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algol13
3 years ago
9

How much glycerol ( is liquid supplied at 100%) would you need to make 200 mL of 20% v/v (volume/volume) glycerol solution?

Chemistry
1 answer:
RSB [31]3 years ago
6 0

Answer:

40mL of glycerol are needed to make a 20% v/v solution

Explanation:

This problem can be solved with a simple rule of three:

20%  v/v is a sort of concentration. In this case, 20 mL of solute are contained in 100 mL of solution.

Therefore, in 100 mL of solution you have 20 mL of solvent (glycerol)

In 200 mL, you would have,  (200 .20)/ 100 = 40 mL

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Mike mixes two chemicals in a container. The container quickly becomes
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it is an exothermic reaction

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Carbon dioxide is released in________and taken in by plants in ______
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3 0
3 years ago
Need help please?!!!!!​
puteri [66]

(a) 33.6 L of oxygen would be produced.

(b) 106 grams of  Na_2CO_3 would be needed

<h3>Stoichiometric calculations</h3>

1 mole of gas = 22.4 L

(a) From the equation, 2 moles of KClO_3 produce 3 moles of O_2. 1 mole of  KClO_3 will, therefore, produce 1.5 moles of  O_2.

1.5 moles of oxygen = 22.4 x 1.5 = 33.6 L

(b) 22.4 L of CO_2 is produced at STP. This means that 1 mole of the gas is produced.

From the equation, 1 mole of  CO_2 requires 1 mole of Na_2CO_3.

Molar mass of  Na_2CO_3 = (23x2)+ (12)+(16x3) = 106 g/mol

Mass of 1 mole  Na_2CO_3 = 1 x 106 = 106 grams

More on stoichiometric calculations can be found here: brainly.com/question/27287858

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6 0
2 years ago
A cell consists of a gold wire and a saturated calomel electrode (S.C.E.) in a 0.150 M AuNO 3 solution at 25 °C. The gold wire i
almond37 [142]

The half reaction that occurs at the Au electrode is  1.64

<u>Explanation:</u>

Half cell reaction at 'Au' electrode

We have the equation,

Au + (aq) + e-  ---> Au(s)

Given the concenration of AuNO3=0.150 M

[Au +] =0.150 m

From the equation,

Au + (aq) + e-  ---> Au(s)

Standard electric potential= Eo= 1.69 volt

Solving the problem using the Nerst equation

E cell= E0 cell - 2.303 RT/ nF  log Qc

Where,

T = 298 K

n= no of electron lost or gained

F= faraday's constant= 965000/mole

R= universal constant= 8.314 J/ K/ Mole

Substitue the values we get

E cell = 1.69 volt - 0.05 g/n log (1/0.150M)

E cell = 1.69 volt - 0.05 g/1  0.824

E cell= 1.64

The half reaction that occurs at the Au electrode is  1.64

<u />

6 0
3 years ago
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