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3 years ago

What volume of a 0.33-M C12H22O11 solution can be diluted to prepare 25 mL of a solution with a concentration of 0.025 M?

Chemistry

1 answer:

Vsevolod [243]3 years ago

7 0

The dilution formula can be used to find the volume needed

c1v1 = c2v2

Where c1 is concentration and v1 is volume of the concentrated solution

And c2 is concentration and v2 is volume of the diluted solution to be prepared

c1 - 0.33 M

c2 - 0.025 M

v2 - 25 mL

Substituting these values in the equation

0.33 M x v1 = 0.025 M x 25 mL

v1 = 1.89 mL

Therefore 1.89 mL of the 0.33 M solution needs to be diluted up to 25 mL to make a 0.025 M solution

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Design an experiment to collect data that supports the claim that a 1.0M NaCl solution is a homogenous mixture.

Alika [10]

The claim is that NaCl mixture is a homogeneous mixture.
Homogeneous mixture means that the components of the mixtures cannot be determined or separated by the naked eye. However, these components can be separated using physical means, such as boiling, evaporation and condensation which will be used in this experiment.

First, we need to prepare one molar solution of NaCl. To do so, we will dilute a mass of 58.44 grams (molar mass of NaCl) in 1 liter of water.
By this, we will have NaCl solution.

We can notice that once the NaCl is diluted in water, all what you can see is a clear solution. You cannot see the separate particles of NaCl in water.
..............> observation I

Now, we will heat this solution until it boils and water starts evaporating. We will place a cold surface above the steam coming out from the boiling solution.

What we will observe is that when all the water evaporates, we can see white precipitate of NaCl in the bottom of the container. Examining the cold surface placed above the steam, we can see that the water has condensed on this surface.
.........>observation II

Based on this, we managed to use boiling, evaporation and condensation (physical methods) to restore the components of the solution separately.
.............>conclusion

Based on observation I, observation II and the conclusion. we were able to prove that NaCl solution is a homogeneous mixture.

7 0

3 years ago

Write 151.5 cm to meters in the correct significant figures

iogann1982 [59]

Answer:

1.515m

hope that helps uh :)

3 0

2 years ago

PLEASE HELPPPP! HOW DO I DO THIS???

kow [346]

Answer:

1552.83J Released

Explanation:

1. mass/m=225

Initial temp:86C, final:32.5C

Changed Temp: 32.5-86= -53.5C

s=0.129 J/gC

Formula: q= m times s times changed Temp.

q=(225)(0.129)(-53.5)

q= -1552.83 J

q=1552.83 J Released

8 0

3 years ago

About 6 × 109 g of gold is thought to be dissolved in the oceans of the world. If the total volume of the oceans is 1.5 × 1021 L

Lelechka [254]

First, determine the number of moles of gold.

Number of moles = $\frac{given mass in g}{molar mass}$

Given mass of gold = $6 \times 10^{9} g$

Molar mass of gold = 196.97 g/mol

Put the values,

Number of moles of gold = $\frac{6 \times 10^{9} g}{196.97 g/mol}$

= $0.03046\times 10^{9} mole$ or $3.046\times 10^{7} moles$

Now, molarity = $\frac{moles of solute}{volume of the solution in liters}$

Put the values, volume of ocean = $1.5 \times 10^{21} L$

Molarity = $\frac{3.046\times 10^{7} moles}{1.5 \times 10^{21} L}$

= $2.03\times 10^{-14} M$

Thus, average molar concentration = $2.03\times 10^{-14} M$

6 0

3 years ago

Two different bromide solutions are mixed with each other: Solution 1 is an aqueous solution of 4.85 g aluminum bromidein 150. m

erma4kov [3.2K]

Answer:

M=0.380 M.

Explanation:

Hello there!

In this case, given those two solutions of aluminum bromide and zinc bromide, it is firstly necessary to compute the moles of bromide ions in each solution as shown below:

$n_{Br^-}^{in\ AlBr_3}=4.85 gAlBr_3*\frac{1molAlBr_3}{266.69gAlBr_3}*\frac{3molBr^-}{1molAlBr_3} =0.05456molBr^-\\\\n_{Br^-}^{in\ ZnBr_2}=7.75gZnBr_2*\frac{1molZnBr_2}{225.22gZnBr_2}*\frac{2molBr^-}{1molZnBr_2} =0.06882molBr^-$

Now, we compute the total moles of bromide:

$n_{Br^-}=0.05456mol+0.06882mol\\\\n_{Br^-}=0.12338mol$

Then, the total volume in liters:

$150mL+175mL=325mL*\frac{1L}{1000mL} \\\\=0.325L$

Therefore, the concentration of total bromide is:

$M=\frac{0.12338mol}{0.325L}\\\\M=0.380M$

Best regards!

8 0

2 years ago