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3 years ago

13

23.9 km + 15 km

1 answer:

Ugo [173]3 years ago

4 0

Answer:

the original answer is 38.9km (3sf)

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PLEASE HELP:<br><br> iron reacts with nitrogen to form iron(III) nitride

lutik1710 [3]

Answer:

$2Fe +N_{2} -> 2FeN$

Explanation:

Iron is Fe, nitrogen is N. Nitrogen is diatomic, which means it occurs as a molecular pair by itself. Iron III nitride has a chemical formula of FeN because nitrogen has a charge of 3-, and iron III tells us the iron has a charge of 3+ so you just need one of each to make the charges balance and the compound neutral.

3 0

2 years ago

A system gains 687 kJ of heat, resulting in a change in internal energy of the system equal to 156 kJ. How much work is done?

Maslowich

Answer:

w = -531 kJ

1. Work was done by the system.

Explanation:

Step 1: Given data

Heat gained by the system (q): 687 kJ (By convention, when the system absorbs heat, q > 0).

Change in the internal energy of the system (ΔU°): 156 kJ

Step 2: Calculate the work done (w)

We will use the following expression.

ΔU° = q + w

w = ΔU° - q

w = 156 kJ - 687 kJ

w = -531 kJ

By convention, when w < 0, work is done by the system on the surroundings.

4 0

2 years ago

2. Explain why sedimentary rock layers of the North and South Rim(side) of the Grand Canyon are the same age, using Steno’s Prin

natali 33 [55]

Answer:

Law of Lateral Continuity The Grand Canyon.

and

he same rock layers on opposite sides of the canyon. The matching rock layers were deposited at the same time, so they are the same age.

3 0

2 years ago

A concentration cell is constructed using two Ni electrodes with Ni2+ concentrations of 1.0 M and 1.00 � 10�4 M in the two half-

Komok [63]

<u>Answer:</u> The cell potential of the cell is +0.118 V

<u>Explanation:</u>

The half reactions for the cell is:

<u>Oxidation half reaction (anode):</u> $Ni(s)\rightarrow Ni^{2+}+2e^-$

<u>Reduction half reaction (cathode):</u> $Ni^{2+}+2e^-\rightarrow Ni(s)$

In this case, the cathode and anode both are same. So, $E^o_{cell}$ will be equal to zero.

To calculate cell potential of the cell, we use the equation given by Nernst, which is:

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Ni^{2+}_{diluted}]}{[Ni^{2+}_{concentrated}]}$

where,

n = number of electrons in oxidation-reduction reaction = 2

$E_{cell}$ = ?

$[Ni^{2+}_{diluted}]$ = $1.00\times 10^{-4}M$

$[Ni^{2+}_{concentrated}]$ = 1.0 M

Putting values in above equation, we get:

$E_{cell}=0-\frac{0.0592}{2}\log \frac{1.00\times 10^{-4}M}{1.0M}$

$E_{cell}=0.118V$

Hence, the cell potential of the cell is +0.118 V

5 0

3 years ago

What is the correct name for OF2?

Gre4nikov [31]

OF2 is Oxygen difluoride

7 0

2 years ago

Read 2 more answers