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2 years ago

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2. (4 pts) There are N, molecules in the air. There is also liquid N2. Draw a molecular level

1 answer:

AlekseyPX2 years ago

3 0

CHY3 I’m sure it’s that

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26.8 moles of argon contain how many atoms?

Maru [420]

<span>1.61 × 1023 Multiply by 26.8 to get the answer.161.33 x 10 ^23 </span>

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3 years ago

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Which substance has the highest viscosity?

Lemur [1.5K]

Room temperature has the highest viscosity

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2 years ago

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A teacher makes the following statement.

melamori03 [73]

The correct answer among the choices given is the first option.The teacher most likely is talking about distillation of a mixture. Distillation is a unit operation that separates component substances from a liquid mixture which is shown by the teacher. Also, the most common purifying technique in the production of gasoline is by this process.

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2 years ago

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5 H2O2(aq) + 2 MnO4-(aq) + 6 H+(aq) → 2 Mn2+(aq) + 8 H2O(l) + 5 O2(g) In a titration experiment, H2O2(aq) reacts with aqueous Mn

Tomtit [17]

Answer:

Oxygen in hydrogen peroxide oxidizes from -1 to 0.

Explanation:

Oxidation is the loss of electrons. The specie which is oxidized has has elevation in its oxidation state as compared in the reactant and the products.

The given reaction is shown below as:

$5 H_2O_2_{(aq)} + 2 MnO_4^-_{(aq)} + 6 H^+_{(aq)}\rightarrow 2 Mn^{2+}_{(aq)} + 8 H_2O_{(l)} + 5 O_2_{(g)}$

Manganese in $MnO_4^-$ has oxidation state of +7

Manganese in $Mn^{2+}$ has an oxidation state of +2

It reduces from +7 to +2

Oxygen in hydrogen peroxide has an oxidation state of -1.

Oxygen in molecular oxygen has an oxidation of 0.

Thus, oxygen in hydrogen peroxide oxidizes from -1 to 0.

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2 years ago

How many sulfur atoms are present in 25.6 g of Al2(S2O3)3

IgorC [24]

Given the mass of $Al_{2}(S_{2}O_{3})_{3}$ =25.6 g

The molar mass of $Al_{2}(S_{2}O_{3})_{3}$ =390.35g/mol

Converting mass of $Al_{2}(S_{2}O_{3})_{3}$ to moles:

$25.6 g Al_{2}(S_{2}O_{3})_{3}*\frac{1molAl_{2}(S_{2}O_{3}}{390.35 gAl_{2}(S_{2}O_{3}} =0.0656molAl_{2}(S_{2}O_{3}$

Converting mol $Al_{2}(S_{2}O_{3})_{3}$ to mol S:

$0.0656mol Al_{2}(S_{2}O_{3})_{3}*\frac{6molS}{1mol Al_{2}(S_{2}O_{3})_{3}}=0.3936 molS$

Converting mol S to atoms of S using Avogadro's number:

1 mol = $6.022*10^{23}atoms$

$0.3936mol S *\frac{6.022*10^{23}atoms S}{1 mol S}=2.37*10^{23} S atoms$

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2 years ago