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3 years ago

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What are the effects of the collision forces when objects of the same mass collide?

1 answer:

Ad libitum [116K]3 years ago

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Answer:

Newton's third law of motion is naturally applied to collisions between two objects. In a collision between two objects, both objects experience forces that are equal in magnitude and opposite in direction. Such forces often cause one object to speed up (gain momentum) and the other object to slow down (lose momentum)

Explanation:

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In which 2 states of matter can a substance take the shape of its container?

Natasha2012 [34]

Answer:

solid and liquid

Explanation:

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3 years ago

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A 51.24-g sample of Ba(OH)2 is dissolved in enough water to make 1.20 liters of solution. How many mL of this solution must be d

g100num [7]

Answer:

0.40 L

Explanation:

Calculation of the moles of $Ba(OH)_2$ as:-

Mass = 51.24 g

Molar mass of $Ba(OH)_2$ = 171.34 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{51.24\ g}{171.34\ g/mol}$

$Moles= 0.2991\ mol$

Volume = 1.20 L

The expression for the molarity is:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$Molarity=\frac{0.2991\ mol}{1.20\ L}=0.24925\ M$

Thus,

Considering

$Molarity_{working\ solution}\times Volume_{working\ solution}=Molarity_{stock\ solution}\times Volume_{stock\ solution}$

Given that:

$Molarity_{working\ solution}=0.100\ M$

$Volume_{working\ solution}=1\ L$

$Volume_{stock\ solution}=?$

$Molarity_{stock\ solution}=0.24925\ M$

So,

$0.100\ M\times 1\ L=0.24925\ M\times Volume_{stock\ solution}$

$Volume_{stock\ solution}=\frac{0.100\times 1}{0.24925}\ L=0.40\ L$

<u>The volume of 0.24925M stock solution added = 0.40 L </u>

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3 years ago

ANSWER FAST <br> is water wet? <br> a.) yes <br> b.) no<br> c.) a baby goat is a kid

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Answer:

c

Explanation:

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How to do my homework? i’ll give 20 points to anyone who will do it!

tensa zangetsu [6.8K]

what is it ??? ill edit this

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3 years ago

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A sodium ion has a radius of 1.16 x 10^-10 m and a nearby fluoride ion has a radius of 1.9 x 10^-10 m. Determine the distance be

Ann [662]

The answer is: the distance between two nuclei is 2.35×10⁻¹⁰ m.

r(Na⁺) = 1.16×10⁻¹⁰ m; radius of sodium cation.

r(F⁻) = 1.9×10⁻¹⁰ m; radius of fluoride anion.

d(NaF) = r(Na⁺) + r(F⁻).

d(NaF) = 1.16×10⁻¹⁰ m + 1.9×10⁻¹⁰ m.

d(NaF) = 2.35×10⁻¹⁰ m; distance between two nuclei.

The sum of ionic radii of the cation and anion gives the distance between the ions in a crystal lattice.

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3 years ago