How many liters of solution can be produced from 2.5 moles of solute if a 2.0 M

Draw a negative relationship between variables

ankoles [38]

I hope this helped♡ I drew the realationship of variables

5 0

3 years ago

Given the reaction:

dezoksy [38]

Answer:

V = 22.34 L

Explanation:

Given data:

Volume of O₂ needed = ?

Temperature and pressure = standard

Number of molecules of water produced = 6.0× 10²³

Solution:

Chemical equation:

2H₂ + O₂ → 2H₂O

Number of moles of water:

1 mole contain 6.022× 10²³ molecules

6.0× 10²³ molecules × 1 mole / 6.022× 10²³ molecules

0.99 mole

Now we will compare the moles of oxygen and water.

H₂O : O₂

2 : 1

0.996 : 0.996

Volume of oxygen needed:

PV = nRT

V = nRT/P

V = 0.996 mol × 0.0821 atm.L/mol.K × 273.15 K / 1 atm

V = 22.34 L

3 0

3 years ago

Ephedrine, a central nervous system stimulant, is used in nasalsprays as a decongestant. this compound is a weak organic base: {

sashaice [31]

Ephedrine, a central nervous system stimulant, is used in nasal sprays as a decongestant. This compound is a weak organic base:

C10H15ON (aq) + H2O (l) -> C10H15ONH+ (aq) + OH- (aq)

A 0.035 M solution of ephedrine has a pH of 11.33.

a) What are the equilibrium concentrations of C10H15ON, C10H15ONH+, and OH-?

b) Calculate Kb for ephedrine.

c(C₁₀H₁₅NO) = 0,035 M.
pH = 11,33.
pOH = 14 - 11,33 = 2,67.
[OH⁻] = 10∧(-2,67) = 0,00213 M.
[OH⁻] = [C₁₀H₁₅NOH⁺] = 0,00213 M.
[C₁₀H₁₅NO] = 0,035 M - 0,00213 M = 0,03287 M.
Kb = [OH⁻] · [C₁₀H₁₅NOH⁺] / [C₁₀H₁₅NO].
Kb = (0,00213 M)² / 0,03287 M = 1,38·10⁻⁴.

3 0

3 years ago

Which of the following would be more likely to conduct electrons? Strontium or Iodine? Defend your choice.

Alchen [17]

Strontium because it is a metal

6 0

3 years ago

The heats of combustion of ethane (C2H6) and butane (C4H10) are 52 kJ/g and 49 kJ/g, respectively. We need to produce 1.000 x 10

LekaFEV [45]

Answer :

(1) The number of grams needed of each fuel $(C_2H_6)\text{ and }(C_4H_{10})$ are 19.23 g and 20.41 g respectively.

(2) The number of moles of each fuel $(C_2H_6)\text{ and }(C_4H_{10})$ are 0.641 moles and 0.352 moles respectively.

(3) The balanced chemical equation for the combustion of the fuels.

$C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O$

$C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O$

(4) The number of moles of $CO_2$ produced by burning each fuel is 1.28 mole and 1.41 mole respectively.

The fuel that emitting least amount of $CO_2$ is $C_2H_6$

Explanation :

Part 1 :

First we have to calculate the number of grams needed of each fuel $(C_2H_6)\text{ and }(C_4H_{10})$ .

As, 52 kJ energy required amount of $C_2H_6$ = 1 g

So, 1000 kJ energy required amount of $C_2H_6$ = $\frac{1000}{52}=19.23g$

and,

As, 49 kJ energy required amount of $C_4H_{10}$ = 1 g

So, 1000 kJ energy required amount of $C_4H_{10}$ = $\frac{1000}{49}=20.41g$

Part 2 :

Now we have to calculate the number of moles of each fuel $(C_2H_6)\text{ and }(C_4H_{10})$ .

Molar mass of $C_2H_6$ = 30 g/mole

Molar mass of $C_4H_{10}$ = 58 g/mole

$\text{ Moles of }C_2H_6=\frac{\text{ Mass of }C_2H_6}{\text{ Molar mass of }C_2H_6}=\frac{19.23g}{30g/mole}=0.641moles$

and,

$\text{ Moles of }C_4H_{10}=\frac{\text{ Mass of }C_4H_{10}}{\text{ Molar mass of }C_4H_{10}}=\frac{20.41g}{58g/mole}=0.352moles$

Part 3 :

Now we have to write down the balanced chemical equation for the combustion of the fuels.

The balanced chemical reaction for combustion of $C_2H_6$ is:

$C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O$

and,

The balanced chemical reaction for combustion of $C_4H_{10}$ is:

$C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O$

Part 4 :

Now we have to calculate the number of moles of $CO_2$ produced by burning each fuel to produce 1000 kJ.

$C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O$

From this we conclude that,

As, 1 mole of $C_2H_6$ react to produce 2 moles of $CO_2$

As, 0.641 mole of $C_2H_6$ react to produce $0.641\times 2=1.28$ moles of $CO_2$

and,

$C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O$

From this we conclude that,

As, 1 mole of $C_4H_{10}$ react to produce 4 moles of $CO_2$

As, 0.352 mole of $C_4H_{10}$ react to produce $0.352\times 4=1.41$ moles of $CO_2$

So, the fuel that emitting least amount of $CO_2$ is $C_2H_6$

5 0

3 years ago