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2 years ago

9

Two rocks one weighing 100m and the other weiging 200n are drpped from a 50-m cliff at the same time when both rocks are 10m fro

m the ground and air friction is ignored which is the same

1 answer:

Ratling [72]2 years ago

8 0

Answer:

Ignore all friction

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What dissolved solid has the highest concentration in the ocean? sodium chloride

jeyben [28]

Answer:

chloride and sodium.

Explanation:

These two make up over 90% of all dissolved ions in seawater.

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3 years ago

A potassium bromide solution is 8.30% potassium bromide by mass and its density is 1.03 g/ml. what mass of potassium bromide is

pshichka [43]

<u>Answer:</u> The mass of potassium bromide present in 41.2 mL of solution will be 3.522 grams.

<u>Explanation:</u>

We are given that KBr is present in 8.3% KBr solution, which means that 8.3 grams of potassium bromide is present in 100 gram of the solution.

To calculate the volume of KBr, we use the formula:

$Density=\frac{Mass}{Volume}$

Mass of the solution = 100 grams

Density of KBr solution = 1.03g/mL

Volume of the solution = ? mL

Putting values in above equation, we get:

$1.03g/mL=\frac{100g}{Volume}\\\\Volume=97.08mL$

Now, to calculate the mass of KBr in 41.2mL of the solution, we use unitary method.

In 97.08 mL of solution, mass of KBr present is 8.3 grams.

So, 41.2 mL of solution will contain = $\frac{8.3g}{97.08mL}\times 41.2mL=3.522g$ of KBr.

Hence, the mass of potassium bromide present in 41.2 mL of solution will be 3.522 grams.

7 0

3 years ago

After performing a dilution calculation, you determine you need 25.0 milliliters of an aqueous stock solution to make 100.0 mill

IRINA_888 [86]

A is obviously out because it leads to a volume of 125.0 milliliters of the new solution and gives you a lower concentration than you were aiming for.

D is out because you are adding 75 milliliters of the stock solution, so your concentration would be too high. You only need 25.0 milometers of stock solution per 100 milliliters of the new solution.

C is also out because it leads to 50.0 milliliters stock solution per 100 milliliters of the new solution and hence the wrong concentration.

B is by default the correct answer. It also details the correct technique. First you add the stock solution (This you know from your calculations to be 25 milliliters.) then you add the water up to the volume you needed. (Because the calculations only tell you the total volume of water not what you need to add) You also add the water last so you can rinse the neck of the flask to make sure you also get all the stock solution residue into the stock solution.

I would add the final step of stirring, but B is the only answer that can be correct.

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3 years ago

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The reaction A → products is first order. If the initial concentration of A is 0.646 M and, after 72.8 seconds have elapsed, the

mario62 [17]

Answer: 0.00867 moldm-3

Explanation:

Since the reaction is 1st order,

Rate of reaction=∆[A]÷t

0.646-0.0146/72.8= 0.00867

Remember that in a first order reaction, the rate of reaction depends on change in the concentration of only one of the reaction species, A in the problem above.

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3 years ago