = 1 x 24 = 24 g

Molar mass of MgCl2 = 24 + (2x35.5) = 95 g/mol

Mass of MgCl2 from the balanced equation = 1 x 95 = 95 g

From the balanced equation above,

24 g of Mg reacted to produce 95 g of MgCl2.

Finally, we shall determine the mass of MgCl2 produced by reacting 35.2 g of Mg.

This can be obtained as follow:

From the balanced equation above,

24 g of Mg reacted to produce 95 g of MgCl2.

Therefore, 35.2 g of Mg will react to produce = (35.2 x 95)/24 = 139.33 g of MgCl2.

From the calculations made above, 139.33 g of magnesium chloride, MgCl2 were produced.

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