How many grams of CuSO4 would be needed to make 150 mL of 0.300 M CuSO4 solution?

Answers

Answer 1
Hey there!

Molar mass CuSO4 = 159.609 g/mol

Note that the molarity is in the enunciation of the issue, then the problem asks the mass of the solute :

Volume in liters of solution:

150 mL / 1000 => 0.15 L

Calculation of quantity in Moles of the solute :

1 L ----------------- 0.300 M
0.15 L ------------  ( moles )

Moles CuSO4 = 0.15 * 0.300 / 1

Moles CuSO4 = 0.045 moles

Therefore :

1 mole CuSO4 --------------------- 159.609 g
0.045 moles CuSO4-------------- mass

mass = 0.045 * 159.609 / 1

mass = 7.182405 g of CuSO4 


Answer 2

[tex]\boxed{{\text{7}}{\text{.182 g}}}[/tex] of [tex]{\text{CuS}}{{\text{O}}_{\text{4}}}[/tex] would be needed to make 150 mL of 0.300 M [tex]{\text{CuS}}{{\text{O}}_4}[/tex] solution.

Further Explanation:

The proportion of substance in the mixture is called concentration. The most commonly used concentration terms are as follows:

1. Molarity (M)

2. Molality (m)

3. Mole fraction (X)

4. Parts per million (ppm)

5. Mass percent ((w/w) %)

6. Volume percent ((v/v) %)

Molarity is a concentration term that is defined as the number of moles of solute dissolved in one litre of the solution. It is denoted by M and its unit is mol/L.

The formula to calculate the molarity of [tex]{\text{CuS}}{{\text{O}}_4}[/tex] solution is as follows:

[tex]{\text{Molarity of CuS}}{{\text{O}}_4}\;{\text{solution}}=\dfrac{{{\text{Moles}}\;{\text{of}}\;{\text{CuS}}{{\text{O}}_4}}}{{{\text{Volume }}\left( {\text{L}}\right){\text{ of}}\;{\text{CuS}}{{\text{O}}_4}\;{\text{solution}}}}[/tex]   ......(1)

Rearrange equation (1) to calculate the moles of [tex]{\text{CuS}}{{\text{O}}_4}[/tex].

[tex]{\text{Moles}}\;{\text{of}}\;{\text{CuS}}{{\text{O}}_4}= \left( {{\text{Molarity of HN}}{{\text{O}}_3}\;{\text{solution}}} \right)\left( {{\text{Volume of}}\;{\text{CuS}}{{\text{O}}_4}\;{\text{solution}}} \right)[/tex]   ......(2)

The volume of [tex]{\text{CuS}}{{\text{O}}_4}[/tex] solution is to be converted into L. The conversion factor for this is,

[tex]{\text{1 mL}} = {10^{ - 3}}\;{\text{L}}[/tex]

So the volume of [tex]{\text{CuS}}{{\text{O}}_4}[/tex] solution is calculated as follows:

[tex]\begin{aligned}{\text{Volume of CuS}}{{\text{O}}_4}\;{\text{solution}}&= \left( {{\text{150 mL}}}\right)\left({\frac{{{{10}^{ - 3}}\;{\text{L}}}}{{{\text{1 mL}}}}} \right)\\&= 0.15{\text{ L}}\\\end{aligned}[/tex]

The molarity of [tex]{\text{CuS}}{{\text{O}}_4}[/tex] solution is 0.300 M.

The volume of [tex]{\text{CuS}}{{\text{O}}_4}[/tex] solution is 0.15 L.

Substitute these values in equation (2).

