All categories

1 year ago

How many grams of oxygen gas occupy 12.3 L of space at 109.4 kPa and 15.4oC?

1 answer:

3 0

17.93 grams of oxygen gas occupy 12.3L of space at 109.4 kPa and 15.4°C. Details about how to calculate mass can be found below.

<h3>How to calculate mass?</h3>

The mass of a given gas can be calculated by multiplying the number of moles of the substance by its molar mass.

However, the number of moles of the gas must be calculated first as follows:

PV = nRT

Where;

P = pressure = 1.0796941atm
V = volume = 12.3L
n = number of moles
T = temperature = 288.4K
R = gas law constant = 0.0821 Latm/molK

1.079 × 12.3 = n × 0.0821 × 288.4

13.27 = 23.68n

n = 13.27/23.68

n = 0.56mol

Mass = 0.56 × 32

mass of oxygen gas = 17.93g

Therefore, 17.93 grams of oxygen gas occupy 12.3L of space at 109.4 kPa and 15.4°C.

Learn more about mass at: brainly.com/question/19694949

You might be interested in

Halppp can anyone come up with a slogan for anti smoking

PtichkaEL [24]

Answer:

"Your Choice: Liberation From Your Own Choices or Continue the Choice That Decreases Your Liberation."

Explanation:

Liberation is the freeing or release from something restraining you. In this case, smoking holds you back from doing the things you used to do.

It's not that creative, but I hope this is good enough for you! :D (don't smoke)

5 0

2 years ago

Which answer choice correctly describes the difference between ionic and covalent bonds?

Stels [109]

Answer:

B) Ionic bonds are formed when one atom takes another atom's electrons, while covalent bonds are formed when two atoms share electrons.

Explanation:

7 0

3 years ago

A solution contains 0.0440 M Ca2 and 0.0940 M Ag. If solid Na3PO4 is added to this mixture, which of the phosphate species would

Olenka [21]

Answer:

C. $Ca_3(PO_4)_2$ will precipitate out first

the percentage of $Ca^{2+}$ remaining = 12.86%

Explanation:

Given that:

A solution contains:

$[Ca^{2+}] = 0.0440 \ M$

$[Ag^+] = 0.0940 \ M$

From the list of options , Let find the dissociation of $Ag_3PO_4$

$Ag_3PO_4 \to Ag^{3+} + PO_4^{3-}$

where;

Solubility product constant Ksp of $Ag_3PO_4$ is $8.89 \times 10^{-17}$

Thus;

$Ksp = [Ag^+]^3[PO_4^{3-}]$

replacing the known values in order to determine the unknown ; we have :

$8.89 \times 10 ^{-17} = (0.0940)^3[PO_4^{3-}]$

$\dfrac{8.89 \times 10 ^{-17}}{(0.0940)^3} = [PO_4^{3-}]$

$[PO_4^{3-}] =\dfrac{8.89 \times 10 ^{-17}}{(0.0940)^3}$

$[PO_4^{3-}] =1.07 \times 10^{-13}$

The dissociation of $Ca_3(PO_4)_2$

The solubility product constant of $Ca_3(PO_4)_2$ is $2.07 \times 10^{-32}$

The dissociation of $Ca_3(PO_4)_2$ is :

$Ca_3(PO_4)_2 \to 3Ca^{2+} + 2 PO_{4}^{3-}$

Thus;

$Ksp = [Ca^{2+}]^3 [PO_4^{3-}]^2$

$2.07 \times 10^{-33} = (0.0440)^3 [PO_4^{3-}]^2$

$\dfrac{2.07 \times 10^{-33} }{(0.0440)^3}= [PO_4^{3-}]^2$

$[PO_4^{3-}]^2 = \dfrac{2.07 \times 10^{-33} }{(0.0440)^3}$

$[PO_4^{3-}]^2 = 2.43 \times 10^{-29}$

$[PO_4^{3-}] = \sqrt{2.43 \times 10^{-29}$

$[PO_4^{3-}] =4.93 \times 10^{-15}$

Thus; the phosphate anion needed for precipitation is smaller i.e $4.93 \times 10^{-15}$ in $Ca_3(PO_4)_2$ than in $Ag_3PO_4$ $1.07 \times 10^{-13}$

Therefore:

$Ca_3(PO_4)_2$ will precipitate out first

To determine the concentration of $[Ca^+]$ when the second cation starts to precipitate ; we have :

$Ksp = [Ca^{2+}]^3 [PO_4^{3-}]^2$

$2.07 \times 10^{-33} = [Ca^{2+}]^3 (1.07 \times 10^{-13})^2$

$[Ca^{2+}]^3 = \dfrac{2.07 \times 10^{-33} }{(1.07 \times 10^{-13})^2}$

$[Ca^{2+}]^3 =1.808 \times 10^{-7}$

$[Ca^{2+}] =\sqrt[3]{1.808 \times 10^{-7}}$

$[Ca^{2+}] =0.00566$

This implies that when the second cation starts to precipitate ; the concentration of $[Ca^{2+}]$ in the solution is 0.00566

Therefore;

the percentage of $Ca^{2+}$ remaining = concentration remaining/initial concentration × 100%

the percentage of $Ca^{2+}$ remaining = 0.00566/0.0440 × 100%

the percentage of $Ca^{2+}$ remaining = 0.1286 × 100%

the percentage of $Ca^{2+}$ remaining = 12.86%

5 0

3 years ago

For a different reaction, the plot of the reciprocal of concentration versus time in seconds was linear with a slope of 0.056 M-

Bond [772]

Answer:

$C_t=0.165M$

Explanation:

From the question we are told that:

Slope $K=0.056 M-1 s -1$

initial Concentration $C_1=2.2M$

Time $t=100$

Generally the equation for Raw law is mathematically given by

$\frac{1}{C}_t=kt+\frac{1}{C}_0$

$\frac{1}{C}_t=0.056*100+\frac{1}{2.2}_0$

$C_t=0.165M$

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

2 years ago