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Marina86 [1]
3 years ago
11

What is the difference between acids and strong acids?

Chemistry
1 answer:
mylen [45]3 years ago
7 0

Answer:

A and b are the both the answers

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What is the heat in joules required to melt 25 grams of ice?
AfilCa [17]

Answer:

8,350 Joules

Hope this helps

7 0
2 years ago
PLs heLp Me
olchik [2.2K]

Answer:

Opaque → a.no light travels through the material; all light is reflected or absorbed

Translucent → b.some light travels through the material and the image cannot be seen clearly

Transparent → c.all light travels through the material and the image can be seen clearly

Explanation:

Hope this helps :)

7 0
3 years ago
Read 2 more answers
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
A student places a starch and water solution into a bag of dialysis tubing and then ties off the ends. Next, he places the fille
Zina [86]

Answer:

After 30 minutes have passed, the solution inside the dialysis tubing and the solution in the beaker will be tested for glucose and starch

-The presence of glucose will be tested with Benedict's solution, Testape®, or Clinistix

4 0
2 years ago
I RLLY NEED HELP. Use Image B (picture above) and calculate the density of a ring that has a mass of 32 grams. Read Page 11 "Mea
dimaraw [331]

Answer:

Mass of ring = 32 g

Volume of ring = 4 mL

Density of ring = 8 g/mL

Explanation:

From the question given above, the following data were obtained:

Mass of ring = 32 g

Volume of water = 64 mL

Volume of water + ring = 68 mL

Density of ring =?

Next, we shall determine the volume of the ring. This can be obtained as follow:

Volume of water = 64 mL

Volume of water + ring = 68 mL

Volume of ring =?

Volume of ring= (Volume of water + ring) – (Volume of water)

Volume of ring = 68 – 64

Volume of ring = 4 mL

Finally, we shall determine the density of the ring. This can be obtained as follow:

Mass of ring = 32 g

Volume of ring = 4 mL

Density of ring =?

Density = mass / volume

Density of ring = 32 / 4

Density of ring = 8 g/mL

8 0
3 years ago
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