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romanna [79]
3 years ago
8

Caffeine (C8H10N4O2) is a stimulant found in coffees and teas. When dissolved in water, it can accept a proton from a water mole

cule. Write the chemical equation for this process and identify the Brønsted-Lowry acid and base

Chemistry
1 answer:
Olenka [21]3 years ago
3 0

Answer:

See figure 1

Explanation:

If we want to find the acid and the Brønsted-Lowry base, we must remember the definition for each of these molecules:

-) Acid: hydrogen donor

-) Base: hydrogen acceptor

In the <u>caffeine structure,</u> we have several atoms of nitrogen. These nitrogen atoms have the ability to <u>accept</u> hydronium ions (H^+). Therefore the caffeine molecule will be the base since it can accept

If caffeine is the base, the water must be the acid. So, the water in this reaction donated a hydronium ion.

<u>Thus, caffeine is the base and water the acid. (See figure 1)</u>

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What is the mole to mole relationship between copper and zinc? Cu2+(aq)+-&gt;Cu(s)+Zn2+(aq)
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The reaction is missing the Zn(s) in the reactants. The stoichiometry of the copper/zinc is 1 mole to 1 mole

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3 years ago
How many moles of NaCl are needed to make 10.0L of a 5 M solution of<br> Salt water?*
vesna_86 [32]

Answer: 50 mol

Explanation:

Molarity=\frac{Moles -of -solute}{liters -of -solution}

M=\frac{mol}{L}

mol=M*L

mol=5*10\\mol=50mol

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3 years ago
What is a glass of iced tea containing excess sugar at the bottom said to be?
Pavel [41]
The answer is saturated
5 0
3 years ago
Water is poured into a conical container at the rate of 10 cm3/sec. The cone points directly down, and it has a height of 20 cm
8090 [49]

Answer:

\frac{dh}{dt}_{h=2cm} =\frac{40}{9\pi}\frac{cm}{2}

Explanation:

Hello,

The suitable differential equation for this case is:

\frac{dV}{dt}=10\frac{cm^3}{s}

As we're looking for the change in height with respect to the time, we need a relationship to achieve such as:

\frac{dh}{dt} = ?*\frac{dV}{dt}

Of course, ?=\frac{dh}{dV}.

Now, since the volume of a cone is V=\pi r^2h/3 and the ratio r/h=15/20=3/4 or r=3/4h, the volume becomes:

V=\pi (\frac{3}{4} h)^2h/3= \frac{3}{16}\pi h^3

We proceed to its differentiation:

\frac{dV}{dh} =\frac{9}{16} \pi h^2\\\frac{dh}{dV} =\frac{16}{9 \pi h^2}

Then, we compute \frac{dh}{dt}

\frac{dh}{dt} = \frac{16}{9 \pi h^2}*\frac{dV}{dt}\\\frac{dh}{dt} = \frac{16}{9\pi h^2}*10\frac{cm^3}{s} =\frac{160}{9 \pi h^2}

Finally, at h=2:

\frac{dh}{dt}_{h=2cm} =\frac{160}{9\pi 2^2}\\\frac{dh}{dt}_{h=2cm} =\frac{40}{9\pi}\frac{cm}{s}

Best regards.

4 0
3 years ago
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