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3 years ago

Why is basalt a sign that the moon once had magma at its surface14 points

1 answer:

7 0

The basalt is a sign that the Moon once had magma at its surface because the basalt is a type of rock that forms directly from the cooling off of the magma. The basalt is part of the igneous type of rocks, which are the type of rocks that form from the cooling of the magma, be it on the surface or inside the ground, so having basalt on the Moon is a sure sign that our natural satellite once had magma on it.

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How many kinds of atoms are in a pure substance?

stepan [7]

Single type of atom builds a single atom

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3 years ago

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What will be the boiling point of a solution of 8 moles of sodium dichromate (Na2Cr2O7) dissolved in 8 kg of water? Use the foll

nexus9112 [7]

Answer:

1. 374.69 K

Explanation:

Hello,

In this case, pure water's boiling point is 373.15 K, thus by considering the boiling point increase equation:

$\Delta T=imKf$

Whereas i=2 since two ionic species are formed,actually, the experimental value is 2.42 so better work with it, thus:

$\Delta T=2.42*(\frac{8}{8}m)*0.512 Km^{-1}=1.239K$

Thus, the required boiling point is:

$T_b=373.15K+1.239K\\T_b=374.69K$

Regards.

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3 years ago

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mrs_skeptik [129]

b or d

Explanation:

I think it's either b or d sorry if I'm wrong

7 0

3 years ago

Explain your reasoning to why the rusting of iron is a chemical change (evidence not needed but may add it)

Tju [1.3M]

When an iron object is left in damp air for a considerable time, it gets covered with a reddish- brown flaky substance called rust. ... Rusting of iron is a chemical change because a new substance iron oxide is formed. The presence of oxygen and water or water vapour is essential for rusting.

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3 years ago

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How do I do this question?The aluminum cup inside your calorimeter weighs 39.78 g. You add 50.01 g of ice cold water to the calo

konstantin123 [22]

Answer:

$Cp_{metal}=0.922\frac{J}{g\°C}$

Explanation:

Hello.

In this problem we must realize that there is heat flow that moves from the hot metal object and the hot water to the cold water and the cold aluminum cup, which allows us to write:

$Q_{cup}+Q_{cold,w}=-(Q_{metal}+Q_{hot,w})$

Which means that the heat lost be the hot metal object and the hot water is gained by both the cold water and the cold aluminum cup, which can be written in terms of mass, specific heats and change in temperature towards the equilibrium temperature (35.9 °C):

$m_{cup}Cp_{cup}(T_{eq}-T_{cup})+m_{cold,w}Cp_{cold,w}(T_{eq}-T_{cold,w})=-(m_{metal}Cp_{metal}(T_{eq}-T_{metal})+m_{hot,w}Cp_{hot,w}(T_{eq}-T_{hot,w})$

We need to solve for the specific heat of the metal as shown below:

$Cp_{metal}=\frac{m_{cup}Cp_{cup}(T_{eq}-T_{cup})+m_{cold,w}Cp_{cold,w}(T_{eq}-T_{cold,w})+m_{hot,w}Cp_{hot,w}(T_{eq}-T_{hot,w})}{-m_{metal}(T_{eq}-T_{metal})} \\\\Cp_{metal}=\frac{39.78g*0.903\frac{J}{g\°C}(35.9-0.5)\°C+50.01g*4.184\frac{J}{g\°C}(35.9-0.5)\°C +50.72g*4.184\frac{J}{g\°C}(35.9-69.5)\°C }{-49.98g(35.9-69.5)\°C } \\\\Cp_{metal}=\frac{1271.6J+7407.2J-7130.3J}{-1679.3g\°C} \\\\Cp_{metal}=0.922\frac{J}{g\°C}$ Best regards.

3 0

3 years ago