[tex]\begin{aligned}{\text{Moles}}\;{\text{of}}\;{\text{CuS}}{{\text{O}}_4}&=\left( {{\text{0}}{\text{.300 M}}} \right)\left( {0.15{\text{ L}}}\right)\\&= 0.04{\text{5 mol}}\\\end{aligned}[/tex]

The formula to calculate the mass of [tex]{\text{CuS}}{{\text{O}}_4}[/tex] is as follows:

[tex]{\text{Mass of CuS}}{{\text{O}}_{\text{4}}}=\left( {{\text{Moles of CuS}}{{\text{O}}_{\text{4}}}} \right)\left( {{\text{Molar mass of CuS}}{{\text{O}}_{\text{4}}}} \right)[/tex]       ......(3)

Substitute 0.045 mol for the moles of [tex]{\text{CuS}}{{\text{O}}_4}[/tex] and 159.609 g/mol for the molar mass of [tex]{\text{CuS}}{{\text{O}}_4}[/tex] in equation (3).

[tex]\begin{aligned}{\text{Mass of CuS}}{{\text{O}}_{\text{4}}}&= \left({{\text{0}}{\text{.045 mol}}} \right)\left( {\frac{{{\text{159}}{\text{.609 g}}}}{{{\text{1 mol}}}}}\right)\\&= 7.1824{\text{05 g}}\\&\approx {\text{7}}{\text{.182 g}}\\\end{aligned}[/tex]

Therefore the mass of [tex]{\mathbf{CuS}}{{\mathbf{O}}_{\mathbf{4}}}[/tex]  is 7.182 g.

Learn more:

1. Calculation of volume of gas: https://brainly.com/question/3636135

2. Determine the moles of water produced: https://brainly.com/question/1405182

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Concentration terms

Keywords: molarity, CuSO4, dilution, moles of CuSO4, volume, solution, 0.045 mol, 150 mL, 0.300 M, 7.182 g, 159.609 g/mol, 0.15 L.


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Answers

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Final answer:

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Explanation:

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Final answer:

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Explanation:

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https://brainly.com/question/18771645

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Answers

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Explanation:

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Answers

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Answers

Answer : The number of atoms in a 4.0 mole sample of argon has greater number of atoms or four times as compared to the number of atoms in a 1.0 mole sample of neon.

Explanation :

As we know that the 1 mole of substance contains [tex]6.022\times 10^{23}[/tex] number of atoms.

As per question,

The 1 mole sample of neon contains [tex]6.022\times 10^{23}[/tex] number of atoms.

And,

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Answers

Do you want to be a scientist or not because I can help you with that it's so easy

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Answers

To find out how many moles of lead iodide are produced from 246.2 g of potassium iodide, first convert the mass of KI to moles using its molar mass, then divide by the stoichiometric ratio from the reaction. It is found that 246.2 g of KI will produce 0.7415 moles of lead iodide.

To determine how many moles of lead iodide are produced from 246.2 g of potassium iodide (KI), it is necessary to consider the stoichiometry of the reaction given by:

[tex]Pb(NO_3)_2 (aq) + 2 KI (aq) \rightarrow PbI_2 (s) + 2 KNO_{3} (aq)[/tex]

First, we calculate the number of moles of KI using its molar mass:

1 mole KI = 166.0028 g/mol. Therefore,

Moles of KI = 246.2 g

166.0028 g/mol = 1.483 moles of KI

Since the reaction requires 2 moles of KI for every mole of [tex]PbI_2[/tex], we divide the moles of KI by 2 to find the moles of [tex]PbI_2[/tex]:

[tex]PbI_2[/tex]moles = 1.483 moles / 2 = 0.7415 moles of [tex]PbI_2[/tex]

Therefore, 246.2 g of KI will produce 0.7415 moles of lead iodide.

in which situation can a mixture always be called a solution

Answers

A solution is when a solute is dissolved by a solvent. Mixture is practically when one doesn't dissolve or doesn't combine chemically and thus becomes a cloudy or foggy.
If I'm not wrong... A mixture is never a solution.

