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3 years ago

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5. What are the two types of pure substances ?

1 answer:

Anna11 [10]3 years ago

5 0

Answer:

diamond and pure sugar.

Explanation:

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A scientist has two samples of gas: The first sample contains one mole of argon atoms and has a mass of 39.948 g; the second sam

V125BC [204]

There are 6.022 × 10²³ atoms in 39.948 g of argon and 4.0026 g of helium.

Explanation:

39.945 g/mole is the molar mass of argon so 39.948 g of argon are equal to 1 mole of argon.

4.0026 g/mole is the molar mass of helium so 4.0026 g of helium are equal to 1 mole of helium.

We know that Avogadro's number tell us the number of particles in 1 mole of substance which is 6.022 × 10²³.

So in 39.948 g of argon and 4.0026 g of helium contains the same number of atoms, 6.022 × 10²³.

Learn more about:

Avogadro's number

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4 0

4 years ago

In which reaction does the oxidation number of hydrogen change? In which reaction does the oxidation number of hydrogen change?

dedylja [7]

<u>Answer:</u> The correct answer is $2Na(s)+2H_2O(l)\rightarrow 2NaOH(aq.)+H_2(g)$

<u>Explanation:</u>

Oxidation number is defined as the number which is given to an atom when it looses or gains electron. When an atom looses electron, it attains a positive oxidation state. When an atom gains electron, it attains a negative oxidation state.

Oxidation state of the atoms in their elemental state is considered as 0. Hydrogen is present as gaseous state.

For the given chemical reactions:

<u>Reaction 1:</u> $2HClO_4(aq.)+CaCO_3(s)\rightarrow Ca(ClO_4)_2(aq.)+H_2O(l)+CO_2 (g)$

Oxidation state of hydrogen on reactant side: +1

Oxidation state of hydrogen on product side: +1

Thus, the oxidation state of hydrogen is not changing.

<u>Reaction 2:</u> $CaO(s)+H_2O(l)\rightarrow Ca(OH)_2(s)$

Oxidation state of hydrogen on reactant side: +1

Oxidation state of hydrogen on product side: +1

Thus, the oxidation state of hydrogen is not changing.

<u>Reaction 3:</u> $HCl(aq.)+NaOH(aq.)\rightarrow NaCl(aq.)+H_2O(l)$

Oxidation state of hydrogen on reactant side: +1

Oxidation state of hydrogen on product side: +1

Thus, the oxidation state of hydrogen is not changing.

<u>Reaction 4:</u> $2Na(s)+2H_2O(l)\rightarrow 2NaOH(aq.)+H_2(g)$

Oxidation state of hydrogen on reactant side: +1

Oxidation state of hydrogen on product side: 0

Thus, the oxidation state of hydrogen is changing.

<u>Reaction 5:</u> $SO_2(g)+H_2O(l)\rightarrow H_2SO_3(aq.)$

Oxidation state of hydrogen on reactant side: +1

Oxidation state of hydrogen on product side: +1

Thus, the oxidation state of hydrogen is not changing.

Hence, the correct answer is $2Na(s)+2H_2O(l)\rightarrow 2NaOH(aq.)+H_2(g)$

6 0

3 years ago

The reaction H2O+CH3Br=CH3OH+HBr.<br> H2O acts as what

uysha [10]

Answer:

H2O acts as an oxidizing agent

3 0

2 years ago

What is the process by which molecules of water vapor in the air become liquid water?

mote1985 [20]

That is called condensation

5 0

4 years ago

The balanced half-reaction in which chlorine gas is reduced to the aqueous chloride ion is a ________ process. one-electron two-

AnnZ [28]

Answer:

Two-electron.

Explanation:

Hello!

In this case, since the half-reaction by which chlorine is reduced to aqueous chloride ion is:

$Cl^0_2\rightarrow 2Cl^-$

We can see each chlorine undergo a decrease of electrons by 1, it means that, since there are two chlorine atoms undergoing an electron decrease, we write:

$Cl^0_2+2e^-\rightarrow 2Cl^-$

Which means this is a two-electron process.

Best regards!

6 0

3 years ago