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Kruka [31]
3 years ago
6

How can you demonstrate waves?​

Chemistry
1 answer:
Lady bird [3.3K]3 years ago
7 0
Shaking a phone cord, strumming a guitar string, playing a trumpet
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Which of the following always occurs in a
vova2212 [387]

Answer:

C

Explanation:All the other answers are wrong

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Calculate the average net charge on phenylalanine if it is in a solution that has a ph of 8.70.
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Phenylalanine has two known pKa charges: 1.83 and 9.13. To determine their individual ionic dissociation charges at pH 10, the equation to be used is

pH = pKa + log [α/(1-α)]

At pKa 1.83:
10 = 1.83 + log [α₁/(1-α₁)]
α₁ = 0.99999

At pKa 9.13:
10 = 9.13 + log [α₂/(1-α₂)]
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Average Net Charge =  0.99999 + 0.88114 = 1.88
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4 years ago
Why doesn’t silicone (IV) oxide conduct electricity
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What volume of oxygen at STP will be produced if the following reaction absorbed 275 kJ of heat?
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3 years ago
Convert 6.35 grams of aluminum sulfate to moles​
vampirchik [111]

Answer:

There are 0.0186 moles of formula units in 6.35 grams of aluminum sulfate \rm Al_2(SO_4)_3.

Explanation:

What's the empirical formula of aluminum sulfate?

Sulfate is an anion with a charge of -2 per ion. When sulfate ions are bonded to metals, the compound is likely ionic.

Aluminum is a group III metal. Its ions tend to carry a charge of +3 per ion.

The empirical formula of an ionic compound shall balance the charge on ions with as few ions as possible.

The least common multiple of 2 and 3 is 6. That is:

  • Three sulfate ions \rm {SO_4}^{2-} will give a charge of -6.
  • Two aluminum ions \rm Al^{3+} will give a charge of +6.

Pairing three \rm {SO_4}^{2-} ions with two \rm Al^{3+} will balance the charge. Hence the empirical formula: \rm Al_2(SO_4)_3.

What's the mass of one mole of aluminum sulfate? In other words, what's the formula mass of \rm Al_2(SO_4)_3?

Refer to a modern periodic table for relative atomic mass data:

  • Al: 26.982;
  • S: 32.06;
  • O: 15.999.

There are

  • two Al,
  • three S, and
  • twelve O

in one formula unit of \rm Al_2(SO_4)_3.

Hence the formula mass of \rm Al_2(SO_4)_3:

\underbrace{2\times 26.982}_{\rm Al} + \underbrace{3\times 32.06}_{\rm S} + \underbrace{12\times 15.999}_{\rm O} = \rm 342.132\;g\cdot mol^{-1}.

How many moles of formula units in 6.35 grams of \rm Al_2(SO_4)_3?

\displaystyle n = \frac{m}{M} = \rm \frac{6.35\;g}{342.132\;g\cdot mol^{-1}} = 0.0186\;mol.

8 0
3 years ago
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