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erik [133]
2 years ago
13

On a shelf sits a bottle of NaCl solutions with a molar concentration of 2.50 M and a total volume of 300 mL. If 15.0 mL of the

solution is added to 985 mL of pure water, what is the molar concentration of NaCl in the new solution
Chemistry
1 answer:
Stella [2.4K]2 years ago
8 0

Given :

On a shelf sits a bottle of NaCl solutions with a molar concentration of 2.50 M and a total volume of 300 mL. If 15.0 mL of the solution is added to 985 mL of pure water .

To Find :

The molar concentration of NaCl in the new solution.

Solution :

Molarity M is given by :

M=\dfrac{\text{Total number of moles}}{\text{Volume in liters}}

Now , 1000 ml of NaCl contains 2.5 moles of NaCl .

So , 15 ml of NaCl contains :

n=\dfrac{15\times 2.5}{1000}\\\\n=3.75\times 10^{-2}\ moles

New , volume V = 15 + 985 = 1000 ml = 1 L .

So , putting value of n and V in above equation :

M=\dfrac{3.75\times 10^{-2}}{1}\ M\\\\M=0.0375\ M

Therefore ,  the molar concentration of NaCl in the new solution is 0.0375 M .

Hence , this is the required solution .

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Answer:

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Explanation:

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2 years ago
Formula for zinc + nitric acid = zinc nitrate
Maslowich

Answer:

<em>Zinc nitrate is an inorganic chemical compound with the formula Zn(NO3)2. This white, crystalline salt is highly deliquescent and is typically encountered as a hexahydrate Zn(NO3)2•6H2O. It is soluble in both water and alcohol.</em>

Explanation:

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Meteorologists use weather balloons to carry weather instruments high into the atmosphere. When it is first released at Earth’s
inessss [21]

The pressure of the gas used in the weather balloon increases to expand the balloon.

Explanation:

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8 0
3 years ago
Find the density of a cube on Earth that weighs 1.5 kg and has a side-length of 10 cm.
Inga [223]

Answer:

1.5g/cm³

Explanation:

density=mass÷volume

mass= 1.5kg (<em>c</em><em>h</em><em>a</em><em>n</em><em>g</em><em>e</em><em> </em><em>i</em><em>n</em><em>t</em><em>o</em><em> </em><em>g</em>) = 1500g

volume of the cube = 10×10×10 = 1000cm³

density= divide 1500g÷1000cm = 1.5g/cm³

<h2>Density= 1.5g/cm³</h2>

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4 0
3 years ago
Determine the empirical formula of a compound containing 1.71 g of silicon and 8.63 g of chlorine.
Basile [38]

Answer:

The answer to your question is: SiCl₄

Explanation:

Data

amount of Si      1.71 g

amount of Cl     8.63 g

MW Si = 28 g

MW Cl = 35.5

Process (rule of three)

For Si                                                        For Cl

        28 g of Si ------------------ 1 mol                      35.5 g of Cl --------------- 1 mol

          1.71g of Si  ---------------   x                              8.63 g of Cl --------------  x

         x = 1.71 x 1 / 28 = 0.06 mol                          x = 8.63 x 1 / 35.5 = 0.24 mol

Now, divide both results by the lowest of them.

Si = 0.06 mol / 0.06 = 1 molecule of Si     Cl = 0.24 / 0.06 = 4 molecules of Cl

Finally

                     Si₁ Cl₄ or SiCl₄

8 0
3 years ago
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