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WARRIOR [948]
3 years ago
15

Newton's Third Law: "For every action, there is an equal and opposite reaction." ... The forces are the result of

Chemistry
1 answer:
Whitepunk [10]3 years ago
8 0

Answer:

The answer to your question is simultaneous. Hope it helps!

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A pH change can be evidence that​
drek231 [11]

Answer:

Explanation:

The acid level has changed

3 0
3 years ago
Identify this reaction <br><br> HBr+ Al(OH)3 —-&gt; H2O + AlBr3
Vikki [24]
<h2><em><u>Answer</u></em></h2>

Neutralization reaction

5 0
3 years ago
.7 liters of CO represents how many molecules?
Ronch [10]

Answer:

               1.88 × 10²² Molecules of CO

Explanation:

At STP for an ideal gas,

Volume = Mole × 22.4 L/mol

Or,

Mole = Volume / 22.4 L/mol

Mole = 0.7 L / 22.4 L/mol

Mole = 0.03125 moles

Now,

No. of Molecules = Moles × 6.022 × 10²³ Molecules/mol

No. of Molecules = 0.03125 × 6.022 × 10²³ Molecules/mol

No. of Molecules = 1.88 × 10²² Molecules of CO

7 0
2 years ago
Calculate the number of moles of ethanol acid in 30 cm³of 1 mol dm-³ solution ​
muminat

Answer:

May I assume "ethanol acid is just ethanol (it has one slightly acidic H atom).  If so, the molar mass is 46.02 g/mole.

Explanation:

We have 30 cm^3 [30 ml] of 1.0 M (1 mole/liter)  [1 dm³ = 1 liter].

That is 1 mole/liter.  30 ml would contain (0.030 liter)*(1 mole/1 liter) = 0.03 moles.

7 0
3 years ago
If the distance between two objects is decreased to - of the original
Charra [1.4K]

Answer:

The new force will be \frac{1}{100} of the original force.

Explanation:

In the context of this problem, we're dealing with the law of gravitational attraction. The law states that the gravitational force between two object is directly proportional to the product of their masses and inversely proportional to the square of a distance between them.

That said, let's say that our equation for the initial force is:

F = G\frac{m_1m_2}{R^2}The problem states  that  the distance decrease to 1/10 of the original distance, this means:[tex]R_2 = \frac{1}{10}R

And the force at this distance would be written in terms of the same equation:

F_2 = G\frac{m_1m_2}{R_2^2}

Find the ratio between the final and the initial force:

\frac{F_2}{F} = \frac{G\frac{m_1m_2}{R_2^2}}{G\frac{m_1m_2}{R^2}}

Substitute the value for the final distance in terms of the initial distance:

\frac{F_2}{F} = \frac{G\frac{m_1m_2}{(\frac{R}{10})^2}}{G\frac{m_1m_2}{R^2}}

Simplify:

\frac{F_2}{F} = \frac{\frac{1}{100R^2}}{\frac{1}{R^2}}=\frac{1}{100}

This means the new force will be \frac{1}{100} of the original force.

8 0
3 years ago
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