The volume of 0.160 m Li2S solution required to completely react with 130 ml of 0.160 CO(NO3)2 is calculated as below

write the reacting equation

Co(NO3)2 + Li2S = 2LiNO3 + COS

find the moles of CO(NO3)2 = molarity x volume

= 130 ml x 0.160=20.8 moles

since the reacting moles between CO(NO3)2 to LiS is 1:1 the moles of LiS is also 20.8 moles

volume of Lis is therefore = moles of Lis/ molarity of LiS

= 20.8/0.160 = 130 Ml

3 0

3 years ago

The electronic configuration of an element is given below.

Amiraneli [1.4K]

The statement which is true about the reactivity of element with 1S²2S²2P⁶3S¹ is

it is reactive because it has to lose one electron to have a full outermost energy level.

Explanation

Element with 1S²2S²2P⁶3S¹ electron configuration is a sodium metal.

sodium has one electron in the outermost energy level.

for sodium to have a full outermost energy level ( 8 electrons) it loses the 1 electron in 3S¹ to form a positively charged ion. (Na⁺)

4 0

3 years ago

Draw a lewis structure for ketene, c2h2o, which has a carbon–carbon double bond

MrMuchimi

Visual representation of covalent bonding indicating the valence shell electrons in the molecule, lines represents the shared pair of electron and pair of electrons that are not involved in bonding are represented as dots(lone pairs) are known as Lewis structures.

Compound formation takes place in order to complete the octet of each element that is according to octet rule, each atom forms bond with other atom in order to complete their octet that is to get eight electrons in its valence shell and attain stability.

An organic compound of the form $R^{'}R^{''}C=C=O$ is known as ketene.

The given ketene is $C_2H_2O$ .

The number of valence electron of:

$C = 4$

$H = 1$

$O = 6$

The number of valence electrons in $C_2H_2O$ = $2\times 4+2\times 1+1\times 6 =16$

2 electrons are involved in each single bond between carbon and hydrogen and 4 electrons are involved in each double bond formed between carbon-carbon and carbon-oxygen. Hence, the total number of electrons involved in bond formation are 12 and rest 2 pair of electrons are present on oxygen as lone pair of electrons.

Therefore, the attached image is the Lewis structure of $C_2H_2O$ .

8 0

3 years ago