All categories

3 years ago

2.00g of Copper wire was added to 50.0 mL of a 0.050M solution of Silver Nitrate. The reaction

Chemistry

1 answer:

Svet_ta [14]3 years ago

4 0

Answer:

See explanation

Explanation:

Now, the equation of the reaction is;

Cu(s) + 2AgNO3(aq) --------> Cu(NO3)3 (aq) + 2Ag(s)

Number of moles of silver precipitate = 0.24/108 g/mol= 0.0022 moles

Number of moles of AgNO3 = 50/1000 * 0.050 = 0.0025 Moles

If 2 moles of AgNO3 yields 2 moles of Ag

Then 0.0025 moles of AgNO3 yields 0.0025 * 2/2 = 0.0025 moles of Ag

Number of moles of Cu = 2.00g/63.5 g/mol = 0.03 moles

If 1 mole of Cu yields 2 moles of Ag

0.03 moles of Cu yields 0.03 moles * 2/1 = 0.06 moles of Ag

Hence AgNO3 is the limiting reactant

Theoretical yield of Ag = 0.0025 moles of Ag * 108 = 0.27 g

Actual yield of Ag = 0.24 g

Percent yield = actual yield/theoretical yield * 100

Percent yield = 0.24/0.27 * 100

Percent yield = 88.8%

You might be interested in

Classify the following change as physical or chemical: A match lights when struck.

mr_godi [17]

I think the answer is:
B. Chemical Change.

8 0

3 years ago

gulaghasi [49]

Taking into account the change of units, the size of the tank is 54.88 liters.

In first place, the rule of three is a tool used to quickly solve problems involving a proportional relationship between two variables.

The rule of three is a way of solving problems of proportionality between three known values and an unknown value, establishing a relationship of proportionality between all of them.

If the relationship between the magnitudes is direct, that is, when one magnitude increases, so does the other (or when one magnitude decreases, so does the other), the direct rule of three should be applied.

The rule of three is applied as follows, with A, B and C being known values and D being an unknown value: if A maintains a certain relationship with B, and it is known that C and D have the same relationship, it is possible to calculate the unknown value D using the following expression:

$D=\frac{CxB}{A}$

In this case, the rule of three can be applied as follows: if 1 gallon is equal to 3.785 liters, 14.5 gallons are equal to how many liters?

$D=\frac{14.5 gallonsx3.785 liters}{1 gallon}$

Solving:

D= 54.88 liters

In summary, the size of the tank is 54.88 liters.

Learn more with this example:

7 0

2 years ago

Read 2 more answers

Calculate the number of moles in 24.18 liters of neon gas at STP.

uranmaximum [27]

The number of moles of Neon gas : 1.079

<h3>Further explanation</h3>

Given

24.18 Liters of Neon gas at STP

Required

The number of moles

Solution

There are 2 conditions that are usually used as a reference in chemical calculations (mainly for determining the volume per mole of a gas or the molar volume), STP and RTP

Conditions at T 0 ° C and P 1 atm are stated by STP (Standard Temperature and Pressure). At STP, Vm is 22.4 liters / mol.

So mol for 24.18 L :

= 24.18 : 22.4

= 1.079 moles

8 0

3 years ago

Define motion of particles in matter

Wittaler [7]

Answer:

Explanation:

The states that all of the particles that make up matter are constantly in motion. As a result, all particles in matter have kinetic energy. The kinetic theory of matter helps explain the different states of matter—solid, liquid, and gas. ... Particles do not always move at the same speed.

5 0

3 years ago

A 0.1510 gram sample of a hydrocarbon produces 0.5008 gram CO2 and 0.1282 gram H2O in combustion analysis. Its

Over [174]

In a combustion of a hydrocarbon compound, 2 reactions are happening per element:

C + O₂ → CO₂
2 H + 1/2 O₂ → H₂O

Thus, we can determine the amount of C and H from the masses of CO₂ and H₂O produced, respectively.

1.) Compute for the amount of C in the compound. The data you need to know are the following:
Molar mass of C = 12 g/mol
Molar mass of CO₂ = 44 g/mol
Solution:
0.5008 g CO₂*(1 mol CO₂/ 44 g)*(1 mol C/1 mol CO₂) = 0.01138 mol C
0.01138 mol C*(12 g/mol) = 0.13658 g C

Compute for the amount of H in the compound. The data you need to know are the following:
Molar mass of H = 1 g/mol
Molar mass of H₂O = 18 g/mol
Solution:
0.1282 g H₂O*(1 mol H₂O/ 18 g)*(2 mol H/1 mol H₂O) = 0.014244 mol H
0.014244 mol H*(1 g/mol) = 0.014244 g H

The percent composition of pure hydrocarbon would be:
Percent composition = (Mass of C + Mass of H)/(Mass of sample) * 100
Percent composition = (0.13658 g + 0.014244 g)/(<span>0.1510 g) * 100
</span>Percent composition = 99.88%

2. The empirical formula is determined by finding the ratio of the elements. From #1, the amounts of moles is:

Amount of C = 0.01138 mol
Amount of H = 0.014244 mol

Divide the least number between the two to each of their individual amounts:
C = 0.01138/0.01138 = 1
H = 0.014244/0.01138 = 1.25

The ratio should be a whole number. So, you multiple 4 to each of the ratios:
C = 1*4 = 4
H = 1.25*4 = 5

Thus, the empirical formula of the hydrocarbon is C₄H₅.

3. The molar mass of the empirical formula is

Molar mass = 4(12 g/mol) + 5(1 g/mol) = 53 g/mol
Divide this from the given molecular weight of 106 g/mol
106 g/mol / 53 g/mol = 2
Thus, you need to multiply 2 to the subscripts of the empirical formula.

Molecular Formula = C₈H₁₀

4 0

3 years ago