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Nitella [24]
3 years ago
6

Prepare 15 mg/dl working standard solution from a stock solution of 20 mg/dl. State the volume of diluent and dilution.

Chemistry
1 answer:
Anastasy [175]3 years ago
4 0

Preparing 15 mg/gl working standard solution from a 20 mg/dl stock solution will require the application of the dilution principle.

Recalling the principle:

initial volume x initial molarity = final volume x final molarity

Since we were not given any volume to work with, we can as well just take an arbitrary volume to be prepared. Let's assume that the stock solution is 10 mL and we want to prepare 15 mg/gl from it:

Applying the dilution principle:

10 x 20 = final volume x 15

final volume = 200/15

                       = 13.33 mL

This means that in order to prepare 13.33 mL, 15 mg/l working standard solution from 10 ml, 20 mg/dl stock solution, 3.33 mL of the diluent must be added to the stock solution.

More on dilution principle can be found here: brainly.com/question/11493179

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If 0.680 kg of copper(I) sulfide reacts with excess oxygen, what mass of copper metal may be produced ? A) 0.680 kg B) 0.136 kg
Katen [24]

Answer:

D. 0.543kg of copper metal is produced from 0.680kg of copper 1 sulphide.

Explanation:

First write the equation for the reaction:

Cu2S + O2 ------> 2Cu + SO2

Determine the mole ratio of the two substances:

I mole of Cu2O forms 2 moles of Copper metal

The number of moles of copper 1 sulphide used is;

n = mass of Cu2S / molar mass of Cu2S

Mass = 0.680kg = 680g

Molar mass = 159.16g/mol

n = 680g / 159.16g/mol

n = 4.272moles

Determine the number of mole of copper:

Number of moles of copper metal produced from 4.272moles of copper 1 sulphide is therefore:

n of copper = 2 * 4.272 Moles

n = 8.544moles.

Determine the mass copper:

The mass of copper metal produced is therefore = number of moles of copper * molar mass of copper

mass = 8.544 moles * 63.55g/mol

mass = 542.97grams

Mass = 0.543kg

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What is the molar mass of H2O?
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Balance this equation please!!! ASAP
Delvig [45]

Answer:

C2H4O2 + 2O2 —> 2CO2 + 2H20

The coefficients are 1, 2, 2, 2

Explanation:

C2H4O2 + O2 —> CO2 + H20

To balance the above, do the following

Since there are 2 carbon on the left, put 2 in front of CO2, we have

C2H4O2 + O2 —> 2CO2 + H20

We have 4 hydrogen on the left, to balance it put 2 in front of H2O i.e

C2H4O2 + O2 —> 2CO2 + 2H20

Now, we have a total of 6 oxygen on the right side. To balance it, put 2 in front of O2:

C2H4O2 + 2O2 —> 2CO2 + 2H20

The coefficients are 1, 2, 2, 2

3 0
3 years ago
The activation energy of a reaction is 56.9 kj/mol and the frequency factor is 1.5×1011/s. Part a calculate the rate constant of
Vanyuwa [196]

We will use Arrehenius equation

lnK = lnA  -( Ea / RT)

R = gas constant = 8.314 J / mol K

T = temperature = 25 C = 298 K

A = frequency factor

ln A = ln (1.5×10 ^11) = 25.73

Ea = activation energy = 56.9 kj/mol = 56900 J / mol

lnK = 25.73 - (56900 / 8.314 X 298) = 2.76

Taking antilog

K = 15.8



5 0
4 years ago
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