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3 years ago

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Part c how long does it take to raise the temperature of the air in a good-sized living room (3.00m×5.00m×8.00m) by 10.0∘c? note

that the specific heat of air is 1006 j/(kg⋅∘c) and the density of air is 1.20kg/m3.

1 answer:

postnew [5]3 years ago

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Derive third law of thermodynamics equation

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The Third Law of thermodynamics states that the entropy of a pure substance in a perfect crystalline state at zero temperature is zero.

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Explanation:

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What is the average atomic mass of all of the naturally occurring isotopes of nickel in amu?

My name is Ann [436]

In general chemistry, isotopes are substances that belong to one specific element. So, they all have the same atomic numbers. But they only differ in the mass numbers, or the number of protons and neutrons in the nucleus. In a nutshell, they only differ in the number of neutrons.

For Nickel, there are 5 naturally occurring isotopes. Their identities, masses and relative abundance are listed below

Isotope Abundance Atomic Mass
Ni-58 68.0769% <span>57.9353 amu
Ni-60 </span>26.2231% <span>59.9308 amu
Ni-61 </span>1.1399 % <span>60.9311 amu
Ni-62 </span>3.6345% <span>61.9283 amu
Ni-64 </span>0.9256% <span>63.9280 amu

To determine the average atomic mass of Nickel, the equation would be:
Average atomic mass = </span>∑Abundance×Atomic Mass

Using the equation, the answer would be:
Average atomic mass = 57.9353(68.0769%) + 59.9308(26.2231%) + 60.9311(1.1399%) + 61.9283(3.6345%) + 63.9280(0.9256%)

Average atomic mass = 58.6933 amu

3 0

4 years ago

How many moles of Cu(OH)2 are soluble in 1L of sodium hydroxide (NaOH) when the pH is 8.23?

Morgarella [4.7K]

Answer:

4.96E-8 moles of Cu(OH)2

Explanation:

Kps es the constant referring to how much a substance can be dissolved in water. Using Kps, it is possible to know the concentration of weak electrolytes. Then, pKps is the minus logarithm of Kps.

Now, we know that sodium hydroxide (NaOH) is a strong electrolyte, who is completely dissolved in water. Therefore the pH depends only on OH concentration originating from NaOH. Let us to figure out how much is that OH concentration.

$pH= -log[H]\\pH= -log (\frac{kw}{[OH]})$

$8.23 = - log(\frac{Kw}{[OH]} \\10^{-8.23} = Kw/[OH]\\ [OH] = Kw/10^{-8.23}$

$[OH]=1.69E-6$

This concentration of OH affects the disociation of Cu(OH)2. Let us see the dissociation reaction:

$Cu(OH)_2 -> Cu^{2+} + 2OH^-$

In the equilibrum, exist a concentration of OH already, that we knew, and it will be added that from dissociation, called "s":

The expression for Kps is:

$Kps= [Cu^{2+}] [OH]^2$

The moles of (CuOH)2 soluble are limitated for the concentration of OH present, according to the next equation.

$Kps= s*(2s+1.69E-6)^2$

"s" is the soluble quantity of Cu(OH)2.

The solution for this third grade equation is $s=4.96E-8 mol/L$

Now, let us calculate the moles in 1 L:

$moles Cu(OH)_2 = 4.96E-8 mol/L * 1 L = 4.96E-8 moles$

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Anna71 [15]

The answer to this question I think would be 2: Better

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