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4 years ago

MAXIMUM POINTS!! PLEASE HELP!

Chemistry

1 answer:

Salsk061 [2.6K]4 years ago

5 0

1.The balanced chemical equation is as below

Cu + 2 Ag(NO)₃ → Cu(NO₃)₂ +2 Ag

2.The much silver produced is 50.9 g

Step 1: find the moles of Cu

moles = mass÷ molar mass

from periodic table the molar mass of Cu = 63.5 g/mol

moles is therefore = 15.00 g÷ 63.5 g/mol = 0.236 moles

Step 2: use the mole ratio to determine the moles of Ag

from equation above Cu:Ag is 1:2

therefore the moles of Ag =0.236 moles x 2/1 =0.472 moles

Step 3: find the mass of Ag

mass = moles x molar mass

from periodic table the molar mass of Ag =107.87 g/mol

mass = 0.472 moles x 107.87 g/mol = 50.9 g

3. The much silver collected worth is 1.797 oz

convert 50.9 g to oz

that is 1 gram = 0.0353 oz

50.9 grams =? oz

by cross multiplication

={(50.9 grams x 0.0353 oz) / 1 gram} =1.797 oz

You might be interested in

How many moles are in 25.3 g of a sample of potassium nitrate (KNO3)?

aksik [14]

Answer:

25.3 g of KNO₃ contain 0.25 moles.

Explanation:

Given data:

Number of moles of KNO₃ = ?

Mass of KNO₃ = 25.3 g

Solution:

Formula:

Number of moles = mass/ molar mass

Molar mass of KNO₃:

KNO₃ = 39 + 14+ 16×3

KNO₃ = 101 g/mol

Now we will put the values in formula.

Number of moles = 25.3 g/ 101 g/mol

Number of moles = 0.25 mol

Thus, 25.3 g of KNO₃ contain 0.25 moles.

8 0

3 years ago

Question 1

sukhopar [10]

Answer:

if I'm not mistaken, it is called a mixture

4 0

4 years ago

If 10.4 grams of iron metal react with 28.4 grams of silver nitrate, how many grams of iron nitrate can be formed and how many g

Trava [24]

Answer:

71.1

Explanation:

1 mol Fe = 10.4 g/55.85 g/mol = 0.186

1 mol AgNo3 = 28.4 g/169.87 g/mol = 0.178 mol AgNo3

then since Ag:Fe is 1:3, AgNo3 is the limiting reactant

So now

0.178 moles * 1/3 * 241.83 g/mol Fe(NO3)3 = 14.35 g Fe(NO3)3

Excess reactant: 0.178 moes AgNO3 * 1/3 = 0.059

0.186 - 0.059 = 0.127 moles Fe * 55.85 g/mol Fe = 7.1 g Fe excess

8 0

3 years ago

4 ) A sample of gas has a pressure of 1 atm while occupying 300 What volume will the gas occupy at 2 atm ? 10 points total . Sho

Gnesinka [82]

Answer:

It would be 600

Explanation:

300 x 2 = 600

4 0

3 years ago

Calculate the standard cell emf for the reaction: ClO3−(aq)+3Cu(s)+6H+(aq)→Cl−(aq)+3Cu2+(aq)+3H2O(l) Pt is used as an inert elec

Rama09 [41]

Answer: The standard EMF of the cell is 1.11 V

Explanation:

For the given chemical equation:

$ClO_3^-(aq)+3Cu(s)+6H^+(aq.)\rightarrow Cl^-(aq.)+3Cu^{2+}(aq.)+3H_2O(l)$

The half reaction follows:

Oxidation half reaction: $Cu(s)\rightarrow Cu^{2+}+2e^-;E^o_{Cu^{2+}/Cu}=0.34V$ ( × 3)

Reduction half reaction: $ClO_3^-(aq.)+6H^+(aq.)+6e^-\rightarrow Cl^-(aq.)+3H_2O(l);E^o=1.45V$

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

Calculating the $E^o_{cell}$ using above equation, we get:

$E^o_{cell}=1.45-0.34=1.11V$

Hence, the standard EMF of the cell is 1.11 V

6 0

3 years ago