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2 years ago

50 Points

Chemistry

2 answers:

natta225 [31]2 years ago

8 0

Answer:

The answer is Nuclear

grin007 [14]2 years ago

6 0

Answer:

nuclear

Explanation:

This is because when the protons and neutrons split the energy being released is heat, also known as nuclear fission.

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A change in matter that produces one or more new substances is a ???

Dvinal [7]

Chemical reaction, generally speaking. It's a vague definition, but it's easy to remember.

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2 years ago

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The radius of a hydrogen atom is 37 pm and its mass is 1.67 × 10-24 g. What is the density of the atom in grams per cubic centim

RoseWind [281]

Answer:

V = 4/3 * 3.1416 * (37x10-10)3

V = 2.12x10-25 cm3

d = m/V

d = 1.67x10-24 / 2.12x10-25 = 7.87 g/cm3

The difference in temperature, let's convert F to ºC:

ºC = -80-32/1.8 = -62.22 ºC

dT = -92.6 + 62.2 = -30.4 ºC

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3 years ago

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How many moles are there in 7.4 X 1023 molecules of AgNO3?

vichka [17]

There are 209 g of moles

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2 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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3 years ago

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2 years ago

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