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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

You might be interested in

Compare the mass number and atomic number for isotopes of an element

Zanzabum

An isotope is when there is same amount of atomic number but different mass number. It also mean that only the number of neutrons changes if there is an isotope present.

3 0

3 years ago

In an experiment you find the density of water to water to be 1.23 g.Ml.The theoritical value of the water's density is 1. 00 g/

Pepsi [2]

Answer:

The percent error is 23%

Explanation:

To find the percent error for the experiment carried out,

Percent error is given by the formula below

$PE = \frac{AE}{TV}$ × $100$ %

Where PE is the percent error

AE is the Absolute error

TV is the Theoretical value

Also, Absolute error (AE) can be determined from

Absolute error (AE) = /Theoretical value(TV) - Experimental value (EV)/

Now, from the question,

Experimental value (EV) = 1.23 g/mL

Theoretical value (TV) = 1.00 g/mL

From,

Absolute error (AE) = /Theoretical value(TV) - Experimental value (EV)/

Absolute error (AE) = /1.00 g/mL - 1.23 g/mL/

Absolute error (AE) = / - 0.23 g/mL/

Absolute error (AE) = 0.23 g/mL

This is the Absolute error

Now, for the Percent error

$PE = \frac{AE}{TV}$ × $100$ %

$PE = \frac{0.23}{1.00}$ × $100$ %

$PE = 23$ %

Hence, the percent error (PE) is 23%

5 0

3 years ago

For which of the following molecules or ions does the following description apply? "The bonding can be explained using a set of

Marina86 [1]

Answer:

O₃

Explanation:

Consider the molecule CO₂. The carbon is sp hybridized. Carbon has 4 valence electrons and oxygen contributes 2 electrons, 1 for each C=O which indicates that there are 8 electrons around the carbon. Since there are 4 bonds all of them are bond pairs. Each C=O double bond uses 2 bond pairs which are considered as single unit. These two double bond units try to get as far apart as possible making the molecule adopt a linear geometry.

Considering the H₂S molecule both oxygen and sulfur are the in the same group, which means both have a valence of 6. The four valence orbitals of sulfur, one 1 s orbital and three 3p orbitals mix together and forms four sp³ hybridized orbitals. Of the four hybridized orbitals, two overlaps with the 1s orbital of hydrogen forming 2 (S - H) bonds while the other two sp³ orbitals remain on sulfur which has lone pair of electrons. Because of the presence of lone pair, the angle between H-S-H bond is slightly less than the ideal tetrahedral bond angle. Thus, H2S having 2 bonding electron pair and 2 lone pairs has a bent shape.

Considering O₃ and according to the VSEPR theory ozone molecule must have a trigonal – planar geometry. It has a total of 18 valence electrons. From the resonance structure given below it is clear the 4 pairs of electrons exit as bonding pair, sp² or σ- bond and the remaining 10 electrons exit as lone pair. Of the three un- hybridized p orbitals one is anti – bonding and remains empty. In ozone the π bond is distributed between the two bonds, and each receives half a π bond. For this molecule the electron pair geometry is trigonal planar but the molecular geometry is bent. The presence of lone pair exerts slight repulsion on the bonding oxygen atoms and a slight compression of the bond angle greater than 120°.

In carbonate ion, <u>the carbon is sp² hybridized</u>. The carbon has 4 valence electrons and there are four bonds to the oxygen which add another 4 making a total of 8. There are 4 pairs of bonding electrons and no lone pair. Of the 4 bond pairs, 2 pairs are used in forming double bond C=O and 2 bond pairs in forming the two C-O single bonds., Thus CO₃²⁻ adopts a trigonal planar geometry.

Of the two molecules only ozone and carbonate ion, have sp2 hybridized central atoms. In ozone the central atoms have lone pair of electrons the hybridization around is sp². Hence the correct option is O₃

7 0

3 years ago

Read 2 more answers

Identify the atoms that are oxidized and reduced, the change in the oxidation state for each, and oxidising and reducing agents

KATRIN_1 [288]

Answer:

Explanation:

a) $Mg + NiCl_2 \rightarrow MgCl_2 + Ni$

Oxidation state of Mg changes to 0 to +2.

Oxidation state of Ni changes to +2 to 0.

Oxidation state of Cl does not change.

So, Mg is oxidized, Ni is reduced.

Mg acts as reducing agent and $NiCl_2$ acts as oxidizing agent.

b) $PCl_3 + Cl_2 \rightarrow PCl_5$

Oxidation state of P changes to +3 to +5.

Oxidation state of Cl changes to 0 to -1.

So,

P is oxidized and Cl is reduced.

$PCl_3$ acts as reducing agent and $Cl_2$ acts as oxidizing agent.

c) $C_2H_4 + 3O_2 \rightarrow 2CO_2 + 2H_2O$

Oxidation state of C changes to -2 to +4.

Oxidation state of O changes to 0 to -2.

Oxidation state of H does not change.

So,

C is oxidized and O is reduced.

$C_2H_4$ acts as reducing agent and $O_2$ acts as oxidizing agent.

d) $Zn + H_2SO_4 \rightarrow ZnSO_4 + H_2$

Oxidation state of Zn changes to 0 to +2.

Oxidation state of H changes to +1 to 0.

Oxidation state of S and O do not change.

So, Zn is oxidized, H is reduced.

Zn acts as reducing agent and $H_2SO_4$ acts as oxidizing agent

e) $K_2S_2O_3 + I_2 \rightarrow K_2S_4_O6 + 2KI$

Oxidation state of S changes to +2 to +2.5.

Oxidation state of I changes to 0 to -1.

Oxidation state of K and O do not change.

So, S is oxidized, I is reduced.

$K_2S_2O_3$ acts as reducing agent and $I_2$ acts as oxidizing agent.

f) $3Cu + 8HNO_3 \rightarrow 3Cu(NO_3)_2 + 4H_2O+2NO$

Oxidation state of Cu changes to 0 to +2.

Oxidation state of N changes to N +5 to +2.

Oxidation state of H and O do not change.

So, Cu is oxidized, N is reduced.

Cu acts as reducing agent and $HNO_3$ acts as oxidizing agent

8 0

2 years ago

How many moles are in 3.01 x 10^23 atoms of zinc?

Lerok [7]

Answer:

<h2>0.5 moles</h2>

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

$n = \frac{N}{L} \\$

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

$n = \frac{3.01 \times {10}^{23} }{6.02 \times {10}^{23} } = \frac{3.01}{6.02} \\ = 0.5$

We have the final answer as

<h3>0.5 moles</h3>

Hope this helps you

6 0

2 years ago