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kogti [31]
3 years ago
9

How do you find the number of moles of copper in copper gluconate? In my experiment I used 1.4 g of copper gluconate, and ended

up with 0.1 g of copper.
Chemistry
1 answer:
Kipish [7]3 years ago
7 0

<u>Answer:</u> 0.0016 moles of copper are present in copper gluconate

<u>Explanation:</u>

The number of moles is defined as the ratio of the mass of a substance to its molar mass.

The equation used is:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}} ......(1)

Given mass of copper = 0.1 g

Molar mass of copper = 63.55 g/mol

Plugging values in equation 1:

\text{Moles of copper}=\frac{0.1g}{63.55g/mol}=0.0016 mol

Hence, 0.0016 moles of copper are present in copper gluconate

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Answer:

If something is in a solid state of matter, it has a definite shape and volume. The volume of an object is the amount of space it occupies. A block of wood placed on a table retains its shape and volume, therefore, it is an example of a solid. If a liquid is poured on that same table, there are very different results

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A 155.0 g piece of copper at 128 oC is dropped into 250.0 g of water at 17.9 oC. (The specific heat of copper is 0.385 J/goC.) C
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Answer:

T_{eq}=23.85^oC

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In this case, as the copper's heat loss is gained by the water, the following energetic relationship is:

\Delta H_{Cu}=-\Delta H_{H_2O}

Therefore the equilibrium temperature shows up as:

m_{Cu}Cp_{Cu}(T_{Cu}-T{eq}) = m_{H_2O}Cp_{H_2O}(T_{eq}-T_{H_2O})\\\\T_{eq}=\frac{m_{Cu}Cp_{Cu}T_{Cu}-m_{H_2O}Cp_{H_2O}T_{H_2O}}{m_{Cu}Cp_{Cu}-m_{H_2O}Cp_{H_2O}} \\

Thus, by knowing that water's heat capacity is 4.18J/g°C, one obtains:

T_{eq}=\frac{155.0g*0.385\frac{J}{g^oC}*128^oC+250.0g*4.18\frac{J}{g^oC}*17.9^oC}{155.0g*0.385\frac{J}{g^oC}+250.0g*4.18\frac{J}{g^oC}}=23.85^oC

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For positive ions, the removal of electron increases the nuclear charge for an outermost electron because the outermost electrons are more strongly attracted by the nucleus. So, the effective nuclear charge increases for cations and thus, the size of the cation will be smaller than that of the corresponding atom.

For negative ions, the addition of electron decreases the nuclear charge for an outermost electron because the outermost electrons are less strongly attracted by the nucleus. So, the effective nuclear charge decreases for anions and thus, the size of the anion will be larger than that of the corresponding atom.

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