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Serga [27]
3 years ago
13

A central atom has two lone pairs on opposite sides and four single bonds. What is the molecule geometry of the result?

Chemistry
1 answer:
ryzh [129]3 years ago
8 0

The molecular geometry is square planar.

According to the Valence Shell Electron Pair Repulsion theory (VSEPR), the shape of a molecule is determined by the number of valence electrons surrounding the outermost shell of the central atom in the molecule.

In this case, the expected geometry based on VSEPR theory is octahedral. However, the lone pairs on opposite sides of the four single bonds leads to a square planar molecular geometry.

Learn more; brainly.com/question/24396703

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The diagram below represents a plan cell Letter X represents a structure in the cell ​
weeeeeb [17]

the cell structure represented by x is the nucleus

4 0
3 years ago
Assume that all of this fossil fuel is in the form of octane (C8H18) and calculate how much CO2 in kilograms is produced by worl
AnnZ [28]

The given question is incomplete. the complete question is:

The world burns the fossil fuel equivalent of approximately 9.50\times 10^{12} kg of petroleum per year. Assume that all of this petroleum is in the form of octane. Calculate how much CO2 in kilograms is produced by world fossil fuel combustion per year.( Hint: Begin by writing a balanced equation for the combustion of octane.)

Answer: 29\times 10^{12}kg

Explanation:

Combustion is a chemical reaction in which hydrocarbons are burnt in the presence of oxygen to give carbon dioxide and water.

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

2_C8H_{18}+17O_2\rightarrow 16CO_2+18H_2O

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of octane}=\frac{9.50\times 10^{12}\times 10^3g}{114g/mol}=0.083\times 10^{15}moles

According to stoichiometry :

As 2 moles of octane give = 16 moles of CO_2

Thus 0.083\times 10^{15}moles of octane give =\frac{16}{2}\times 0.083\times 10^{15}=0.664\times 10^{15}moles  of CO_2

Mass of CO_2=moles\times {\text {Molar mass}}=0.664\times 10^{15}moles\times 44g/mol=29.2\times 10^{15}g=29.2\times 10^{12}kg

Thus 29\times 10^{12}kg of CO_2 is produced by world fossil fuel combustion per year.

5 0
3 years ago
a concentration solution of H2so4 is 59.4% by mass (m/m) and has a density of 1.83 g/mL. How many mL of the solution would be re
Blababa [14]

Answer: 41.5 mL

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

Molarity=\frac{n}{V_s}

where,

n = moles of solute

V_s = volume of solution in L

Given : 59.4 g of H_2SO_4 in 100 g of solution  

moles of H_2SO_4=\frac{\text {given mass}}{\text {molar mass}}=\frac{59.4g}{98g/mol}=0.61

Volume of solution =\frac{\text {mass of solution}}{\text {density of solution}}=\frac{100g}{1.83g/ml}=54.6ml

Now put all the given values in the formula of molality, we get

Molality=\frac{0.61\times 1000}{54.6ml}=11.2M

To calculate the volume of acid, we use the equation given by neutralisation reaction:

M_1V_1=M_2V_2

where,

M_1\text{ and }V_1 are the molarity and volume of stock acid which is H_2SO_4

M_2\text{ and }V_2 are the molarity and volume of dilute acid which is H_2SO_4

We are given:

M_1=11.2M\\V_1=mL\\M_2=0.30M\\V_2=1550mL

Putting values in above equation, we get:

11.2\times V_1=0.30\times 1550\\\\V_1=41.5mL

Thus 41.5 mL of the solution would be required to prepare 1550 mL of a .30M solution of the acid

4 0
3 years ago
Please help! Help me solve problems about naming structures with IUPAC rules
lianna [129]
A. The longest carbon chain is eight, and it has two methyl groups attached to carbon three, and a special group attached to carbon five. Its two names could be:

3-dimethyl-5-(1-methylethyl)octane
3-dimethyl-5-isopropyloctane

Both of these are correct. This is an alkane, because it has all single bonds.

B. This has a triple bond contained between carbons 2 and 3, and has a methyl group off carbon 4. The longest chain is 5. It’s name is:

4-methyl-2-pentyne

This is an alkene, because of the double bond.

C. This has a double bond contained between carbons 2 and 3, and has a methyl off of four and an methyl off of six. The longest chain is eight (follow the longest chain of carbons).

4,6-dimethyl-2-octene

This is an alkene, because of the double bond.

D. This has an ethyl group at 1 and a methyl group at 2 (rotate the compound to make it as clean as possible, in this case, the ring is flipped and rotated to make it alphabetical with the smallest numbers possible). The two names are:

1-ethyl-2-methylbenzene
ortho-ethylmethylbenzene

Both are correct, the ortho prefix telling the location of the ethyl and methyl groups. This is an aromatic structure because of its double bonded ring.

E. The longest chain is nine, and has methyls at three, five, and seven, along with a propyl at five. The name is:

3,5,7-trimethyl-5-propylnonane

This is an alkane, due to the single bonds.

Hope this helps!
7 0
3 years ago
In poor quality fireworks what will you notice and why
anastassius [24]
The best answer I could find was when you Google it, that the fuse is of poor quality. I cannot leave you a link, but you can find it for yourself. Put in poor quality fireworks and all sorts of things will pop up. No pun intended.
8 0
3 years ago
Read 2 more answers
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