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3 years ago

True or false solar power take power from wind and uses it to make electricity

Chemistry

1 answer:

quester [9]3 years ago

3 0

The answer is false

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Balancing oxidation-reduction reactions <br> Mg+ N2—>Mg3N2

BartSMP [9]

Answer:

${ \sf{3Mg_{(s)} + N_{2(g)} →Mg _{3}N_{2(s)}}}$

3 0

2 years ago

A gas sample has a temperature of 22c with an unknown volume. The same gas has a volume of 456 mL when the temperature is 86c wi

Mkey [24]

The initial volume is 116.65 mL

<u>Explanation:</u>

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Given:

Temperature, T₁ = 22°C

T₂ = 86°C

Volume, V₂ = 456 m

V₁ = ?

According to Charle's law:

$\frac{V_1}{T_1} = \frac{V_2}{T_2}$

Substituting the values:

$\frac{V_1}{22} =\frac{456}{86} \\\\V_1 = \frac{456 X 22}{86} \\\\V_1 = 116.65 mL$

Therefore, the initial volume is 116.65 mL

3 0

3 years ago

Why do different solvents affect adsorbate-surface interactions?

Effectus [21]

Answer:

so that the can fiuger out what it is for the answer

Explanation:

3 0

2 years ago

What is the name of the molecule shown below?

xxTIMURxx [149]

Answer:

B. 1-heptene

Explanation:

The 1-HEPTENE molecule contains a total of 20 bond(s) There are 6 non-H bond(s), 1 multiple bond(s), 4 rotatable bond(s) and 1 double bond(s).

6 0

2 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