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Pavel [41]
3 years ago
7

What is the molecular formula of the following compound: NH_CI, molar mass = 51.5 g/mol?

Chemistry
2 answers:
Helen [10]3 years ago
6 0

Answer:

Explanation:

Explanation:

As you know, the empirical formula tells you what the smallest whole number ratio that exists between the atoms that make up a compound is.

In your case, you know that the empirical formula is

NH Cl

    2

, which means that the regardles of how many atoms of each element you get in the actual compound, the ratio that exists between them will always be

1:2:1.

What you actually need to determine is how many empirical formulas are needed to get to the molecular formula.

Notice that the problem provides you with the molar mass of the compound. This means that you can use the molar mass of the empirical formula to determine exactly how many atoms you need to form the compound's molecule.

molar mass empirical formula×n=molar mass compound

To get the molar mass of the empirical formula, use the molar masses of its constituent atoms

14.0067 g/mol+2×1.00794 g/mol+35.453 g/mol=51.48 g/mol≈

51.5 g/mol

This means that you have

51.5g/mol×n=51.5g/mol

As you can see, you have

n=1.

This means that the empirical formula and the molecular formula are equivalent,

NH Cl.

    2

weqwewe [10]3 years ago
4 0

Answer:

NH_Cl 2 is the answer to What is the molecular formula of the following compound: NH_CI, molar mass = 51.5 g/mol?

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<u>Answer:</u>

<u>For a:</u> The standard Gibbs free energy of the reaction is -347.4 kJ

<u>For b:</u> The standard Gibbs free energy of the reaction is 746.91 kJ

<u>Explanation:</u>

Relationship between standard Gibbs free energy and standard electrode potential follows:

\Delta G^o=-nFE^o_{cell}           ............(1)

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The given chemical equation follows:

2Ce^{4+}(aq.)+3I^{-}(aq.)\rightarrow 2Ce^{3+}(aq.)+I_3^-(aq.)

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<u>Reduction half reaction:</u>   3I^_(aq.)+2e^-\rightarrow I_3^-(aq.)

We are given:

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  • <u>For b:</u>

The given chemical equation follows:

6Fe^{3+}(aq.)+2Cr^{3+}+7H_2O(l)(aq.)\rightarrow 6Fe^{2+}(aq.)+Cr_2O_7^{2-}(aq.)+14H^+(aq.)

<u>Oxidation half reaction:</u>   Fe^{3+}(aq.)\rightarrow Fe^{2+}(aq.)+e^-       ( × 6)

<u>Reduction half reaction:</u>   2Cr^{2+}(aq.)+7H_2O(l)+6e^-\rightarrow Cr_2O_7^{2-}(aq.)+14H^+(aq.)

We are given:

n=6\\E^o_{cell}=-1.29V\\F=96500

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3 years ago
The normal freezing point of a certain liquid Xis-7.30°C but when l02. g of iron(III) chloride (FeCl3) are dissolved in 650. g o
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Answer:

2.7 °C.kg/mol

Explanation:

Step 1: Calculate the freezing point depression (ΔT)

The normal freezing point of a certain liquid X is-7.30°C and the solution freezes at -9.9°C instead. The freezing point depression is:

ΔT = -7.30 °C - (-9.9 °C) = 2.6 °C

Step 2: Calculate the molality of the solution (b)

We will use the following expression.

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Answer: The moles of carbon needed will be, 13.8 moles

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3 years ago
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