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Komok [63]
2 years ago
8

If 200.4g of water is mixed with 101.42g of salt the mass of the final solution would be reported as

Chemistry
1 answer:
d1i1m1o1n [39]2 years ago
8 0

Answer:

301.8 g

Explanation:

We prepare a solution with 200.4 g of water (solvent) and 101.42 g of salt (solute). The mass of the solution is equal to the sum of the mass of the solvent and the mass of the solute.

m(solution) = m(solute) + m(solvent)

m(solution) = 200.4 g + 101.42 g

m(solution) = 301.8 g (we round-off to one decimal according to the significant figures rules)

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Solid iron (III) oxide reacts with hydrogen gas to form iron and water. How many grams of iron are produced when 440.23 grams of
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When 440.23 grams of iron(III) oxide are reacted with hydrogen gas, the amount of iron produced will be 307.66 grams

<h3>Stoichiometric calculation</h3>

From the equation of the reaction:

Fe_2O_3 + 3H_2 --- > 2Fe + 3H_2O

The mole ratio of iron(III) oxide to produced iron is 1:2.

Mole of 440.23 iron(III) oxide = 440.23/159.69 = 2.76 moles

Equivalent mole of produced iron = 2.76 x 2 = 5.52 moles

Mass of 5.52 moles of iron = 5.52 x 55.8 = 307.66 grams

More on stoichiometric calculations can be found here; brainly.com/question/27287858

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What is the difference between a carnivore and a herbivore?? ​
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If an ultraviolet photon has a wavelength of 77.8 nm calculate the energy of one mole ultraviolet photon.
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Answer:

Explanation:

E = (hc)/(λ)

E = (6.624x10^(-27))Js x ((3×10^8)ms^(-1)) /

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3 years ago
Which of these half-reactions represents reduction?
gogolik [260]

Answer: The half-reactions represents reduction are as follows.

  • Cr_{2}O^{2-}_{7} \rightarrow Cr^{3+}
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Explanation:

A half-reaction where addition of electrons take place or a reaction where decrease in oxidation state of an element  takes place is called reduction-half reaction.

For example, the oxidation state of Cr in Cr_{2}O^{2-}_{7} is +6 which is getting converted into +3, that is, decrease in oxidation state is taking place as follows.

Cr_{2}O^{2-}_{7} + 3 e^{-} \rightarrow Cr^{3+}

Similarly, oxidation state of Mn in MnO^{-}_{4} is +7 which is getting converted into +2, that is, decrease in oxidation state is taking place as follows.

MnO^{2-}_{4} + 5 e^{-} \rightarrow Mn^{2+}

Thus, we can conclude that half-reactions represents reduction are as follows.

  • Cr_{2}O^{2-}_{7} \rightarrow Cr^{3+}
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