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Alecsey [184]
3 years ago
6

HELP please quick HELP

Chemistry
1 answer:
stiks02 [169]3 years ago
4 0
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A solution of potassium cyanide, KCN, is made by dissolving 3.2 g of KCN in water. When this solution reacts with excess hydroch
Rama09 [41]

Answer:

22.7%

Explanation:

We must first put down the equation of reaction to guide our work while solving the problem.

KCN(aq) + HCl (aq)--> KCl(aq) + HCN(aq)

The questions specifically state that HCl is in excess so KCN is the limiting reactant.

Number of moles of KCN reacted= mass of KCN reacted / molar mass of KCN

Mass of KCN reacted= 3.2 g

Molar mass of KCN= 65.12 g/mol

Number of moles of KCN= 3.2/65.12 g/mol= 0.049 moles

Theoretical yield of HCN is obtained thus;

From the reaction equation;

1 mol of KCN produced 1 mole of HCN thus 0.049 moles of KCN will produce 0.049 moles of HCN.

Mass of HCN = number of moles ×molar mass

Molar mass of HCN= 27.0253 g/mol

Hence mass of HCN produced= 0.049mol × 27.0253 g/mol= 1.32g of HCN

Actual yield of HCN= 0.30g

% yield= actual yield/ theoretical yield ×100

% yield= 0.30/1.32 ×100

%yield= 22.7%

3 0
3 years ago
Read 2 more answers
How many grams are in 1.23 x 1020 atoms of arsenic?
Alina [70]

Given the data from the question, the mass of arsenic that contains 1.23×10²⁰ atoms is 0.0153 g

<h3>Avogadro's hypothesis </h3>

6.02×10²³ atoms = 1 mole of arsenic

But

1 mole of arsenic = 75 g

Thus, we can say that:

6.02×10²³ atoms = 75 g of arsenic

<h3>How to determine the mass that contains 1.23×10²⁰ atoms</h3>

6.02×10²³ atoms = 75 g of arsenic

Therefore,

1.23×10²⁰ atoms = (1.23×10²⁰ × 75) / 6.02×10²³ atoms)

1.23×10²⁰ atoms = 0.0153 g of arsenic

Thus, 1.23×10²⁰ atoms is present in 0.0153 g of arsenic

Learn more about Avogadro's number:

brainly.com/question/26141731

6 0
1 year ago
Can someone tell me the parts and the functions ?
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Each hemisphere has four sections, called lobes: frontal, parietal, temporal and occipital. Each lobe controls specific functions. For example, the frontal lobe controls personality, decision-making and reasoning, while the temporal lobe controls, memory, speech, and sense of smell.
3 0
3 years ago
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Be sure to answer all parts. Express the rate of reaction in terms of the change in concentration of each of the reactants and p
Tanzania [10]

Answer :  The [H] is increasing at the rate of 0.36 mol/L.s

Explanation :

The general rate of reaction is,

aA+bB\rightarrow cC+dD

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}

\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}

\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}

\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}

Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

2D(g)+3E(g)+F(g)\rightarrow 2G(g)+H(g)

The expression for rate of reaction :

\text{Rate of disappearance of }D=-\frac{1}{2}\frac{d[D]}{dt}

\text{Rate of disappearance of }E=-\frac{1}{3}\frac{d[E]}{dt}

\text{Rate of disappearance of }F=-\frac{d[F]}{dt}

\text{Rate of formation of }G=+\frac{1}{2}\frac{d[G]}{dt}

\text{Rate of formation of }H=+\frac{d[H]}{dt}

\text{Rate of reaction}=-\frac{1}{2}\frac{d[D]}{dt}=-\frac{1}{3}\frac{d[E]}{dt}=-\frac{d[F]}{dt}=+\frac{1}{2}\frac{d[G]}{dt}=+\frac{d[H]}{dt}

Given:

-\frac{d[D]}{dt}=0.18mol/L.s

As,  

-\frac{1}{2}\frac{d[D]}{dt}=+\frac{d[H]}{dt}=0.18mol/L.s

and,

+\frac{d[H]}{dt}=2\times 0.18mol/L.s

+\frac{d[H]}{dt}=0.36mol/L.s

Thus, the [H] is increasing at the rate of 0.36 mol/L.s

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