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3 years ago

Write an equation that shows the formation of a copper (I) ion from a neutral copper atom

Chemistry

2 answers:

e-lub [12.9K]3 years ago

7 0

Answer: The equation is written below.

Explanation:

Copper is the 29th element of the periodic table having electronic configuration of $[Ar]3d^{10}4s^1$

When an atom looses an electron to form positive ion, it is called an oxidation reaction.

Copper will loose 1 electron to form +1 ion. The equation for the formation of copper (I) ion from neutral copper atom follows:

$Cu\rightarrow Cu^++1e^-$

Hence, the equation is written above.

kumpel [21]3 years ago

5 0

Cu ----> Cu(+) + 1 e-

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Answer:

The percent yield of a reaction is 48.05%.

Explanation:

$WO_3+3H_2\rightarrow W+3H_2O$

Volume of water obtained from the reaction , V= 5.76 mL

Mass of water = m = Experimental yield of water

Density of water = d = 1.00 g/mL

$M=d\times V = 1.00 g/mL\times 5.76 mL=5.76 g$

Theoretical yield of water : T

Moles of tungsten(VI) oxide = $\frac{51.5 g}{232 g/mol}=0.2220 mol$

According to recation 1 mole of tungsten(VI) oxide gives 3 moles of water, then 0.2220 moles of tungsten(VI) oxide will give:

$\frac{3}{1}\times 0.2220 mol=0.6660 mol$

Mass of 0.6660 moles of water:

0.666 mol × 18 g/mol = 11.988 g

Theoretical yield of water : T = 11.988 g

To calculate the percentage yield of reaction , we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

$=\frac{m}{T}\times 100=\frac{5.76 g}{11.988 g}\times 100=48.05\%$

The percent yield of a reaction is 48.05%.

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