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Roman55 [17]
3 years ago
13

A(n) __________ eclipse might occur when the Earth, Moon, and Sun are in the positions shown in the picture.

Chemistry
2 answers:
m_a_m_a [10]3 years ago
8 0
I believe the answer is letter D.
zheka24 [161]3 years ago
3 0

Answer:

The answer is most likely D

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geniusboy [140]
Li because its charge is +1.
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Which of the following describes an endothermic reaction?
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The endothermic reaction would be D. Batter becomes a pancake when heated. This is because the batter is absorbing the heat in order to turn into a pancake.

Answer is D. Batter becomes a pancake when heated.


Please vote brainliest thank you!
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3 years ago
[Rn] 7s25f6d4<br> Which element is denoted
Colt1911 [192]

Answer:

Explanation:

[Rn] 7s25f6d4

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4 0
3 years ago
Magnesium and oxygen balanced
11Alexandr11 [23.1K]
2Mg+ O₂= 2MgO
The charge of Mg is+2 and the charge of Oxygen is -2. 2 oxygen atoms will have a charge of -4 so the Mg atoms have to equal +4. A coefficient of 2 for the Mg atoms will balance it out 
8 0
2 years ago
Three kilograms of steam is contained in a horizontal, frictionless piston and the cylinder is heated at a constant pressure of
lakkis [162]

Answer:

Final temperature: 659.8ºC

Expansion work: 3*75=225 kJ

Internal energy change: 275 kJ

Explanation:

First, considering both initial and final states, write the energy balance:

U_{2}-U_{1}=Q-W

Q is the only variable known. To determine the work, it is possible to consider the reversible process; the work done on a expansion reversible process may be calculated as:

dw=Pdv

The pressure is constant, so:  w=P(v_{2}-v_{1} )=0.5*100*1.5=75\frac{kJ}{kg} (There is a multiplication by 100 due to the conversion of bar to kPa)

So, the internal energy change may be calculated from the energy balance (don't forget to multiply by the mass):

U_{2}-U_{1}=500-(3*75)=275kJ

On the other hand, due to the low pressure the ideal gas law may be appropriate. The ideal gas law is written for both states:

P_{1}V_{1}=nRT_{1}

P_{2}V_{2}=nRT_{2}\\V_{2}=2.5V_{1}\\P_{2}=P_{1}\\2.5P_{1}V_{1}=nRT_{2}  

Subtracting the first from the second:

1.5P_{1}V_{1}=nR(T_{2}-T_{1})

Isolating T_{2}:

T_{2}=T_{1}+\frac{1.5P_{1}V_{1}}{nR}

Assuming that it is water steam, n=0.1666 kmol

V_{1}=\frac{nRT_{1}}{P_{1}}=\frac{8.314*0.1666*373.15}{500} =1.034m^{3}

T_{2}=100+\frac{1.5*500*1.034}{0.1666*8.314}=659.76 ºC

7 0
3 years ago
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