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4 years ago

The valence shell holds up to two electrons in which 2 elements? a. H and C b. H and Li c. H and He d. He and Be

Chemistry

2 answers:

tatuchka [14]4 years ago

6 0

Answer : The correct option is, (d) He and Be

Explanation :

First we have to determine the electronic configuration of the following elements.

The electronic configuration of hydrogen (H) is:

$1s^1$

The electronic configuration of carbon (C) is:

$1s^22s^22p^2$

The electronic configuration of lithium (Li) is:

$1s^22s^1$

The electronic configuration of helium (He) is:

$1s^2$

The electronic configuration of beryllium (Be) is:

$1s^22s^2$

From the electronic configuration of the elements we conclude that, the hydrogen element has '1' valence electron, carbon element has '4' valence electrons, lithium element has '1' valence electron, helium element has '2' valence electrons and beryllium element has '2' valence electrons.

Thus, the helium and beryllium are the elements that holds up to two electrons in their outermost shell.

Hence, the correct option is, (d) He and Be

Mariulka [41]4 years ago

5 0

The answer is C. H and HE

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juin [17]

2Al + 3Br2 -------------> 2AlBr3

3 g Al = 0.11 mol Al.

6 g Br2 = 0.0375 mole bromine (it is diatomic). 

moles of aluminium will take part in reaction = 0.0375 X (2/3) = 0.025. 

Gram-mole of AlBr3 will be produced = 0.025 mole = 6.6682 g.
moles of Al left = 0.11 - 0.025 = 0.086

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4 years ago

A gas has a volume of 27.5 L at 302 K and 1.40 atm. How many moles are in the sample of gas?

ioda

Answer:

1.55 mol

Step-by-step explanation:

We can use the Ideal Gas Law to solve this problem.

pV = nRT Divide both sides by RT

n = (pV)/(RT)

Data:

p = 1.40 atm

V = 27.5 L

R = 0.082 06 L·atm·K⁻¹mol⁻¹

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n = (1.40 × 27.5)/(0.082 06 × 302)

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3 years ago

Which aqueous solution will have the highest boiling point temperature? A. 0.100 molal NiBr2(aq) B. 0.250 molal CH3OH(aq) C. 0.1

zaharov [31]

Answer: 0.150 m $Na_2SO_4(aq)$ will have highest boiling point.

Explanation:

Formula used for Elevation in boiling point :

$\Delta T_b=i\times k_b\times m$

where

$\Delta T_b=T_b-T^o_b$ = elevation in boiling point

[tex[k_b[/tex] = boiling point constant

m = molality

i = Van't Hoff factor

A) 0.100 m $NiBr_2$

i = 3 as $NiBr_2\rightarrrow Ni^{2+}+2Br^-$

concentration will be $3\times 0.100=0.300$

B) 0.250 m $CH_3OH$

i = 1 as $CH_3OH$ is a non electrolyte

concentration will be $1\times 0.250=0.250$

C) 0.100 molal $MgSO_4(aq)$

i = 2 as $MgSO_4\rightarrrow Mg^{2+}+SO_4^{2-}$

concentration will be $2\times 0.100=0.200$

D. 0.150 molal $Na_2SO_4(aq)$

i = 3 as $Na_2SO_4\rightarrrow 2Na^{+}+SO_4^{2-}$

concentration will be $3\times 0.150=0.450$

E. 0.150 molal $NH_4NO_3(aq)$

i = 2 as $NH_4NO_3\rightarrrow NH_4^{+}+NO_3^{-}$

concentration will be $2\times 0.150=0.300$

The solution having the highest concentration of ions will have the highest boiling point and thus 0.150 m $Na_2SO_4(aq)$ will have highest boiling point.

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bazaltina [42]

Answer:

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