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Karo-lina-s [1.5K]
3 years ago
6

If I use the dropped to add 10 mL of drink mix to 50 mL of water, what is the percent by volume of the drink mix in the solution

?
Chemistry
1 answer:
deff fn [24]3 years ago
8 0

The percent by volume of the drink mix in the solution is 16.67%.

Explanation:

The formula used to determine the percentage by volume of the solute present in the solution is by finding the ratio of the volume of solute to the total volume of the solution multiplied by 100.

Percentage by volume = \frac{Volume of solute}{Total volume of solution} \times 100

As in the present case, the solute is the drink mix and its volume is given as 10 mL. And the solution is the water. So the total volume of water is 60 mL.

Then the percentage by volume of the drink mix in the solution will be

Percentage by volume = \frac{10}{60} \times 100 =20

So, the percent by volume of the drink mix in the solution is 16.67%.

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A sample of As weighs 73.2 grams. Will a sample of Kr that contains the same number of atoms weigh more or less than 73.2 grams?
My name is Ann [436]

Answer:

This means a sample of 73.2 grams As atoms weighs less than the same amount of Kr atoms

Explanation:

Step 1: Data given

Mass of As = 73.2 grams

Molar mass As = 74.92 g/mol

Molar mass of Kr = 83.80 g/mol

Step 2: Calculate moles As

Moles As = Mass As / molar mass As

Moles As = 73.2 grams . 74.92 g/mol

Moles As = 0.977 moles

Step 3: Calculate As atoms

As atoms = moles As * number of Avogadro

As atoms = 0.977 moles * 6.02 * 10^23

As atoms = 5.88 *10^23 As atoms

Step 4: Calculate moles Kr

Moles Kr = Atoms Kr / number of Avogadro

Moles Kr = 5.88 * 10^23 Kr atoms / 6.02 *10^23

Moles Kr = 0.977 moles

Step 5: Calculate mass Kr

Mass Kr = moles Kr * molar mass Kr

Mass Kr = 0.977 moles * 83.80 g/mol

Mass Kr = 81.9 grams

This means a sample of 73.2 grams As atoms weighs less than the same amount of Kr atoms

8 0
3 years ago
A 100 W light bulb is placed in a cylinder equipped with a moveable piston. The light bulb is turned on for 2.0×10−2 hour, and t
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Answer:

(a) ΔU = 7.2x10²

(b) W = -5.1x10²

(c) q = 5.2x10²

Explanation:

From the definition of power (p), we have:

p = \frac {\Delta W}{\Delta t} = \frac {\Delta U}{\Delta t} (1)

<em>where, p: is power (J/s = W (watt)) W: is work = ΔU (J) and t: is time (s) </em>  

(a) We can calculate the energy (ΔU) using equation (1):

\Delta U = p \cdot \Delta t = 100 \frac{J}{s} \cdot 2.0\cdot 10^{-2} h \cdot \frac{3600s}{1h} = 7.2 \cdot 10^{2} J  

(b) The work is related to pressure and volume by:

\Delta W = -p \Delta V

<em>where p: pressure and ΔV: change in volume = V final - V initial      </em>

\Delta W = - p \cdot (V_{fin} - V_{ini}) = - 1.0 atm (5.88L - 0.85L) = - 5.03 L \cdot atm \cdot \frac{101.33J}{1 L\cdot atm} = -5.1 \cdot 10^{2} J

(c) By the definition of Energy, we can calculate q:

\Delta U = \Delta W + \Delta q

<em>where Δq: is the heat transfer </em>

\Delta q = \Delta U - \Delta W = 7.2 J - (-5.1 \cdot 10^{2} J) = 5.2 \cdot 10^{2} J    

I hope it helps you!  

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