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2 years ago

Which of the following best defines Avogadro's number? (5 points)

Chemistry

2 answers:

qaws [65]2 years ago

4 0

Answer:

Number of particles in exactly one mole of a substance.

Explanation:

Avogardo's number is used as a constant to determine how many atoms, or subatomic particles, are in an element. We use it only when determining atoms, molar mass has its own equation.

Likurg_2 [28]2 years ago

3 0

Answer:

Avogadro's number is 6.022* 10^23

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Chemical weathering takes place fastest in a _______ and _________ climate

ser-zykov [4K]

Possibly wet and unstable

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3 years ago

when you put a book on a table the table pushes on the book is that newton's 1st,2nd,or 3rd law of motion ?

ohaa [14]

The table pushes the book and the book pushes the table
It's 3rd law

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2 years ago

Which term is de ned as the region in an atom where an electron is most likely to be located?

charle [14.2K]

Orbital
All atoms have the same number of electrons as protons. Negative electrons are attracted to the positive nucleus. This force of attraction keeps electrons constantly moving around the nucleus. The region where an electron is most likely to be found is called an orbital.
Here you go :)

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2 years ago

24) What is momentum? *

sergiy2304 [10]

Answer:

Momentum is the measure of the motion of an object found by multiplying the objects mass and velocity.

Symbol: p

Units: kg x m/s

Explanation:

......

6 0

2 years ago

If a gas occupies 1532.7 mL at standard temperature, what volume does it occupy at 49.4 ºC if the pressure remains constant?

ElenaW [278]

Answer:

a. 1810mL

Explanation:

When conditions for a gas change under constant pressure (and the number of molecules doesn't change), it follows Charles' Law:

$\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}$ where the temperatures must be measured in Kelvin

To convert from Celsius to Kelvin, add 273, or use the equation: $T_C+273=T_K$

For this problem, one must also recall that standard temperature is 0°C (or 273K).

So, $T_1 = 273[K]$ , and $T_2 = (49.4+273)[K]=322.4[K]$ .

$\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}$

$\dfrac{(1532.7[mL])}{(273[K])}=\dfrac{V_2}{(322.4[K])}$

$\dfrac{(1532.7[mL])}{(273[K\!\!\!\!\!{-}])}(322.4[K\!\!\!\!\!{-}] )=\dfrac{V_2}{(322.4[K]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{----})}(322.4[K]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{----})$

$1810.04571428[mL]=V_2$

Adjusting for significant figures, this gives $V_2=1810[mL]$

4 0

1 year ago