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3 years ago

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Imagine that all microorganism suddenly disappeared from earth from what u have learned why do u think that animals would eventu

ally disappeared from earth? Why would plant disappeared

1 answer:

weeeeeb [17]3 years ago

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Answer:

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The process in which organisms grow and replace worn-out cells is called: A. Cell regeneration B. Cell division C. Mitosis D. Bo

djyliett [7]

The mitosis is the cell division so it both B and C

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3 years ago

What is easier to observe about an object,it’s physical properties or it’s chemical properties. Explain why.

Sladkaya [172]

Answer:

Physical Properties

Explanation:

You are able to see physical properties but are unable to see chemical properties.

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3 years ago

3. Sand mixed with water

Alekssandra [29.7K]

Answer:

<h3>Sand cannot mixed on water not float on water .</h3>

<h3>When we mix sand and water , No reaction take place . The sand simply settles down at the bottom of the water container . This is why sand is heavier than water and therefore cannot float in water .</h3>

<h2>Hope this helps you ✌️</h2>

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2 years ago

What is a nonpolar bond?

Ierofanga [76]

Answer:

Nonpolar covalent bonds are a type of bond that occurs when two atoms share a pair of electrons with each other. These shared electrons glue two or more atoms together to form a molecul

Explanation:

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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