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krok68 [10]
2 years ago
15

What is the percent of Cu in CuSO4?

Chemistry
1 answer:
hoa [83]2 years ago
7 0
Anhydrous Copper sulfate is 39.81 percent copper and 60.19 percent sulfate by mass, and in its blue, hydrous form, it is 25.47% copper, 38.47% sulfate (12.82% sulfur) and 36.06% water by mass.
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My teacher never taught me how to do this helpppp!!!!!!!
djverab [1.8K]
Just search up how much the weight is equal to another, then multiply it, thats what i do lol sorry if im no help
6 0
2 years ago
What is the name of this alkane?
Otrada [13]
It's a cyclohexane ring with an ethyl group at 1 and a methyl group at 3. The Ethyl group is bigger and more important group get's the first position.

1-ethyl-3-methylcyclohexane
3 0
3 years ago
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A saturated solution is made by dissolving 0.327 g of a polypeptide (a substance formed by joining together in a chainlike fashi
faust18 [17]

Answer: The approximate molecular mass of the polypeptide is 856 g/mol

Explanation:

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

\pi=iMRT

Or,

\pi=i\times \frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution in L)}}\times RT

where,

\pi = osmotic pressure of the solution = 4.19 torr

i = Van't hoff factor = 1 (for non-electrolytes)

Mass of solute (polypeptide) = 0.327 g

Volume of solution = 1.70 L

R = Gas constant = 62.364\text{ L.torr }mol^{-1}K^{-1}

T = temperature of the solution = 26^oC=[273+26]K=299K

Putting values in above equation, we get:

4.19torr=1\times \frac{0.327}{\text{Molar mass of solute}\times 1.70}\times 62.364\text{ L.mmHg }mol^{-1}K^{-1}\times 299K\\\\\text{molar mass of solute}=856g/mol

Hence, the molar mass of the polypeptide is 856 g/mol

8 0
2 years ago
Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in aqueous solution. What should be the molarity of
dsp73
The first thing we need to do here is to recognize the unit of molarity and the units of the given percentage of nitric acid.

Molarity is mol HNO3 / L of solution. This is our aim

The given percentage is 0.68 g HNO3/ g solution

multiplying this with density to convert g solution into mL solution and dividing with the molecular weight of HNO3 (63 g/mol) to convert g HNO3 to mol. Therefore we obtain

0.016 mol/ mL or 16.23 mol/ L (M)

6 0
3 years ago
Which product forms from a reaction between calcium (Ca) and phosphorus
IrinaVladis [17]

Answer:

A

Explanation:

3 0
2 years ago
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