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ludmilkaskok [199]
3 years ago
7

How many atoms of each element are chemical formula: 3K2O3.

Chemistry
2 answers:
hammer [34]3 years ago
4 0

Answer:

In 3K2O3 - 6 K atoms, 9 O atoms

In 2H3NCH3Br - 12 H atoms, 2 N atoms, 2 C atoms, 2 Br atoms

In 4 Ar2CO3 - 8 Ar atoms, 4 C atoms, 12 O atoms

Explanation:

All you need to do is multiply the coefficient by the number of each atom in the compound. For example, H3NCH3Br has 1 nitrogen, 1 carbon, 1 bromine, and 6 hydrogens. If you have 2H3NCH3Br, just multiply all those numbers by 2.

aksik [14]3 years ago
4 0
3k2o3 = 15

2h3nch3br = 17

4ar2co3 = 24
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Which compound is an isomer of C2H5OC2H5?
Anastaziya [24]

Answer:

D) C₄H₉OH

Explanation:

In chemistry, isomers are chemical substances with the same formula (That is, same atoms but in different structures).

For the compound C₂H₅OC₂H₅ there are 4 atoms of C, 10 atoms of H and 1 atom of O.

A) CH₃COOH . This compound have 2 atoms of C, 4 atoms of H and 2 atoms of O. Thus, isn't an isomer.

B) C₂H₅COOCH₃. There are 4 atoms of C but 8 atoms of H and 2 atoms of O. Isn't an isomer

C) C₃H₇COCH₃. There are 5 atoms of C, 10 atoms of H and 1 atom of O. Thus, isn't an isomer

D) C₄H₉OH. Here, you have 4 atoms of C, 10 atoms of H and 1 atom of O. That means this structure is the isomer of C₂H₅OC₂H₅

6 0
3 years ago
What would you observe when zinc is added to solutions of iron II sulphate? write the chemical reaction that takes place.​
kompoz [17]

Answer:

When zinc is added to the solution of iron sulphate, the colour of iron sulphate solution changes. It is because zinc is more reactive than iron, it displaces iron from its solution of iron sulphate and a grey precipitate of iron and a colourless solution of zinc sulphate is formed.

Explanation:

Zn(s) + FeSO4(aq) -----> ZnSO4(aq) + Fe(ppt.)

8 0
3 years ago
Calculate the molarity of 0.289 moles of FeC13 dissolved in 120 ml of solution
yanalaym [24]

Answer:

2.41 M

Explanation:

The molarity is the moles of FeCl3 over the liters of solution. Since you're given mL you need to change it to L which is 0.12 L. 0.289 divided by 0.12 is your answer

5 0
4 years ago
Arrange the bromine-containing compounds CaBr2, Br2, NBr3, CBr4 in order of increasing predicted boiling point.
Fudgin [204]

Explanation:

In an ionic compound, there will be strong force of attraction between the combining atoms due to the opposite charges present on them.

For example, CaBr_{2} is an ionic compound where calcium has a +2 charge and each bromine atom has a -1 charge.

Therefore, in order to break the bond we need to provide more heat. Hence, boiling point of calcium bromide will be the highest.

NBr_{3} is a covalent compound and as nitrogen is more electronegative in nature and also has a lone pair of electron hence, there will be a net dipole moment.

CBr_{4} is also a covalent compound. And, as bromine is more electronegative than carbon atom so, dipole moment is in the outwards direction. Hence, in CBr_{4} there will be zero dipole moment.

Therefore, its boiling is less than the boiling point of NBr_{3}.

Br_{2} is a covalent compound and there will be no dipole moment.

Thus, we can conclude that given bromine-containing compounds are placed in their increasing boiling point order as follows.

    Br_{2} < CBr_{4} < NBr_{3} < CaBr_{2}

3 0
4 years ago
1. (a) Write the two electrochemical half reactions for ethanol (C2H5OH) direct electrochemical conversion in a fuel cell with a
Natalka [10]

Answer:

(a) Two electrochemical half reactions for ethanol (C_{2}H_{5}OH) is written below

      C_{2}H_{5} + 60 ⇒ 2CO_{2} + 3H_{2}O + 12e^{-} (anode)

      3O_{2} + 12e^{-} ⇒ 60^{=} (Cathode)

(b) The direct oxidation of the fuel will occur in a solid oxide fuel cell but we have to compete with other sets of chemical electrochemical reaction such as water-gas-shift reaction

      CO + H_{2}O ⇆ CO_{2}  + H_{2}(WGSR)

(c) The electrochemical half reaction that could convert ethanol directly into a fuel cell that conducts hydrogen ions is shown below

C_{2}H_{5}OH + 3H_{2}O  ⇒  2CO_{2} + 12H^{+} + 12e^{-} (anode)

3O_{2} + 12H^{+} + 12e^{-} - 6H_{2}O (Cathode)

C_{2} H_{5} OH + 3O_{2}  ⇒  3H_{2}O + 2CO_{2}

(d) The half reactions that would be required are shown below

C_{2}H_{5} OH + 12OH^{-}  ⇒  CO_{2} + 9H_{2}O  + 12e^{-}  (anode)

3O_{2} + 6H_{2}O + 12e^{-}  ⇒  12OH^{-}  (Cathode)

C_{2}H_{5}OH + 3O_{2}  ⇒  3H_{2}O  + 2CO_{2}

8 0
4 years ago
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