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myrzilka [38]
2 years ago
14

Copper metal (Cu) reacts with silver nitrate (AgNO3) in aqueous solution to form Ag and Cu(NO3)2. An excess of AgNO3 is present.

The balanced chemical equation is shown below. Cu + 2AgNO3 mc021-1.jpg Cu(NO3)2 + 2Ag The molar mass of Cu is 63.5 g/mol. The molar mass of Ag is 107.9 g/mol. What mass, in grams, of Ag is produced from reaction of 31.75 g of Cu? 26.95 107.9 215.91 431.82
Chemistry
2 answers:
Dvinal [7]2 years ago
7 0

Answer: 107.9 grams of Ag will be produced.

Explanation: Reaction of Copper metal with Silver nitrate follows:

Cu(s)+2AgNO_3(aq.)\rightarrow 2Ag(s)+Cu(NO_3)_2(aq.)

As AgNO_3 is present in excess, therefore Copper metal is considered as the limiting reagent because it limits the formation of product.

We are given 31.75 grams of Copper metal are used in the reaction

Molar mass of Copper = 63.5 g/mol.

Moles can be calculated by:

Moles=\frac{\text{Given mass}}{\text{Molar mass}}     ....(1)

\text{Moles of Copper}=\frac{31.75g}{63.5g/mol}=0.5moles

From the reaction, we see that

For every 1 mole of Copper metal, 2 moles of Ag are produced.

For 0.5 moles of Copper, moles of Ag produced = \frac{2}{1}\times 0.5

Moles of Ag produced = 1 mole

Mass of Ag produced can be calculated by using equation 1.

Mass of Ag produced will be = Moles × Molar mass

Molar mass of Ag = 107.9 g/mol

\text{Mass of Ag produced}=1\times 107.9g/mol

Mass of Ag produced = 107.9 g/mol

sweet [91]2 years ago
3 0
Toichiometry time! Remember to look at the equation for your molar ratios in other problems.
31.75 g Cu | 1 mol Cu | 2 mol Ag | 107.9 g Ag         6851.65⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻   →   ⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻  = 107.9 g Ag       ∅        | 63.5 g Cu | 1 mol Cu | 1 mol Ag              63.5

There's also a shorter way to do this: Notice the molar ratio from Cu to Ag, which is 1:2. When you plug in 31.75 into your molar mass for Cu, it equals 1/2 mol. That also means that you have 1 mol Ag because of the ratio, qhich you can then plug into your molar mass, getting 107.9 as well.

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Answer is: A. Chemical energy to electromagnetic energy and thermal energy.  

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2 years ago
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Quinine, an antimalarial drug, is 8.63% nitrogen. There are two nitrogen atoms per molecule. What is the molecular weight of qui
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Answer:

324.18 g/mol

Explanation:

Let the molecular mass of the antimalarial drug, Quinine is x g/mol

According to question,

Nitrogen present in the drug is 8.63% of x

So, mass of nitrogen = \frac {8.63}{100}\times x

Also, according to the question,

2 atoms are present in 1 molecule of the drug.

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So, mass of nitrogen = 14.01×2 = 28.02

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solving for x, we get:

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6 0
2 years ago
Describe how you would prepare exactly 100 mL of 0.109 M picolinate buffer, pH 5.61. Possible starting materials are pure picoli
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Answer:

1.342g of picolinic acid and 6.743mL of 1.0M NaOH diluting the mixture to 100.0mL

Explanation:

<em>The pKa of the picolinic acid is 5.4.</em>

Using Henderson-Hasselbalch formula for picolinic-picolinate buffer:

pH = pKa + log [Picolinate] / [Picolinic]

<em>Where [] could be taken as moles of each species</em>

<em />

5.61 = 5.4 + log [Picolinate] / [Picolinic]

0.21 = log [Picolinate] / [Picolinic]

1.62181 = [Picolinate] / [Picolinic] <em>(1)</em>

<em></em>

Now, both picolinate and picolinic acid will be:

0.100L * (0.109mol / L) =

0.0109 moles = [Picolinate] + [Picolinic] <em>(2)</em>

<em></em>

First, as we will start with picolinic acid, we need add:

0.0109 moles picolinic acid * (123.10g/mol) = 1.342g of picolinic acid

Now, replacing (2) in (1):

1.62181 = 0.0109 moles - [Picolinic] / [Picolinic]

1.62181 [Picolinic] = 0.0109 moles - [Picolinic]

2.62181 [Picolinic] = 0.0109 moles

[Picolinic] = 4.157x10⁻³ moles

And:

[Picolinate] = 0.0109 - 4.157x10⁻³ moles =

<h3>6.743x10⁻³ moles</h3><h3 />

To obtain these moles of picolinate ion we need to make the reaction of the picolinic acid with NaOH:

Picolinic acid + NaOH → Picolinate + Water

<em>That means to obtain 6.743x10⁻³ moles of picolinate ion we need to add 6.743x10⁻³ moles of NaOH</em>

<em />

6.743x10⁻³ moles of NaOH that is 1.0M are, in mL:

6.743x10⁻³ moles * (1L / 1mol) = 6.743x10⁻³L * 1000 =

<h3>6.743mL of the 1.0M NaOH must be added</h3><h3 />

Thus, we obtain the desire moles of picolinate and picolinic acid to obtain the buffer we want, the last step is:

<h3>Dilute the mixture to 100mL, the volume we need to prepare</h3>
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Answer:

The correct order will be

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b. Dilute the solution with water to the 250.0 mL mark.

Explanation:

Preparation of NaCl solution in 250.0 ml volumetric flask:

Add the weighed NaCl directly to volumetric flask and add small amount of water to it and mix it will until all NaCl gets dissolved( if not add small water amount of water more)

After dissolving NaCl add the water upto the mark.

The correct order will be

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e. Dissolve the NaCl in less than 250 mL of water and mix well.

b. Dilute the solution with water to the 250.0 mL mark.

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