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aleksklad [387]
3 years ago
9

In what situation do we use a volumetric flask, conical flask, pipette and graduated cylinder? Explain your answer from accuracy

aspects of these apparatus.
Chemistry
1 answer:
Ad libitum [116K]3 years ago
3 0
A volumetric flask is used to contain a predetermined volume of substance and only measures that volume, for example 250 ml.
Conical flasks can be used to measure the volume of substances but the accuracy they provide is usually up to 10ml. Conical flasks are used in titrations, reactions where the liquid may boil, and reactions which involve stirring. 
Pippettes are of two types, volumetric and graduated. Pippettes are used where high accuracy is required and volumetric pippettes come in as little as 1 ml. Pippettes are usually used in titrations.
Graduated cylinders come in a wide variety of sizes and their accuracy can be down to as much as 1 ml. They are used to contain liquids.
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Adding O2 to the reaction 6CO2 + 6H2O <---> C6H12O6 + 6O2, WHICH WAY WILL THE REACTION SHIFT?
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Increase in Oxygen shift the equilibrium towards reactant side.

<u>Explanation:</u>

6CO₂ + 6H₂O ⇄ C₆H₁₂O₆ + 6O₂

This is the reaction occurs in the photosynthesis of plants by means of sunlight. In this case, if the concentration of Oxygen increases or adding more oxygen to the product side will shift the equilibrium towards the reactant side according to the Le Chatlier's principle, which adjusts the equilibrium by itself for any changes that is increase or decrease in pressure, temperature or concentration of reactants or products.

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Are metallic bonds are found in compounds
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Explanation:

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You have been asked to draw a Bohr model of the element carbon. How would you arrange the dots that represent electrons? They would be embedded in a solid core. They would be in rings around the nucleus

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Sulfur dioxide is a main component of: photochemical smog, thermal inversion, industrial smog.
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Cu(s) + 4 HNO3 (aq) --&gt; Cu(NO3)2 (aq) + 2NO2 (g) + 2H2O(l)
GREYUIT [131]

Answer:

The percent by mass of copper in the mixture was 32%

Explanation:

The ammount of HNO₃ used is:

mol HNO₃ = volume * concentration

mol HNO₃ = 0.015 l * 15.8 mol/l

mol HNO₃ = 0.237 mol

According to the reaction, 4 mol HNO₃ will react with 1 mol Cu and produce 1 mol Cu²⁺. Since we have 0.237 mol HNO₃, the amount of Cu that could react would be (0.237 mol HNO₃ * 1 mol Cu / 4 mol HNO₃) 0.06 mol. This reaction would produce 0.060 mol Cu²⁺, however, only 0.010 mol Cu²⁺ were obtained, indicating that only 0.010 mol Cu were present in the mixture. This means that the acid was in excess, so we can assume that all copper present in the mixture has reacted.

Since 0.010 mol of Cu²⁺ were produced, the amount of Cu was 0.01 mol.

1 mol of Cu has a mass of 63.5 g, then 0.01 mol has a mass of:

0.01 mol Cu * 63.5 g / 1 mol = 0.635 g.

Since this amount was present in 2.00 g mixture, the amount of copper in 100 g of the mixture will be:

100 g(mixture) * 0.635 g Cu / 2 g(mixture) = 32 g

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