Sassdspioewqhfoirwebfoeubeowiwfnbrufjiwrjf

PLZZ HELPPP FRLLL IM GONNA GET MY LIFE ENDED

olga_2 [115]

Not completely sure but I think it's

Answer: Local winds affect a smaller area while global winds affect a larger area

Explanation: This is because local winds affect certain areas and global winds go all around the planet

7 0

3 years ago

Read 2 more answers

2-Ethoxy-2,3-dimethylbutane reacts with concentrated aqueous HI to form two initial organic products (A and B). Further reaction

Leno4ka [110]

Answer:

A = 2-iodo-2,3-dimethylbutane

B = Ethanol

C = Iodoethane (also called ethyl-iodide)

Explanation:

2-Ethoxy-2,3-dimethylbutane reacts with conc. HI to cleave the oxy-functional group.

On one end, ethanol is formed and on the other hand, 2-iodo-2,3-dimethylbutane is formed.

But ethanol reacts further with conc HI to give iodoethane.

Therefore,

A = 2-iodo-2,3-dimethylbutane

B = Ethanol

C = Iodoethane (also called ethyl-iodide)

This is all shown in the attached image.

Hope this Helps!!!

6 0

4 years ago

It takes 24.56 mL of a sodium hydroxide solution to precisely neutralize 2.023 g of potassium hydrogen phthalate (F.W. = 204.22g

Alinara [238K]

Answer:

0.4033 M is the concentration of the sodium hydroxide solution.

Explanation:

Mass of potassium hydrogen phthalate = 2.023 g

Molar mass of potassium hydrogen phthalate = 204.22 g/mol

Moles of potassium hydrogen phthalate :

= $\frac{2.023 g}{204.22 g/mol}=0.009906 mol$

$C_8H_5KO_4+NaOH\rightarrow C_8H_4KO_4Na+H_2O$

According to reaction, 1 mole of potassium hydrogen phthalate reacts with 1mole of NaOH. Then 0.009906 moles of potassium hydrogen phthalate will recat with:

$\frac{1}{1}\times 0.009906 mol=0.009906 mol$ of NaOH

Moles of NaOH = 0.009906 mol

Volume of NaOH solution = 24.56 mL = 0.02456 L

$Molarity=\frac{Moles}{Volume (L)}$

$[NaOH]=\frac{0.009906 mol}{0.02456 L}=0.4033 M$

0.4033 M is the concentration of the sodium hydroxide solution.

8 0

3 years ago

HELP ASAP

Artemon [7]

10 minutes hope it helps

3 0

3 years ago

In the following chemical reaction between H_2 and Cl_2 to produce HCl, what is the mass of HCl produced and leftover reactants

ira [324]

Answer: The total amount of leftover reactants and HCl is 12.79 grams

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For hydrogen gas:

Given mass of hydrogen gas = 0.36 g

Molar mass of hydrogen gas = 2 g/mol

Putting values in equation 1, we get:

$\text{Moles of hydrogen gas}=\frac{0.36g}{2g/mol}=0.18mol$

For chlorine gas:

Given mass of chlorine gas = 12.41 g

Molar mass of chlorine gas = 71 g/mol

Putting values in equation 1, we get:

$\text{Moles of chlorine gas}=\frac{12.41g}{71g/mol}=0.175mol$

The chemical equation for the reaction of hydrogen gas and chlorine gas is:

$H_2+Cl_2\rightarrow 2HCl$

By Stoichiometry of the reaction:

1 moles of chlorine gas reacts with 1 mole of hydrogen gas

So, 0.175 moles of chlorine gas will react with = $\frac{1}{1}\times 0.175=0.175mol$ of hydrogen gas

As, given amount of hydrogen gas is more than the required amount. So, it is considered as an excess reagent.

Thus, chlorine gas is considered as a limiting reagent because it limits the formation of product.

Moles of excess reactant left (hydrogen gas) = [0.18 - 0.175] = 0.005 moles

By Stoichiometry of the reaction

1 moles of chlorine gas produces 2 moles of HCl

So, 0.175 moles of chlorine gas will produce = $\frac{2}{1}\times 0.175=0.350$ moles of HCl

Now, calculating the mass of hydrogen gas left and HCl from equation 1, we get:

For hydrogen gas:

Molar mass of hydrogen gas = 2 g/mol

Moles of excess hydrogen gas = 0.005 moles

Putting values in equation 1, we get:

$0.005mol=\frac{\text{Mass of excess hydrogen gas}}{2g/mol}\\\\\text{Mass of excess hydrogen gas}=(0.005mol\times 2g/mol)=0.01g$

For HCl:

Molar mass of HCl = 36.5 g/mol

Moles of HCl = 0.350 moles

Putting values in equation 1, we get:

$0.350mol=\frac{\text{Mass of HCl}}{36.5g/mol}\\\\\text{Mass of HCl}=(0.350mol\times 36.5g/mol)=12.78g$

Total mass of HCl and leftover reactants = [12.78 + 0.01] = 12.79 g

Hence, the total amount of leftover reactants and HCl is 12.79 grams

7 0

4 years ago