Answer: when the mixture is homogeneous

Explanation:

complete the tables of value for four aqueous solutions at 25 celcius

Answers

There are a total of four parameters you need to know to determine the pH of the solution. The pH and the pOH are negative logarithms of the concentrations of the ions H+ and OH-, respectively. These ions are the solvated ions of in an acid-base reaction. The equations are as follows:

pH = -log [H+]
pOH = -log [OH-]

Lastly, the pH range is from 1 to 14. So, it is easier to remember that pH + pOH is equal to 14. So, we use these three equations for the solutions.

Solution A:
pH = -log(6.4×10^-13) = 12.2
pOH = 14 - 12.2 = 1.8
1.8 = -log[OH-]
[OH-] = 0.016 M


Solution B:
pOH = -log(2.7×10^-10) = 9.6
pH = 14 - 9.6 = 4.4
4.4 = -log[H+]
[H+] = 3.98×10^-5 M


Solution C:
8.11 = -log(H+)
H+ = 7.76×10^-9
pOH = 14 - 8.11 = 5.89
5.89 = -log[OH-]
[OH-] = 1.288×10^-6 M


Solution D:
pH = 14 - 4.73 = 9.27
9.27= -log[H+]
[H+] = 5.37×10^-10 M
4.73 = -log[OH-]
[OH-] = 1.862×10^-5 M

The table of values for four aqueous solutions at   is as shown in Table 1.

Further Explanation:

The negative logarithm of hydrogen ion concentration is termed as pH while that of hydroxide ion concentration is pOH. Higher the pH of solution, more will be the alkalinity and vice-versa. Similarly, higher value of pOH indicates solution is acidic in nature .

The expression for pH is mentioned below.

[tex]\text{pH}=-\text{log}\left[\text{H}^+\right][/tex]                                             ...... (1)

Where [tex]\left[\text{H}^+\right][/tex]  is the concentration of hydrogen ion.

The expression for pOH is mentioned below.

[tex]\text{pOH}=-\text{log}\left[\text{OH}^-\right][/tex]                                        ...... (2)

Where [tex]\left[\text{OH}^-\right][/tex]  is the concentration of hydroxide ion.

pH and pOH are related to each other by following expression:

pH + pOH = 14                                               …… (3)

Solution A:

Substitute [tex]6.4\times10^{-13}\text{ M}[/tex]  for [tex]\left[\text{H}^+\right][/tex]  in equation (1).

[tex]\begin{aligned}\text{pH}=&-\text{log}\left[6.4\times10^{-13}\text{ M}\right]\\=&12.2\end{aligned}[/tex]

Rearrange equation (3) for pOH.

pOH = 14 – pH                                                                          …… (4)

Substitute 12.2 for pH in equation (4).

[tex]\begin{aligned}\text{pOH}=&14-12.2\\=&1.8\end{aligned}[/tex]

Rearrange equation (2) for [tex]\left[\text{OH}^-\right][/tex]  .

[tex]\left[\text{OH}^-\right]=10^{-\text{pOH}}[/tex]                                                                        …… (5)

Substitute 1.8 for pOH in equation (5).

[tex]\begin{aligned}\left[\text{OH}^-\right]=&10^{-\text{1.8}}\\=&0.016\text{ M}\end{aligned}[/tex]

Solution B:

Substitute  [tex]2.7\times10^{-10}\text{ M}[/tex] for [tex]\left[\text{OH}^-\right][/tex]  in equation (2).

[tex]\begin{aligned}\text{pOH}=&-\text{log}\left[2.7\times10^{-10}\text{ M}\right]\\=&\ 9.6\end{aligned}[/tex]

Rearrange equation (3) for pH.

pH = 14 – pOH                                                                                 …… (6)

Substitute 9.6 for pOH in equation (6).

[tex]\begin{aligned}\text{pH}=&14-9.6\\=&4.4\end{aligned}[/tex]

Rearrange equation (1) for [tex]\left[\text{H}^+\right][/tex] .

