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2 years ago

Which of the following groups in the periodic

1 answer:

7 0

Elements of Group 1 and group 2 in the periodic table contain elements so reactive that they are never found in the free state

<u>Explanation</u>:

The metals in group 1 of periodic table consisting of 'alkali metals' which include lithium, potassium, sodium, rubidium, Francium and caesium. They are highly reactive because they have low ionisation energy and larger radius. The group 2 metals consist of 'alkaline earth metals' which include calcium, strontium, barium, beryllium, radium and magnesium. These alkaline earth metal have +2 oxidation number, hence are highly reactive.

These both group metals are mostly reactive and so are never found in a free state. When they are exposed to air they would immediately react with oxygen. Hence, are stored in oils to avoid oxidation.

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PLEASE HELP 20 POINTS What is a key event found in CO2 and glucose?

irina [24]

the answer is carbon

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2 years ago

In a constant‑pressure calorimeter, 70.0 mL of 0.350 M Ba(OH)2 was added to 70.0 mL of 0.700 M HCl. The reaction caused the temp

asambeis [7]

Answer:

ΔH = 57 Kj/mole H₂O

Explanation:

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2 years ago

Read 2 more answers

A chemical reaction produces 11.8 moles of tin atoms how many grams of tin are made

valina [46]

Convert mols to grams by multiplying grams of tin by the number of mols.

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2 years ago

. A bright violet line occurs at 435.8 nm in the emission spectrum of mercury vapor. What amount of energy, in joules, must be r

Tom [10]

Answer : The energy released by an electron in a mercury atom to produce a photon of this light must be, $4.56\times 10^{-19}J$

Explanation : Given,

Wavelength = $435.8nm=435.8\times 10^{-9}m$

conversion used : $1nm=10^{-9}m$

Formula used :

$E=h\times \nu$

As, $\nu=\frac{c}{\lambda}$

So, $E=h\times \frac{c}{\lambda}$

where,

$\nu$ = frequency

h = Planck's constant = $6.626\times 10^{-34}Js$

$\lambda$ = wavelength = $435.8\times 10^{-9}m$

c = speed of light = $3\times 10^8m/s$

Now put all the given values in the above formula, we get:

$E=(6.626\times 10^{-34}Js)\times \frac{(3\times 10^{8}m/s)}{(435.8\times 10^{-9}m)}$

$E=4.56\times 10^{-19}J$

Therefore, the energy released by an electron in a mercury atom to produce a photon of this light must be, $4.56\times 10^{-19}J$

3 0

2 years ago

The partial negative charge at one end of a water molecule is attracted to the partial positive

Neporo4naja [7]

Answer:

Hydrogen Bond

Explanation:

Hydrogen bond interactions are formed between the hydrogen atom bonded to most electronegative atoms (i.e. F, O and N) of one molecule and most electronegative atom (i.e. F, O and N) of another molecule.

In this interaction the hydrogen atom has partial positive charge and electronegative atom has partial negative charge.

6 0

1 year ago