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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%

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A 5.325g sample of methyl benzoate, a compound in perfumes , was found to contain 3.758 g of carbon, 0.316 g of hydrogen, and 1.

Alexxandr [17]

Answer: The empirical and molecular formula of the compound is $C_4H_4O$ and $C_8H_8O_2$ respectively

Explanation:

We are given:

Mass of C = 3.758 g

Mass of H = 0.316 g

Mass of O = 1.251 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{3.758g}{12g/mole}=0.313moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.316g}{1g/mole}=0.316moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{1.251g}{16g/mole}=0.078moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.078 moles.

For Carbon = $\frac{0.313}{0.078}=4.01\approx 4$

For Hydrogen = $\frac{0.316}{0.078}=4.05\approx 4$

For Oxygen = $\frac{0.078}{0.078}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 4 : 4 : 1

The empirical formula for the given compound is $C_4H_4O$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

We are given:

Mass of molecular formula = 130 g/mol

Mass of empirical formula = 68 g/mol

Putting values in above equation, we get:

$n=\frac{130g/mol}{68g/mol}=1.9\approx 2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(2\times 4)}H_{(2\times 4)}O_{(2\times 2)}=C_8H_8O_2$

Hence, the empirical and molecular formula of the compound is $C_4H_4O$ and $C_8H_8O_2$ respectively

4 0

3 years ago

2. What is the molarity of 100 mL of a 3.0% H2O2 (mass/volume) solution? What is the molarity of 100 mL of a 2.25% H2O2 solution

KatRina [158]

The 3% mass/volume H₂O₂ means 3 g of H₂O₂ in 100 ml of water.

Now, Molarity (M) = No. of moles of H₂O₂ / Volume of solution in liter

No. of moles of H₂O₂ = Mass / Molar mass = 3 g / 34 g/mol = 0.088 mol

So, molarity = 0.088 × 1000 ml / 100 ml = 0.88 M

In case of 2.25 % H₂O₂,

No of moles = 2.25 g / 34 g/mol = 0.066 mol

Molarity = 0.066 mol / 0.100 L = 0.66 M.

7 0

3 years ago

The density of whole blood is 1.05 g/ml. a typical adult has between 4.7 and 5.5 l of whole blood. what is the mass in pounds of

Kisachek [45]

1.05 g/ml * 1000 ml = 1050g/l because of 1g/ml = 1 kg/l so, a/q mass of 4.7 l of whole blood in pound = 4.7 * 1050 = 4935 g so in pound 4935g = 10.87981p

5 0

3 years ago

What is the % dissociation of a solution of acetic acid if at equilibrium the solution has a pH = 4.74 and a pKa = 4.74?

Ymorist [56]

Answer:

$\% diss = 50\%$

Explanation:

Hello there!

In this case, when considering weak acids which have an associated percent dissociation, we first need to set up the ionization reaction and the equilibrium expression:

$HA\rightleftharpoons H^++A^-\\\\Ka=\frac{[H^+][A^-]}{[HA]}$