Which of the following describes a compound? (Hint: Carbon and oxygen both appear on the periodic table.)

The solubility of CO2 in water is 0.161 g/100 mL at 20oC and a partial pressure of CO2 of 760 mmHg. What partial pressure of CO2

Schach [20]

<u>Answer:</u> The partial pressure of carbon dioxide having solubility 0.886g/100mL is 4182.4 mmHg

<u>Explanation:</u>

Henry's law states that the amount of gas dissolved or molar solubility of gas is directly proportional to the partial pressure of the liquid.

The equation given by Henry's law is:

$C_{CO_2}=K_H\times p_{CO_2}$ ......(1)

where,

$C_{CO_2$ = solubility of carbon dioxide in water = 0.161 g/100 mL

$K_H$ = Henry's constant = ?

$p_{CO_2}$ = partial pressure of carbon dioxide = 760 mmHg

Putting values in equation 1, we get:

$760mmHg=K_H\times 0.161g/100mL\\\\K_H=\frac{760mmHg}{0.161g/100mL}=4720.5g.mmHg/100mL$

Now, calculating the pressure of carbon dioxide using equation 1, we get:

$C_{CO_2$ = solubility of carbon dioxide in water = 0.886 g/100 mL

$K_H$ = Henry's constant = 4720.5 g.mmHg/100 mL

$p_{CO_2}$ = partial pressure of carbon dioxide = ?

Putting values in equation 1, we get:

$p_{CO_2}=4720.5g.mmHg/100mL\times 0.886g/100mL=4182.4mmHg$

Hence, the partial pressure of carbon dioxide having solubility 0.886g/100mL is 4182.4 mmHg

4 0

3 years ago

QUIZ SCIENCE:

vlabodo [156]

1. A soluble salt can be prepared by reacting an acid with a suitable insoluble reactant including:

a metal
a metal oxide
a carbonate
3. I don’t know this one
4. A term base or glossary is a database containing single words or expressions related to a specific subject.
5. Strong acid is an acid that ionizes completely in aqueous solution. It always loses a proton (H+) when dissolved in water. Weak acid is an acid that ionizes partially in a solution. ... Because the rate of reaction depends upon the degree of dissociation αand strong acids have higher degrees of dissociation.

im not sure of the rest

7 0

3 years ago

What expression approximates the volume of O2 consumed, measure at STP, when 55 g of Al reacts completely with excess O2?2 Al(s)

Genrish500 [490]

Answer:

34.28 L ( 1.5*22.4 L)

Explanation:

Calculation of the moles of aluminum as:-

Mass = 55 g

Molar mass of aluminum = 26.981539 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{55\ g}{26.981539\ g/mol}$

$Moles= 2.0384\ mol$

According to the reaction:-

$4Al+3O_2\rightarrow 2Al_2O_3$

4 moles of aluminum react with 3 moles of oxygen gas

1 mole of aluminum react with $\frac{3}{4}$ moles of oxygen gas

2.0384 moles of aluminum react with $\frac{3}{4}\times 2.0384$ moles of oxygen gas

Moles of oxygen gas = 1.5288 moles

At STP,

Pressure = 1 atm

Temperature = 273.15 K

Using ideal gas equation as:

$PV=nRT$

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

1 atm × V = 1.5288 mol × 0.0821 L.atm/K.mol × 273.15 K

⇒V = 34.28 L ( 1.5*22.4 L)

7 0

3 years ago

Analyze the map. How would the temperature on the east coast of South America (near Brazil) differ from the west coast of South

hjlf

Answer: Uruguay on north on the Atlantic side, Chile and north on the Pacific side. The higher up you go the warmer, so Brazil, Ecuador, and Colombia will be the warmest especially in terms of water temperature.

5 0

3 years ago

Read 2 more answers

Iridium has two naturally occurring isotopes, Iridium-191 and Iridium-193. If the average atomic mass of orodum s 192.217, what

Lina20 [59]

Answer:

The percent isotopic abundance of Ir-193 is 60.85 %

The percent isotopic abundance of Ir-191 is 39.15 %

Explanation:

we know there are two naturally occurring isotopes of iridium, Ir-191 and Ir-193

First of all we will set the fraction for both isotopes

X for the isotopes having mass 193

1-x for isotopes having mass 191

The average atomic mass is 192.217

we will use the following equation,

193x + 191(1-x) = 192.217

193x + 191 - 191x = 192.217

193x- 191x = 192.217 - 191

2x = 1.217

x= 1.217/2

x= 0.6085

0.6085 × 100 = 60.85 %

60.85% is abundance of Ir-193 because we solve the fraction x.

now we will calculate the abundance of Ir-191.

(1-x)

1-0.6085 =0.3915

0.3915× 100= 39.15 %

6 0

3 years ago