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fredd [130]
3 years ago
5

Determine the volume of a 75-gram sample of gold at stp

Chemistry
1 answer:
Anuta_ua [19.1K]3 years ago
7 0
Let V = the volume of the sample at STP.

The mass is  m = 75 g
The density of gold is 19.32 g/cm³

Because density = mass/volume, therefore
(19.32 g/cm³) = (75 g)/(V cm³)
V = (75 g)/(19.32 g/cm³) = 3.882 cm³

Answer: 3.882 cm³
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I need help with this
Lelechka [254]

Answer:

The answer is its equal to the volume of its container.

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7 0
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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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7 0
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A person walking covers 5.20 m in 10.4 s. How fast is the person moving?
salantis [7]

Answer:

<h2>0.50 m/s</h2><h2></h2>

Explanation:

Velocity = distance over time

where distance = 5.20 m

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velocity = <u>  5.20  m </u>

                  10.4 s.

            = 0.50 m/s  

8 0
3 years ago
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