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2 years ago

what are the moon phases. but be more detailed and don't put in "the phases of the moon" in your own words.

1 answer:

7 0

As the moon orbits or circles the Earth, the phase changes. We'll start with what is called the New Moon phase. This is where we can't see any of the lit up side of the moon. The moon is between us and the sun (see the picture). As the moon orbits the Earth we can see more and more of the lit up side until finally the moon is on the opposite side of the Earth from the sun and we get a full moon. As the moon continues to orbit the Earth we now see less and less of the lit up side. The phases of the moon starting with the New Moon are: New Moon Waxing Crescent First Quarter Waxing Gibbous Full Waning Gibbous Third Quarter Waning Crescent Dark Moon.

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If 7.84 × 107 J of energy is released from a fusion reaction, what amount of mass in kilograms would be lost? Recall that c = 3

katrin [286]

m = 7.84x107/(3x108)2kg = 7.84x107/9x1016kg = 0.871x10-9 kg = 8.71x10-10 kg

3 0

2 years ago

Read 2 more answers

Old thermometers contained very small amounts of mercury. The mercury in the photo has a melting point of -38.8 degrees Celsius.

Nikolay [14]

Answer:

••It's melting point equals -38.8 degrees Celsius because it is mercury. |••

Explanation:

This one makes the most sense out of the answers.

7 0

3 years ago

Is Cu2O+C= Cu+CO2 a balanced equation

attashe74 [19]

Cuco4 balanced equation

5 0

3 years ago

The reform reaction between steam and gaseous methane (CH4) produces "synthesis gas," a mixture of carbon monoxide gas and dihyd

kenny6666 [7]

Answer:

The answer is " $= 0.078 \ kg \ H_2$ ".

Explanation:

calculating the moles in $CH_4 =\frac{PV}{RT}$

$=\frac{(0.58 \ atm) \times (923 \ L) }{ (0.0821 \frac{L \cdot atm}{K \cdot mol})(232^{\circ} C +273)}\\\\=\frac{(535.34 \ atm \cdot \ L) }{ (0.0821 \frac{L \cdot atm}{K \cdot mol})(505)K}\\\\=\frac{(535.34 \ atm \cdot \ L) }{ (41.4605 \frac{L \cdot atm}{mol})}\\\\= 12.9 \ mol$

Eqution:

$CH_4 +H_2O \to 3H_2+ CO \ (g)$

Calculating the amount of $H_2$ produced:

$= 12.9 \ mol CH_4 \times \frac{3 \ mol \ H_2 }{1 \ mol \ CH_4}\times \frac{2.016 g H_2}{1 \ mol \ H_2}\\\\= 78 \ g \ H_2 \\\\= 0.078 \ kg \ H_2$

So, the amount of dihydrogen produced = $0.078 \frac{kg}{s}$

5 0

2 years ago

The acetic acid/acetate buffer system is a common buffer used in the laboratory. Write the equilibrium equation for the acetic a

hichkok12 [17]

Answer:

CH₃CO₂H + H₂O ⇄ CH₃CO₂⁻ + H₃O⁺

Explanation:

A buffer is defined as the mixture of a weak acid and its conjugate base or vice versa.

For the acetic acid buffer, CH₃CO₂H is the weak acid and its conjugate base is the ion without H⁺, that is CH₃CO₂⁻. The equilibrium equation in water knowing this is:

<em>In the equilibrium, the acid is dissociated in the conjugate base and the hydronium ion.</em>

8 0

3 years ago