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labwork [276]
3 years ago
12

Why do people act like they are your friends, then talk and whisper behind your back to another?

Chemistry
2 answers:
Usimov [2.4K]3 years ago
8 0

Answer:

Because humans are cruel

Explanation:

If they act like you’re friend but talk behind you’re back, don’t waist your time with them. There are better people out there that want to be ur friend too. People that will actually care for you. Trust me.

Burka [1]3 years ago
7 0

Answer:

it's just the way of people, of course there are some that don't but a lot of us have a tendency to want to fit in. If there with a group of friends that don't like you most people will talk behind your back to fit in with that group. Either that or their just fake and simply never liked you to begin with

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mihalych1998 [28]
For this problem, we use the Beer Lambert's Law. Its usual equation is:

A = ∈LC
where
A is the absorbance
∈ is the molar absorptivity
L is the path length
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As you notice, we only have to find the absorbance. But since we are not given with the molar absorptivity, we will have to use the modified equation that relates % transmittance to absorbance:

A = 2 - log(%T)
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2 years ago
Which of the following would have the largest amount of mass?
Klio2033 [76]
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Is cutting your nails a physical or chemical change
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Read 2 more answers
A sample of 0.0860 g of sodium chloride is added to 30.0 mL of 0.050 M silver nitrate, resulting in the formation of a precipita
mestny [16]

Answer:

The answer is:

(a) NaCl(aq)+AgNO_3(aq) \rightarrow AgCl(r) +NaNO_3 (aq)

(b) NaCl

(c) 0.211 g

Explanation:

Given:

The mass of NaCl,

= 0.0860 g

The molar mass of NaCl,

= 58.44 g/mol

The volume of AgNO_3,

= 30.0 ml

or,

= 0.030 L

Molarity of AgNO_3,

= 0.050 M

Moles of NaCl will be:

= \frac{Given \ mass}{Molar \ mass}

= \frac{0.0860}{58.44}

= 0.00147 \ mol

now,

Moles of AgNO_3 will be:

= Molarity\times Volume

= 0.050\times 0.030

=0.0015 \ mol

(a)

The reaction is:

⇒ NaCl(aq)+AgNO_3(aq) \rightarrow AgCl(r) +NaNO_3 (aq)

(b)

1 mole of NaCl react with,

= 1 mol of AgNO_3

0.0015 mol AgNO_3 needs,

= 0.00150 \ mol \ NaCl

Available mol of NaCl < needed amount of NaCl

So,

The limiting reagent is "NaCl".

(c)

The precipitate formed,

= 0.00147\times \frac{1}{1}\times \frac{143.32}{1}

= 0.211 \ g \ AgCl

4 0
3 years ago
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