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Mars2501 [29]
3 years ago
6

What is the oxidation number of Nitrogen in HNO2? +1 -1 +3 -3

Chemistry
1 answer:
WITCHER [35]3 years ago
5 0
I believe the correct answer from the choices listed above is the third option. The <span>oxidation number of Nitrogen in HNO2 would be +3. It is calculated as follows:

1 + x + (-2)(2) = 0
x = +3

Hope this answers the question. Have a nice day.</span>
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Define atomic number ​
tester [92]

Answer:

the number of protons in the nucleus of an atom, which determines the chemical properties of an element and its place in the periodic table.

8 0
3 years ago
Which of the following can explain the daily change in sea level observed along a coast?
salantis [7]
The answer for the question above is A. the gravitational pull of the moon on the water near the coast. The sun and and the moon are responsible for the rising and falling of the ocean tides. The gravitational pull of the moon and the sun makes the water in the oceans bulge, causing a continuous change between high and low tide. 
5 0
3 years ago
If you have 5.42 x 1024 aluminum atoms, approximately how many moles is that
sattari [20]

Answer:

<h2>9.00 moles</h2>

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

n =  \frac{N}{L} \\

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

n =  \frac{5.42 \times  {10}^{24} }{6.02 \times {10}^{23} }  \\  = 9.0033

We have the final answer as

<h3>9.00 moles</h3>

Hope this helps you

4 0
3 years ago
1 mol super cooled liquid water transformed to solid ice at -10 oC under 1 atm pressure.
Arada [10]

Answer:

Explanation:

Given that:

number of moles of super cooled liquid water = 1

Melting enthalpy of ice = 6020 J/mol

Freezing point =0 °C = (0 + 273 K)= 273 K

The decrease in entropy of the system during freezing for 1 mol (i.e during transformation from liquid water to solid ice )  = - 6020 J/mol × 1 mol /273 K = -22.051 J/K

Entropy change during further cooling from 0 °C (273 K) to -10 °C (263 K)

\Delta \ S = \int\limits^{T_2}_{T_1}\dfrac{nC_p(s)dT}{T}

\Delta \ S = {nC_p(s)In \dfrac{T_2}{T_1}

\Delta \ S = {(1*37.7)In \dfrac{263}{273}

Δ S = -1.4 J/K

Total entropy change of the system = - 22.05 J/K - 1.4 J/K = - 23.45 J/K

Entropy change of universe = entropy change of the system+ entropy change of the surrounding

According to the second law of thermodynamics

Entropy change of universe  >0

SO,

Entropy change of the system + entropy change in the surrounding > 0

Entropy change in the surrounding > - entropy change of the system

Entropy change in the surrounding > - (- 23.53 J/K)

Entropy change in the surrounding > 23.53 J/K

b) Make some comments on entropy changes from the obtained data.

From the data obtained; we will realize that the entropy of the system decreases as cooling takes place when water is be convert to ice , randomness of these molecules reduces and as cooling proceeds , hence, entropy reduces more as well and the liberated heat will go into the surrounding due to this entropy of the surrounding increasing.

4 0
3 years ago
The value of the rate constant for a gas phase reaction can be changed by increasing the A. temperature of the reaction vessel.
prohojiy [21]

Answer:

temperature of the reaction vessel

Explanation:

temperature of the reaction vessel

3 0
3 years ago
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