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3 years ago

Explain why the weight of an object is different on earth and on moon even though the objects mass is the same in both place.

2 answers:

5 0

Weight = mass x gravitational field strength

Gravitational field strength is determined by two factors: how large the planet/moon/body is, and how close the object is to that body. Discounting the second factor (which has a minute effect on field strength over distances below 100km or so), the size of the planet/moon/body is proportional to the field strength.

On Earth, the field strength is roughly 9.81N/kg (where kg is the SI unit of mass and Newtons are the SI unit of weight) - this means that every kilo of mass weighs just less than 10N here.

Conversely on the moon, the field strength is roughly 1.6N/kg, meaning that every kilo weighs only 1.6N there. The field strength is so much smaller because the moon has less that 1/81th of the Earth's mass.

In conclusion, take the example of a a 5kg bowling ball: on Earth it would weigh 5 * 9.81 = 53.955N whereas on the moon it would weigh 5 * 1.6 = 8kg. They have the same mass in both places, but weigh very different amounts.

jeka57 [31]3 years ago

4 0

The Mass of the Earth is larger than that of the moon resulting in a larger Fg because the strength of gravity on earth is stronger.

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When 61.6 g of alanine (C3H7NO2) are dissolved in 1150. g of a certain mystery liquid X, the freezing point of the solution is 2

MrRissso [65]

Answer:

Explanation:

From the given information:

TO start with the molarity of the solution:

$= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ C_3H_7 NO_3}{89.1 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}$

= 0.601 mol/kg

= 0.601 m

At the freezing point, the depression of the solution is $\Delta \ T_f = T_{solvent}- T_{solution}$

$\Delta \ T_f = 2.9 ^0 \ C$

Using the depression in freezing point, the molar depression constant of the solvent $K_f = \dfrac{\Delta T_f}{m}$

$K_f = \dfrac{2.9 ^0 \ C}{0.601 \ m}$

$K_f = 4.82 ^0 C / m}$

The freezing point of the solution $\Delta T_f = T_{solvent} - T_{solution}$

$\Delta T_f = 7.3^ 0 \ C$

The molality of the solution is:

$= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ NH_4Cl}{53.5 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}$

Molar depression constant of solvent X, $K_f = 4.82 ^0 \ C/m$

Hence, using the elevation in boiling point;

the Vant'Hoff factor $i = \dfrac{\Delta T_f}{k_f \times m}$

$i = \dfrac{7.3 \ ^0 \ C}{4.82 ^0 \ C/m \times 1.00 \ m}$

$\mathbf {i = 1.51 }$

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2 years ago

If 3.5 grams of NaN3 decomposed, how many grams of N2 would be produced?

Wewaii [24]

Answer:

5.25 moles.

Explanation:

The decomposition reaction of NaN₃ is as follows :

$2NaN_3(s)\rightarrow 2Na(s)+3N_2(g)$

We need to find how many grams of N₂ produced in the process.

From the above balanced chemical reaction, we conclude that the ratio of moles of sodium azide and nitrogen gas are 2 : 3.

2 moles of sodium azide decomposes to give 3 moles of nitrogen gas. So,

3.5 moles of sodium azide decomposes to give $\dfrac{3}{2}\times 3.5=5.25$ moles of nitrogen gas.

Hence, the number of moles produced is 5.25 moles.

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2 years ago

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Marina86 [1]

Answer:

1 tell your family difrent ways to safe water

2 dont leave the water running when you brush your teeth

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Explanation:

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2 years ago