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3 years ago

What is the name of this alkane? mc012-1.jpg

Chemistry

1 answer:

9966 [12]3 years ago

6 0

THe name of this alkane is <span>1-ethyl,3-methylcyclohexane</span>

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Which of these terms refers to matter that could be heterogeneous?

Dafna11 [192]

Answer:

B) mixture

Explanation:

Many homogeneous mixtures are commonly referred to as solutions. A heterogeneous mixture consists of visibly of three phases or states of matter are gas, liquid, and solid.

5 0

3 years ago

A gas mixture of Ne and Ar has a total pressure of 4.00 atm and contains 16.0 mol of gas. If the partial pressure of Ne is 2.75

Stolb23 [73]

Answer:

5 moles of Argon is present in the mixture.

Explanation:

Total pressure of the gaseous mixture = 4 atm

Total number of moles = 16

Partial pressure of Ne = 2.75 atm

By Dalton's law of partial pressure, the total pressure of gaseous mixture is the sum of partial pressures of individual gases which are non-reactive.

Hence:

$P_{total}=P_{Ar}+P_{Ne}\\4=P_{Ar}+2.75\\P_{Ar}=1.25\ atm$

Also :

Partial pressure = mole fraction*total pressure

$P_{Ar}=X_{Ar}P_{total}$

$X_{Ar}=\frac{1.25}{4}=0.3125$

$\frac{n_{Ar}}{n_{total}}=0.3125\\n_{Ar}=5$

∴Number of moles of Argon = 5

4 0

4 years ago

What is a meso compound.

Nataly_w [17]

Answer:

A meso compound is a non-optically active member of a set of stereoisomers, at least two of which are optically active.

3 0

2 years ago

What is the correct formula for iron(III) oxide?

stiv31 [10]

Answer:

Fe 3+

Explanation:

iron is Fe and it's 3 and has positive

7 0

3 years ago

Read 2 more answers

You are given the reaction Cu + HNO3 Cu(NO3)2 + NO + H2O complete the final balanced equation based on half-reactions

Evgen [1.6K]

Answer:

3Cu + 8HNO3 → 3Cu(NO3)2 + 2NO + 4H2O

Explanation:

Cu + HNO3 → Cu(NO3)2 + NO + H2O

The first step is to write the oxidation numbers for each atoms in the given equation

Cu0 + H+1N+5O-23 → Cu+2(N+5O-23)2 + N+2O-2 + H+12O

Identify the oxidizing and reducing agent

OXIDATION --- Cu0 → Cu+2(N+5O-23)2 + 2e-

REDUCTION---H+1N+5O-23 + 3e- → N+2O

Balance equation in half reaction

Cu0 + 2HNO3 → Cu+2(N+5O-23)2 + 2e-

H+1N+5O-23 + 3e- → N+2O

Now balance the charge

Cu0 + 2HNO3 → Cu+2(N+5O-23)2 + 2e- + 2H+

H+1N+5O-23 + 3e- + 3H+ → N+2O

Balance the oxygen atom

Cu0 + 2HNO3 → Cu+2(N+5O-23)2 + 2e- + 2H+

H+1N+5O-23 + 3e- + 3H+ → N+2O-2 + 2H2O

Make electron gain equivalent to electron lost.

3Cu0 + 6HNO3 → 3Cu+2(N+5O-23)2 + 6e- + 6H+

2H+1N+5O-23 + 6e- + 6H+ → 2N+2O-2 + 4H2O

Complete reaction

3Cu0 + 8H+1N+5O-23 + 6e- + 6H+ → 3Cu+2(N+5O-23)2 + 2N+2O-2 + 6e- + 4H2O + 6H+

Simplify the equation

3Cu0 + 8H+1N+5O-23 → 3Cu+2(N+5O-23)2 + 2N+2O-2 + 4H2O

Final equation

3Cu + 8HNO3 → 3Cu(NO3)2 + 2NO + 4H2O

5 0

3 years ago