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3 years ago

If the theoretical yield of a reaction is 0.110 g and the actual yield is 0.104 g , what is the percent yield?

1 answer:

3 0

Theoretical Yield is an Ideal yield with 100 % conversion of reactant to product. It is in fact a paper work.

While,

Actual Yield is the yield which is obtained experimentally. It is always less than theoretical yield because it is not possible to have 100% conversion of reactants into products. Even some amount of product is lost while handling it during the process.

Percentage Yield is Calculated as,

%age Yield = Actual Yield / Theoretical Yield × 100

Data Given:
Actual Yield = 0.104 g

Theoretical Yield = 0.110 g

Putting Values,

%age Yield = 0.104 g / 0.110 g × 100

%age Yield = 94.54 %

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The density for potassium is 0.856 g/cm3. What would be the mass of a 30 cm3 piece of potassium? A. 51.36g B. 25.68g C. 25.5g D.

CaHeK987 [17]

Density = mass / Volume 0.856 g/cm3 = x / 30 cm3(0.856)(30) = xx = 25.68 grams

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4 years ago

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Why is pluto classified as a dwarf planet?

Julli [10]

Answer:

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Explanation:

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3 years ago

Give the nuclear symbol for the isotope of phosphorus for which a=31? enter the nuclear symbol for the isotope (e.g., 42he).

Bogdan [553]

A = number of mass = 31 => number of protons + number of neutrons = 31

Phosphours, Z = atomic number = 15 = number of protons

Symbol of the element phosphorus = P

=> The symbol of the isotope is the symbol of the element, P, with the number of mass, 31, to the left, as a superscript, and the atomic number, 15, to the left, as a suscript.

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3 years ago

A sample of citric acid, C6H8O7, has a mass of 22.9 g. What amount, in moles, does this mass represent?

aliina [53]

Answer

C6H807 = 192.124 g/mol

n = m/M

$\frac{22.9g}{192.124 g/mol}$

n = 0.119 mol

Explanation:

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3 years ago

A 60.0 mL solution of 0.112 M sulfurous acid (H2SO3) is titrated with 0.112 M NaOH. The pKa values of sulfurous acid are 1.857 (

djverab [1.8K]

Answer:

a)4.51

b) 9.96

Explanation:

Given:

NaOH = 0.112M

H2S03 = 0.112 M

V = 60 ml

H2S03 pKa1= 1.857

pKa2 = 7.172

a) to calculate pH at first equivalence point, we calculate the pH between pKa1 and pKa2 as it is in between.

Therefore, the half points will also be the middle point.

Solving, we have:

pH = (½)* pKa1 + pKa2

pH = (½) * (1.857 + 7.172)

= 4.51

Thus, pH at first equivalence point is 4.51

b) pH at second equivalence point:

We already know there is a presence of SO3-2, and it ionizes to form

SO3-2 + H2O <>HSO3- + OH-

$Kb = \frac{[ HSO3-][0H-]}{SO3-2}$

$Kb = \frac{10^-^1^4}{10^-^7^.^1^7^2} = 1.49*10^-^7$

[HSO3-] = x = [OH-]

mmol of SO3-2 = MV

= 0.112 * 60 = 6.72

We need to find the V of NaOh,

V of NaOh = (2 * mmol)/M

= (2 * 6.72)/0.122

= 120ml

For total V in equivalence point, we have:

60ml + 120ml = 180ml

[S03-2] = 6.72/120

= 0.056 M

Substituting for values gotten in the equation $Kb=\frac{[HSO3-][OH-]}{[SO3-2]}$

We noe have:

$1.485*10^-^7=\frac{x*x}{(0.056-x)}$

$x = [OH-] = 9.11*10^-^5$

$pOH = -log(OH) = -log(9.11*10^-^5)$

=4.04

pH = 14- pOH

= 14 - 4.04

= 9.96

The pH at second equivalence point is 9.96

4 0

3 years ago