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3 years ago

The (C6H14)2NH molecule represents a ----- amine.

Chemistry

2 answers:

STALIN [3.7K]3 years ago

7 0

Answer:

Secondary .

Explanation:

Primary amine contains only one alkyl group.

Secondary amine contains two alkyl groups.

Tertiary amine contains three alkyl groups.

Since (C₆H₁₄)₂NH contains 2 alkyl groups.

So, it is secondary amine.

Kipish [7]3 years ago

6 0

Answer:

Secondary

Explanation:

Because of this amount of hydrogens being substituted, we can classify the amines in primary (R-NH2), secondary (R1R2NH) and tertiary (R1R2R3N) amines. In the primary amines, only one of the hydrogens is replaced by an organic radical, for example, in the following CH3 - NH2 molecule, in the secondary we have the substitution of two hydrogen atoms, as in the case of CH3 - NH - CH3 and in the tertiary all are replaced.

In the molecule (C6H14) 2NH, the bold part shows the substitution of two hydrogen, so we can classify the molecule as a secondary amine.

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What element has 1 energy level and 2 valence electrons

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Answer: Hydrogen and helium

Explanation:

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3 years ago

What is the mass of 61.9 L of oxygen gas collected at STP?

Tcecarenko [31]

Answer:

D. 44.2 g O₂

General Formulas and Concepts:
Gas Laws

STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K

Stoichiometry

Dimensional Analysis
Mole Ratio

Explanation:

Step 1: Define

Identify given.

61.9 L O₂ at STP

Step 2: Convert

We know that the oxygen gas is at STP. Therefore, we can set up and solve for how many moles of O₂ is present:

$\displaystyle 61.9 \ \text{L} \ \text{O}_2 \bigg( \frac{1 \ \text{mol} \ \text{O}_2}{22.4 \ \text{L} \ \text{O}_2} \bigg) = 2.76339 \ \text{mol} \ \text{O}_2$

Recall the Periodic Table (Refer to attachments). Oxygen's atomic mass is roughly 16.00 grams per mole (g/mol). We can use a mole ratio to convert from moles to grams:

$\displaystyle 2.76339 \ \text{mol} \ \text{O}_2 \bigg( \frac{16.00 \ \text{g} \ \text{O}_2}{1 \ \text{mol} \ \text{O}_2} \bigg) = 44.2143 \ \text{g} \ \text{O}_2$

Now we deal with sig figs. From the original problem, we are given 3 significant figures. Round your answer to the exact same number of sig figs:

$\displaystyle 44.2143 \ \text{g} \ \text{O} \approx \boxed{ 44.2 \ \text{g} \ \text{O}_2 }$

∴ our answer is letter choice D.

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Topic: AP Chemistry

Unit: Stoichiometry

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