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3 years ago

What are gravitational forces?

Chemistry

2 answers:

harina [27]3 years ago

5 0

Answer:

Good Question!!! No one has found the answer to that question.

Explanation:

Kruka [31]3 years ago

3 0

Answer:

Gravity, or gravitation, is a natural phenomenon by which all things with mass or energy—including planets, stars, galaxies, and even light—are brought toward one another. On Earth, gravity gives weight to physical objects, and the Moon's gravity causes the ocean tides.

Gravity, also called gravitation, in mechanics, the universal force of attraction acting between all matter. ... On Earth all bodies have a weight, or downward force of gravity, proportional to their mass, which Earth's mass exerts on them. Gravity is measured by the acceleration that it gives to freely falling objects

The gravitational force is a force that attracts any two objects with mass. We call the gravitational force attractive because it always tries to pull masses together, it never pushes them apart. In fact, every object, including you, is pulling on every other object in the entire universe!

Gravitational force is the force that attracts to objects with mass. ... This forces tries to pull masses together, that is why attractive in nature. For example: When ball is thrown upwards it always come downwards after a certain height because the ball is pulled or attracted by the Earth's gravitation la force.

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3 years ago

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4 years ago

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3 years ago

A sample of 211 g of iron (III) bromide is reacted with

Alisiya [41]

FeBr₃ ⇒ limiting reactant

mol NaBr = 1.428

<h3>Further explanation</h3>

Reaction

2FeBr₃ + 3Na₂S → Fe₂S₃ + 6NaBr

Limiting reactant⇒ smaller ratio (mol divide by coefficient reaction)

FeBr₃

211 g of Iron (III) bromide(MW=295,56 g/mol), so mol FeBr₃ :

$\tt n=\dfrac{mass}{MW}\\\\n=\dfrac{211}{295,56}\\\\n=0.714$

Na₂S

186 g of Sodium sulfide(MW=78,0452 g/mol), so mol Na₂S :

$\tt n=\dfrac{186}{78.0452}=2.38$

Coefficient ratio from the equation FeBr₃ : Na₂S = 2 : 3, so mol ratio :

$\tt FeBr_3\div Na_2S=\dfrac{0.714}{2}\div \dfrac{2.38}{3}=0.357\div 0.793$

So FeBr₃ as a limiting reactant(smaller ratio)

mol NaBr based on limiting reactant (FeBr₃) :

$\tt \dfrac{6}{3}\times 0.714=1.428$

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3 years ago

How many moles of aspartame are present in 1.00 mg of aspartame?

Diano4ka-milaya [45]

Answer:- There are $3.40*10^-^6$ moles.

Solution:- It is a unit conversion problem where we are asked to convert mg of aspartame to moles. Aspartame is $C_1_4H_1_8N_2O_5$ and it's molar mass is 294.31 grams per mole.

mg are converted to grams and then the grams are converted to moles as:

$1.00mg Aspartame(\frac{1g}{1000mg})(\frac{1mole}{294.31g})$

= $3.40*10^-^6$ moles of aspartame

So, there would be $3.40*10^-^6$ moles of aspartame in 1.00 mg of it.

3 0

4 years ago