[tex]\left[\text{H}^+\right]=10^\text{-pH}[/tex]                                                                                        …… (7)  

Substitute 4.4 for pH in equation (7).

[tex]\begin{aligned}\left[\text{H}^+\right]=&10^\text{-4.4}\\=&0.0000398\text{ M}\end{aligned}[/tex]

Solution C:

Substitute 8.11 for pH in equation (7).

[tex]\begin{aligned}\left[\text{H}^+\right]=&10^\text{-8.11}\\=&0.0000000078\text{ M}\end{aligned}[/tex]

Substitute 8.11 for pH in equation (4).

[tex]\begin{aligned}\text{pOH}=&14-8.11\\=&5.89\end{aligned}[/tex]

Substitute 5.89 for pOH in equation (5).

[tex]\begin{aligned}\left[\text{OH}^-\right]=&10^{-\text{5.89}}\\=&0.000001288\text{ M}\end{aligned}[/tex]

Solution D:

Substitute 4.73 for pOH in equation (5).

[tex]\begin{aligned}\left[\text{OH}^-\right]=&10^{-\text{4.73}}\\=&0.00001862\text{ M}\end{aligned}[/tex]

Substitute 4.73 for pOH in equation (6).

[tex]\begin{aligned}\text{pH}=&14-4.73\\=&9.27\end{aligned}[/tex]

Substitute 9.27 for pH in equation (7).

[tex]\begin{aligned}\left[\text{H}^+\right]=&10^\text{-9.27}\\=&0.0000000005\text{ M}\end{aligned}[/tex]

Learn more:

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Answer details:

Grade: High School

Chapter: Acids, base and salts

Subject: Chemistry

Keywords: pH, pOH, H+, OH-, solution A, solution B, solution C, solution D, hydrogen ion, hydroxide ion, negative logarithm.

A buret containing NaOH solution has an initial volume reading of 0.14 mL.
25.00 mL of 0.1000 M HCl is titrated to a phenolphthalein end point, which occurs at a final reading of 27.16 mL NaOH. What is the molarity of the NaOH titrant?

Answers

IF molarity is in mol/dm3 then
Number of moles of HCL = 25/1000 x 0.1000=0.0025moles
Number of moles of NaOH= 0.0025 moles As 1 is to 1 ratio among them as from equation
      NaOH+HCL=NaCL+H20
Volume of NaOH used=27.16-0.14 ml=27.02ml
moles=volume x molarity
0.0025=27.02/1000 x molarity
ANSWER molarity of NaOH=0.0925 mol/dm3 

Answer:

0.092 M

Explanation:

NaOH(aq) + HCl(aq) ------> NaCl(aq) + H2O(l)

CA= concentration of acid= 0.1000 M

CB= concentration of base= ?????

VA= volume of acid= 25.00 mL

VB= volume of base= 27.16 mL

NA= number of moles of acid= 1

NB= number of moles of base= 1

CAVA/CBVB= nA/nB

CB= CAVAnB/VBnA

CB= 0.1000 M × 25.00 mL ×1/27.16 mL ×1

CB= 0.092 M

Which is a way that the model of the atom became more accurate due to new investigations? Select one: A. Experiments showed that the atom is made of smaller particles. B. Neutrons were detected orbiting the nucleus. C. Cathode-ray tube experiments showed that the atom is indivisible. D. Thomson showed that the atom contains alpha particles.

Answers

A. Experiments showed that the atom is made of smaller particles. Reason being: neutrons don't orbit the nucleus, the atom is not invisible, and alpha particles are larger than hydrogen atoms, which cancels out the possibility of atoms consisting of them (you can't be made of something larger than yourself).

Answer: The correct answer would 100% be A.  Experiments showed that the atom is made of smaller particles.

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Ammonia is a (Lewis) base. Ammonium hydroxide (not ammonia) is a base. Ammonium hydroxide: NH4OH (ammonia gas dissolved in water) Ammonia: NH3
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