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4 years ago

A bottle of coke has a volume of 250.0ml. what is this volume in cubic centimeters

Chemistry

1 answer:

ki77a [65]4 years ago

7 0

Answer:

the answer to your question would be a fraction since we are using cubic centimeters , so the answer would be 2/50.0

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Vanyuwa [196]

Answer:

If the volume of a gas increased from 2 to 6 L while the temperature was held constant, the pressure of the gas decreased by a factor of 3.

Explanation:

Boyle's law that says "The volume occupied by a given gaseous mass at constant temperature is inversely proportional to pressure." This means that if the pressure increases, the volume decreases, while if the pressure decreases, the volume increases.

Boyle's law is expressed mathematically as:

Pressure * Volume = constant

P * V = k

To obtain the proportionality factor k you must make the quotient:

$k=\frac{V2}{V1} =\frac{6 L}{2 L}$

k= 3

This means that if the volume of a gas increased from 2 to 6 L while the temperature was held constant, the pressure of the gas decreased by a factor of 3.

4 0

3 years ago

Which pair of elements are the most similar? A. Ca and F B. Na and Cl C. Ne and Ar D. K and Ca

Vinil7 [7]

Answer:

Explanation:

8 0

3 years ago

Which statement about gases is true

ExtremeBDS [4]

Umm well its particles are always moving and excited so they are not so close

7 0

3 years ago

Read 2 more answers

H2 how many elements?

rjkz [21]

There are two elements, two hydrogen to be exact.

4 0

3 years ago

The rate of decomposition of N2O5 in CCl4 at 317 K has been studied by monitoring the concentration of N2O5 in the solution. 2 N

pav-90 [236]

Answer:

Average rate of reaction is 0.000565 M/min

Explanation:

Applying law of mass action for the given reaction:

Average rate = $-\frac{1}{2}\frac{[N_{2}O_{5}]}{\Delta t}=\frac{1}{4}\frac{\Delta [NO_{2}]}{\Delta t}=\frac{\Delta [O_{2}]}{\Delta t}$

Where, $-\frac{1}{2}\frac{[N_{2}O_{5}]}{\Delta t}$ represents average rate of disappearance of $N_{2}O_{5}$ , $\frac{1}{4}\frac{[NO_{2}]}{\Delta t}$ represents average rate of appearance of $NO_{2}$ and $\frac{[O_{2}]}{\Delta t}$ represents average rate of appearance of $O_{2}$

Here, $-\frac{[N_{2}O_{5}]}{\Delta t}$ = $-\frac{(2.16-2.36)}{(177-0)}M/min=0.00113M/min$

So average rate of reaction = $[tex]-\frac{1}{2}\frac{[N_{2}O_{5}]}{\Delta t}$ [/tex] = $\frac{1}{2}\times (0.00113M/min)=0.000565M/min$

7 0

3 years ago