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4 years ago

how can you make the gravitational force between two 50kg objects less than the gravitational force between two 5kg objects

Chemistry

2 answers:

Norma-Jean [14]4 years ago

6 0

Answer:

By increasing the distance between the two 50 kg objects compared with the distance between the 5 kg objects.

Explanation:

The formula for the force between 2 objects is

Fg = (G m1 m2) / r^2

where Fg is the force, G = gravitational constant, m1 and m2 are the masses of the objects and r = the distance between the objects.

So as r is increased Fg is decreased.

Paraphin [41]4 years ago

3 0

Explanation:

The gravitational force between two 50kg objects could be less than the gravitational force between a 50kg and a 5kg object if the two 50kg objects are much farther apart.

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lesya692 [45]

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Explanation:

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4 years ago

A student placed 10.5 g of glucose (C6H12O6) in a volumetric fla. heggsk, added enough water to dissolve the glucose by swirling

aniked [119]

<u>Answer:</u> The mass of glucose in final solution is 0.420 grams

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$ .........(1)

Initial mass of glucose = 10.5 g

Molar mass of glucose = 180.16 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

$\text{Initial molarity of glucose}=\frac{10.5\times 1000}{180.16\times 100}\\\\\text{Initial molarity of glucose}=0.583M$

To calculate the molarity of the diluted solution, we use the equation:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of the concentrated glucose solution

$M_2\text{ and }V_2$ are the molarity and volume of diluted glucose solution

We are given:

$M_1=0.583M\\V_1=20.0mL\\M_2=?M\\V_2=0.5L=500mL$

Putting values in above equation, we get:

$0.583\times 20=M_2\times 500\\\\M_2=\frac{0.583\times 20}{500}=0.0233M$

Now, calculating the mass of final glucose solution by using equation 1:

Final molarity of glucose solution = 0.0233 M

Molar mass of glucose = 180.16 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

$0.0233=\frac{\text{Mass of glucose in final solution}\times 1000}{180.16\times 100}\\\\\text{Mass of glucose in final solution}=\frac{0.0233\times 180.16\times 100}{1000}=0.420g$

Hence, the mass of glucose in final solution is 0.420 grams

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4 years ago

Determine the number of grams of carbon dioxide that can be formed from 0.500 grams of iron oxide and an excess of carbon.

Nimfa-mama [501]

Taking into account the reaction stoichiometry, 0.2066 grams of CO₂ are formed from 0.500 grams of iron oxide and an excess of carbon.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

2 Fe₂O₃ + 3 C → 4 Fe + 3 CO₂

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

Fe₂O₃: 2 moles
C: 3 moles
Fe:4 moles
CO₂: 3 moles

The molar mass of the compounds is:

Fe₂O₃: 159.7 g/mole
C: 12 g/mole
Fe: 55.85 g/mole
CO₂: 44 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

Fe₂O₃: 2 moles ×159.7 g/mole= 319.4 grams
C: 3 moles ×12 g/mole= 36 grams
Fe: 4 moles ×55.85 g/mole= 223.4 grams
CO₂: 3 moles ×44 g/mole= 132 grams

<h3>Mass of CO₂ formed</h3>

The following rule of three can be applied: if by reaction stoichiometry 319.4 grams of Fe₂O₃ form 132 grams of CO₂, 0.500 grams of Fe₂O₃ form how much mass of CO₂?

$mass of CO_{2} =\frac{0.500 grams of Fe_{2}O_{3}x132 grams of CO_{2} }{319.4 grams of Fe_{2}O_{3}}$

<u><em>mass of CO₂= 0.2066 grams</em></u>

Then, 0.2066 grams of CO₂ are formed from 0.500 grams of iron oxide and an excess of carbon.

Learn more about the reaction stoichiometry:

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In Earth's water cycle, evaporation takes place as solid ice and snow are converted to water vapor. true or false?

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