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3 years ago

Can u see water vapour by a microscope?

Chemistry

2 answers:

pochemuha3 years ago

6 0

Answer:

No. Unlike clouds, fog, or mist which are simply suspended particles of liquid water in the air, water vapour itself cannot be seen because it is in gaseous form

tigry1 [53]3 years ago

3 0

Answer:

It can be formed either through a process of evaporation or sublimation. Unlike clouds, fog, or mist which are simply suspended particles of liquid water in the air, water vapour itself cannot be seen because it is in gaseous form

Explanation:

hope it help

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Which of the following is true for balancing equations?

algol [13]

Answer:

Explanation:

An equation with the equal amount and proportion of atoms of each element on both sides of the reaction is commonly referred to as a balanced chemical equation.

The law of conservation of matter asserts that no observable and empirical change in the amount of matter occurs within a conventional chemical process. As a result, each element in the product would have the same equal amount or numbers of atoms as the reactants.

5 0

3 years ago

A site in Pennsylvania receives a total annual deposition of 2.688 g/mof sulfate from fertilizer and acid rain. The ratio by mas

sertanlavr [38]

According to the statement

2.12 x 10^4 lbs pounds of CaCO₃ are needed to neutralize this acid

<h3>What is neutralization?</h3>

A chemical reaction in which an acid and a base react quantitatively with each other is known as neutralization or neutralization. In a water reaction, neutralization ensures that there is no excess of hydrogen or hydroxide ions in the solution.

<h3>According to the given information:</h3>

The equation of the neutralization reaction between H2SO4 and CaCO3.

CaCO3 + H2SO4 → CaSO4 + H2CO3

H2CO3 dissociate to water and carbon dioxide.

CaCO3 + H2SO4 → CaSO4 + H2O + CO2

Now solving for the mass of CaCO3 needed to neutralize the acid.

mass of CaCO3 = 9460 Kg H2SO4 × $\frac{1000 \mathrm{~g}}{1 \mathrm{~kg}} \times \frac{1 \mathrm{~mol} \mathrm{H}_2 \mathrm{SO} 4}{98.1 \mathrm{gH}_2 \mathrm{SO}_4} \times \frac{1 \mathrm{~mol} \mathrm{CaCO}\left(\mathrm{O}_3\right.}{1 \mathrm{~mol} \mathrm{H}_2 \mathrm{SO}_4}$ $\times \frac{100.1 \mathrm{~g} \mathrm{CaCO}_3}{1 \mathrm{~mol} \mathrm{CaCO}_3} \times \frac{2.205 \mathrm{lb}}{1000 \mathrm{~g}}$

= 21284.56606

mass of CaCO3 = 2.12 x 10^4 lbs

2.12 x 10^4 lbs pounds of CaCO₃ are needed to neutralize this acid.

To know more about neutralization visit:

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4 0

1 year ago

Which of the following are indicators of a chemical change? Select all that apply.

Arturiano [62]

<span>Which of the following are indicators of a chemical change? Select all that apply.
</span>
These are the answers:

<span>color change
temperature change
precipitate formation
gas formation

Hope this helps.
</span>

4 0

2 years ago

Read 2 more answers

Calculate the vapor pressure of a solution of 32.5 g of glycerol (C3H8O3) in 500.0 g of water at 25°C. The vapor pressure of wat

Luba_88 [7]

The vapor pressure is obtained as 23.47 torr.

<h3>What is the vapor pressure?</h3>

Given that; p = x1p°

p = vapor pressure of the solution

x1 = mole fraction of the solvent

p° = vapor pressure of the pure solvent

Δp = p°(1 - x1)

Δp =x2p°

Δp = vapor pressure lowering

x2 = mole fraction of the of the solute

Number of moles of glycerol = 32.5 g/92 g/mol = 0.35 moles

Number of moles of water = 500.0 g/18 g/mol = 27.8 moles

Total number of moles = 0.35 moles + 27.8 moles = 28.15 moles

Mole fraction of glycerol = 0.35 moles/28.15 moles = 0.012

Mole fraction of water = 27.8 moles/28.15 moles =0.99

Δp = 0.012 * 23.76 torr

Δp = 0.285 torr

p1 = p° - Δp

p1 = 23.76 torr - 0.285 torr

p1 = 23.47 torr

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5 0

2 years ago

(c) Polar solutes were separated by hydrophilic interaction chromatography (HILIC) with a strongly polar bonded phase. How would

Lelu [443]

Answer:

Explanation:

For the polar solutes which were separated using the hydrophilic interaction chromatography (HILIC) with a strongly polar bonded phase, the retention time would be higher if eluent were changed from 80 vol% to 90 vol% acetonitrile in water.

However, for the polar solutes which were separated using the normal-phase chromatography on bare silica with methyl t=butyl ether and 2-propanol solvent, the retention time would be lower if the eluent were changed from 40 vol% to 60 vol% 2-propanol.

5 0

3 years ago