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krok68 [10]
3 years ago
15

What is the percent of Cu in CuSO4?

Chemistry
1 answer:
hoa [83]3 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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Sodium metal (Na) reacts with chlorine gas ( Cl2 ) to produce solid sodium chloride (NaCl).
Fynjy0 [20]

The unbalanced equations are the equations with different atomic numbers on the sides of the reaction. The unbalanced reaction is Na + Cl₂ → NaCl

<h3>What are balanced equations?</h3>

Balanced equations are the chemical reaction representation that has an equal number of the atomic number of the same species on the left and the right side of the reaction.

An unbalanced equation between sodium metal and chloride can be shown as:

Na + Cl₂ → NaCl

The equation is unbalanced as the number of chloride ions is more on the reactant side than the product side.

The balanced reaction will be:

2 Na + Cl₂  2NaCl

Learn more about balanced equations here:

brainly.com/question/10306791

#SPJ1

3 0
2 years ago
In which way does a na1+ ion differ from a neutral na atom?
DochEvi [55]
Na 1+ misses 1 electron
8 0
3 years ago
HELP IM TIMEDDDDDDDDDDDD
nadya68 [22]

Answer:

ethier a dessert or a plains

though plains can get rain in the summer

from me living in both

it seems more like a dessert

3 0
3 years ago
Read 2 more answers
Anyone can help me in science assignment
olya-2409 [2.1K]

Answer:

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Explanation:

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6 0
3 years ago
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A buffer solution is made that is 0.347 M in H2C2O4 and 0.347 M KHC2O4.
irga5000 [103]

Answer:

1. pH = 1.23.

2. H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)

Explanation:

Hello!

1. In this case, for the ionization of H2C2O4, we can write:

H_2C_2O_4\rightleftharpoons HC_2O_4^-+H^+

It means, that if it is forming a buffer solution with its conjugate base in the form of KHC2O4, we can compute the pH based on the Henderson-Hasselbach equation:

pH=pKa+log(\frac{[base]}{[acid]} )

Whereas the pKa is:

pKa=-log(Ka)=-log(5.90x10^{-2})=1.23

The concentration of the base is 0.347 M and the concentration of the acid is 0.347 M as well, as seen on the statement; thus, the pH is:

pH=1.23+log(\frac{0.347M}{0.347M} )\\\\pH=1.23+0\\\\pH=1.23

2. Now, since the addition of KOH directly consumes 0.070 moles of acid, we can compute the remaining moles as follows:

n_{acid}=0.347mol/L*1.00L=0.347mol\\\\n_{acid}^{remaining}=0.347mol-0.070mol=0.277mol

It means that the acid remains in excess yet more base is yielded due to the effect of the OH ions provided by the KOH; therefore, the undergone chemical reaction is:

H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)

Which is also shown in net ionic notation.

Best regards!

4 0
3 years ago
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