Chemistry, 08.02.2022 14:00 Hhyyuu

How many grams of copper metal are required to react with 4.00 grams of silver nitrate to produce copper (II) nitrate?

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Answer from: weeman6546

0.7479 grams of copper metal is required to react with 4 grams of silver nitrate...~

How many grams of copper metal are required to react with 4.00 grams of silver

nitrate to produce

Answer from: sassycutie523

0.753 grams of copper metal are required to react with 4.00 grams of silver nitrate to produce copper (II) nitrate

Explanation:

mass = moles * Mr

Equation of this reaction:

Cu(s) + 2AgNO₃(aq) → Cu(NO₃)₂(aq) + 2Ag(s)

---------------------------------------------------------------------------------------------

Find moles of silver nitrate:

4 = moles * 170

→ $\frac{2}{85}$ moles

Here we "2" molar ratio before silver nitrate and "1" before copper metal, so copper metal has half the moles of silver nitrate. copper metal has $\frac{2}{85}$ ÷ 2 = $\frac{1}{85}$ moles

so find the mass of copper metal:

mass = $\frac{1}{85}$ * 64

mass of copper is 0.753 grams.

Answer from: Quest

assuming that your periodic table has eighteen columns, you should highlight the four elements in the third column from the left of the table.

explanation

neutral atoms of elements after calcium contain d electrons. the energy of d orbitals of a particular main shell lies between the energy of the s and p orbitals of the next main shell. for example, electrons in the $3d$ have energy higher than those in the $4 s$ orbital while lower than those of the $4 p$ orbital.

by the aufbau principle, electrons fill orbitals one at a time, starting from those of the lowest energy. the $4s$ orbital is therefore filled right after $3p$ orbitals and before $3 d$ orbitals.

potassium has electron configuration $[\text{ar}] 4 s^{1}$ and calcium $[\text{ar}] 4 s^{2}$ . the s orbital of a main shell holds a maximum of two electrons. the next element would thus have the electron configuration $[\text{ar}] 4 s^{2} \; 3d^{1}$ and satisfy the requirements.

a neutral atom of the element contains three electrons on top of the most recent noble gas element. it should thus be placed in the third column of the periodic table. both scandium and yttrium would satisfy the conditions. an aufbau diagram indicates that the 4f orbital is filled before the 5d. however, it appears that exceptions exists for both lanthanum and actinium. neutral atoms of the two d-block elements also demonstrate the $ns^{2} \; (n-1) p^{1}$ and should be highlighted as well.

Answer from: Quest

Adozen is 12, and your needing to calculate the weight of one medium egg, so i would do 21/12=1.75 per egg, and check your answer by doing 1.75 * 12= 21 : ) : ): ): ): ) i hoped i

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