Which of the foods did you determine to be nutritious snacks?

You drop an unknown substance that weighs 8.3g into a graduated cylinder with 6ml of water. The water rises to 8ml when you drop

valkas [14]

Answer:

<h3>The answer is 4.15 g/mL</h3>

Explanation:

The density of a substance can be found by using the formula

$density = \frac{mass}{volume} \\$

From the question

mass of object = 8.3 g

volume = final volume of water - initial volume of water

volume = 8 - 6 = 2 mL

So we have

$density = \frac{8.3}{2} \\$

We have the final answer as

Hope this helps you

7 0

3 years ago

A salt bridge is used to provide electrons a path to flow from the electrode of one half-cell to another. TRUE FALSE

kicyunya [14]

It is false. The salt bridge is not a path for electrons, but a path for ions to flow from one half-cell to another. It help to balance the charge between the oxidation and reduction vessels.

8 0

3 years ago

Which property describes a mixture? It cannot be separated by physical methods. It has a single chemical composition. It cannot

natima [27]

Answer:

it cannot be separated by physical methods

Explanation:

7 0

3 years ago

Read 2 more answers

Which statement about cellulose is true?

abruzzese [7]

Uhh I think u forgot the picture :,)

3 0

3 years ago

While ethanol (CH3CH2OH) is produced naturally by fermentation, e.g. in beer- and wine-making, industrially it is synthesized by

grin007 [14]

Answer:

$n_{C2H_5OH}^{eq}=14.234mol$

Explanation:

Hello,

In this case, the reaction is:

$C_2H_4+H_2O\rightleftharpoons CH_3CH_2OH$

Thus, the law of mass action turns out:

$Kc=\frac{[CH_3CH_2OH]_{eq}}{[H_2O]_{eq}[CH_2CH_2]_{eq}}$

Thus, since at the beginning there are 29 moles of ethylene and once the equilibrium is reached, there are 16 moles of ethylene, the change $x$ result:

$[CH_2CH_2]_{eq}=29mol-x=16mol\\x=29-16=13mol$

In such a way, the equilibrium constant is then:

$Kc=\frac{\frac{x}{V} }{\frac{16mol}{V}* \frac{3mol}{V}} =\frac{\frac{13mol}{75.0L} }{\frac{16mol}{75.0L}* \frac{3mol}{75.0L}} =20.31$

Thereby, the initial moles for the second equilibrium are modified as shown on the denominator in the modified law of mass action by considering the added 15 moles of ethylene:

$Kc=\frac{\frac{13+x_2}{V} }{\frac{16+15-x_2}{V}* \frac{3-x_2}{V}} =20.31$

Thus, the second change, $x_2$ finally result (solving by solver or quadratic equation):

$x_2=1.234mol$

Finally, such second change equals the moles of ethanol after equilibrium based on the stoichiometry:

$n_{C2H_5OH}^{eq}=x+x_2=13mol+1.234mol\\n_{C2H_5OH}^{eq}=14.234mol$

Best regards.

5 0

3 years ago