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3 years ago

What does the roman numeral stand for in copper(1) oxide should it not be copper(II) oxide

Chemistry

1 answer:

photoshop1234 [79]3 years ago

3 0

Answer:

The roman numeral in copper(I) oxide indicates that the oxidation number of copper in the compound is 1.

Explanation:

Roman numeral is used to indicate the oxidation number of an element in a compound.

The roman numeral in copper(I) oxide indicates that the oxidation number of copper in the compound is 1.

This can be seen from the following illustration:

copper(I) oxide => Cu₂O

Oxidation number of O = –2

Oxidation number of Cu₂O = 0

Oxidation number of Cu =?

Cu₂O = 0

2Cu + O = 0

2Cu – 2 = 0

Collect like terms

2Cu = 0 + 2

2Cu = 2

Divide both side by 2

Cu = 2/2

Cu = 1

Thus, we can see that the oxidation number of Cu in Cu₂O is 1. Hence the name of Cu₂O is copper(I) oxide indicating that the oxidation number of of copper (Cu) in the compound is 1.

For copper(II) oxide, we shall determine the oxidation number of Cu. This can be obtained as follow:

copper(II) oxide, CuO => CuO

Oxidation number of O = –2

Oxidation number of CuO = 0

Oxidation number of Cu =?

CuO = 0

Cu + O = 0

Cu – 2 = 0

Collect like terms

Cu = 0 + 2

Cu = 2

Thus, the oxidation number of Cu in CuO is 2. Hence the name of CuO is copper(II) oxide indicating that the oxidation number of of copper (Cu) in the compound is 2.

From the above illustrations,

We can see that the roman numeral in both copper(I) oxide, Cu₂O and copper(II) oxide, CuO are different because the oxidation number of Cu in both cases are different.

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Anybody know chemistry Grahams law ?

sukhopar [10]

There is two different types but i’ll just do both meanings just incase.

Graham's Law of Diffusion: the rate of diffusion of one gas through another is inversely proportional to the square root of the density of the gas.

Graham's Law of Effusion: the rate of effusion of a gas is inversely proportional to the square root of the density of the gas.

hopes this helps..!

7 0

3 years ago

Sodium sulfate is slowly added to a solution containing 0.0500 M Ca 2 + ( aq ) and 0.0390 M Ag + ( aq ) . What will be the conce

levacccp [35]

The given question is incomplete. The complete question is as follows.

Sodium sulfate is slowly added to a solution containing 0.0500 M $Ca^{2+}(aq)$ and 0.0390 M $Ag^{+}(aq)$ . What will be the concentration of $Ca^{2+}$ (aq) when $Ag_{2}SO_{4}(s)$ begins to precipitate? What percentage of the $Ca^{2+}(aq)$ can be separated from the Ag(aq) by selective precipitation?

Explanation:

The given reaction is as follows.

$Ag_{2}SO_{4} \rightleftharpoons 2Ag^{+} + SO^{2-}_{4}$

$[Ag^{+}]$ = 0.0390 M

When $Ag_{2}SO_{4}$ precipitates then expression for $K_{sp}$ will be as follows.

$K_{sp} = [Ag^{+}]^{2}[SO^{2-}_{4}]$

$1.20 \times 10^{-5} = (0.0390)^{2} \times [SO^{2-}_{4}]$

$[SO^{2-}_{4}]$ = 0.00788 M

Now, equation for dissociation of calcium sulfate is as follows.

$CaSO_{4} \rightleftharpoons Ca^{2+} + SO^{2-}_{4}$

$K_{sp} = [Ca^{2+}][SO^{2-}_{4}]$

$4.93 \times 10^{-5} = [Ca^{2+}] \times 0.00788$

$[Ca^{2+}]$ = 0.00625 M

Now, we will calculate the percentage of $Ca^{2+}$ remaining in the solution as follows.

$\frac{0.00625}{0.05} \times 100$

= 12.5%

And, the percentage of $Ca^{2+}$ that can be separated is as follows.

100 - 12.5

= 87.5%

Thus, we can conclude that 87.5% will be the concentration of $Ca^{2+}(aq)$ when $Ag_{2}SO_{4}(s)$ begins to precipitate.

4 0

3 years ago

2. Matt went to his friend’s party. He ate a big meal and drank a keg of beer. He felt heartburn after the meal and took Tums to

evablogger [386]

Answer:

390.85mL

Explanation:

Step 1:

Data obtained from the question.

Initial pressure (P1) = 780 torr

Initial volume (V1) = 400mL

Initial temperature (T1) = 40°C = 40°C + 273 = 313K

Final temperature (T2) = 25°C = 25°C + 273 = 298K

Final pressure (P2) = 1 atm = 760torr

Final volume (V2) =?

Step 2:

Determination of the final volume i.e the volume of the gas outside Matt's body.

The volume of the gas outside Matt's body can be obtained by using the general gas equation as shown below:

P1V1/T1 = P2V2/T2

780 x 400/313 = 760 x V2 /298

Cross multiply to express in linear form

313 x 760 x V2 = 780 x 400 x 298

Divide both side by 313 x 760

V2 = (780 x 400 x 298) /(313 x 760)

V2 = 390.85mL

Therefore, the volume of the gas outside Matt's body is 390.85mL

3 0

3 years ago

How many dL in 1,000 mL

Anna007 [38]

1,000 mL is the same as 10 dL.

6 0

3 years ago

Read 2 more answers

Match the following with their correct molecular weight. 2-butanone Propyl acetate 4-methyl-2-pentanone Butyl acetate Methanol E

Hatshy [7]

Answer:

2-butanone = 72.11 g/mol (option F)

Propyl acetate = 102.13 g/mol (option C)

4-methyl-2-pentanone = 100.16 g/mol (option D)

Butyl acetate = 116.16 g/mol (option B)

Methanol = 32.04 g/mol (option E)

Ethanol = 46.07 g/mol (option A)

Explanation:

Step 1: Data given

Molar mass of C = 12.01 g/mol

Molar mass of H = 1.01 g/mol

Molar mass of O = 16.00 g/mol

Step 2:

2-butanone = C4H8O

⇒ 4*12.01 + 8*1.01 + 16.00 = 72.11 g/mol (option F)

Propyl acetate = C5H10O2

⇒ 5*12.01 + 10*1.01 + 2*16.00 = 102.13 g/mol (option C)

4-methyl-2-pentanone = C6H12O

⇒ 6*12.01 + 12*1.01 + 16.00 = 100.16 g/mol (option D)

Butyl acetate = C6H12O2

⇒ 6*12.01 + 12*1.01 + 2*16.00 = 116.16 g/mol (option B)

Methanol = CH3OH = CH4O

⇒ 12.01 + 4*1.01 + 16.00 = 32.04 g/mol (option E)

Ethanol = C2H5OH = C2H6O

⇒ 2*12.01 + 6*1.01 + 16.00 = 46.07 g/mol (option A)

4 0

3 years ago