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ivanzaharov [21]
3 years ago
12

Polar winds are usually from the________ direction.

Chemistry
1 answer:
solong [7]3 years ago
6 0
Polar winds are usually from the east direction.
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Given the unbalanced equation: Al + O2 → Al2O3
ArbitrLikvidat [17]
I believe it's 1. 9.
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3 years ago
Pure research becomes ___ when scientists develop a hypothesis based on the data and try to solve a specific problem
Inessa05 [86]

Answer:

The answer is <u>applied research</u>

Explanation:

Pure research becomes <u>applied research</u> when scientists develop a hypothesis based on the data and try to solve a specific problem.

This is because the pure research try to understand, predict or explain the behavior of different phenomena <em>(the data)</em> while the applied research try to develop new technologies or methods (<em>hypothesis)</em> to take part, intervene and/or create changes on these phenomena and solve a <em>specific problem.</em>

5 0
3 years ago
Name this compound Cr2(CO3)3
kiruha [24]

Answer:

Molecular Weight

Explanation:

Chromium(III) Carbonate Cr2(CO3)3 Molecular Weight -- EndMemo.

5 0
3 years ago
How do i found out how animals gain energy and from where​
otez555 [7]

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Animals gain energy from the food they eat. Some animals eat plants while others eat other animals. This is passing of energy from the sun to plants to animal to other animals is called a food chain

6 0
3 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

7 0
3 years ago
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