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3 years ago

How do elements on the right side of the periodic table differ from elements on the left side of the table?

Chemistry

1 answer:

lozanna [386]3 years ago

4 0

The answer is c.

Elements on the left side of the table are metals, such as sodium, lithium, potassium, etc.

Elements on the right side are non metals, such as Chlorine, Fluorine, Bromine, etc.

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Write the molecular formula for the compound that exhibits a molecular ion at M+ = 112.0499. Assume that C, H, N, and O might be

navik [9.2K]

Answer:

C₆H₁₀NO

Explanation:

In order to arrive at a molecular formula we have to make some assumptions and they are

Assuming there is one ( 1 ) N and one ( 1 ) O that is present in the said molecule

Total mass = 29.998

next step: subtract the total mass from 112.0499 = 82.501

next : assume the presence of 6 carbon atoms in said molecule

Total mass = 6 * 12 = 72

Mass of Hydrogens = 82.501 - 72 = 10.501

∴ number of hydrogens = 10.501 / 1.0078 ≈ 10

Hence Total mass = 29.998 + 82.501 ≈ 112.0499

Finally Molecular formula = C₆H₁₀NO

3 0

3 years ago

The ability to travel great distances in the shortest time was achieved by which invention?

Nutka1998 [239]

By space travel.

Or aeroplanes.

Whichever one is in your choices

6 0

3 years ago

Of the groups given, which are the least chemically reactive?

AfilCa [17]

Answer:

Group 8A

Explanation:

This is because the elements in Group 8A is stable, what I mean is this elements have the maximum number of electrons in there last orbit so they dont need to form any compund with any other element in the periodic table.

HOPE THIS WILL HELP YOU❤️

7 0

2 years ago

Under standard conditions, which of the following is the net reaction that occurs in the cell? Cd | Cd2+ | | Cu2+ | Cu

dlinn [17]

Answer:

Cd + Cu2+ -----> Cu + Cd2+

Explanation:

6 0

3 years ago

A sample consisting of 2 moles He is expanded isothermally at 0 degrees from 5.0dm3 to 20.0dm3. Calculate w, q and deltaU for ea

FinnZ [79.3K]

Answer:

i) $\Delta U=0 w=-6293 J q=6293 J$

ii) $\Delta U=0 w=-3404,52 J q=3404,52 J$

ii) $\Delta U=0 w=0 J q=0 J$

Explanation:

As the initial and final states of the sample are the same, the ΔU of the sample is, for the three cases

$\Delta U=n.C_{V}.\Delta T=0$ since $\Delta T=0$

i)Reversibly $P_{ext} =P_{sys}$ so $w$ can be calculated by

$w=-n.R.T.ln(\frac{V_{f}}{V_{i}})=-2 \times 8.314\frac{J}{mol K} \times 273,15K \times ln(\frac{}{5dm^{3}})=-6293 J$

and because of the first law of thermodynamics

$q=-w=6293 J$

ii)Irreversibly with $P_{ext} =P_{f}$

we can calculate $P_{f}$ by the law of ideal gases

$P_{f} =\frac{n\times R\times T}{V_{f}} =\frac{2\times 0.082\frac{dm^{3}atm}{mol K}\times 273,15K}{20dm^{3}} =2,24 atm$

then w can be calculated by

$w=-P_{ext} \times \Delta V=-2,24 atm \times (20-5) dm^{3} \times frac{101.325J }{atm dm^{3}=-3404,52J$

and

$q=-w=3404,52J$

iii)a free expansion $P_{ext}$ so $w=0$ (there's no work at vaccum) and $q=-w=0$

7 0

3 years ago