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3 years ago

What is the molarity of a 10 L solution containing 5.0 miles if solute

1 answer:

3 0

Hey there !

Solution :

Moles of solution = 5.0

Volume of solution = 10 L

Therefore :

M = n / V

M = 5.0 / 10

= 0.5 M

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Watching the assignment without knowing how to solve the questions is the worst thing to experience hope someone can help with t

frez [133]

Answer:

- 622.5kJ

Explanation:

1) 2NH3 + 3N2O ----> 4N2 + 3H2O ΔH° 1= - 1010kJ

- 3 (N2O + 3H2 ---> N2H4 + H2O ΔH° 2 = - 317 kJ)

2) 2NH3 + 3N2O ----> 4N2 + 3H2O ΔH° 1= - 1010kJ

- 3N2O - 9 H2 ----> - 3N2H4 - 3H2O - 3ΔH° 2 = 3*317 kJ

2NH3 + 3N2H4 --->4N2+9H2, ΔH° 1 - 3*ΔH° 2

2NH3 + 3N2H4 --->4N2+9H2, ΔH° 1 - 3*ΔH° 2

3) -(2NH3 +1/2O2 ---> N2H4 + H2O, ΔH°3 = -143 kJ)

-2NH3 - 1/2O2 ---> - N2H4 - H2O, - ΔH°3 = 143 kJ

2NH3 + 3N2H4 --->4N2+9H2, ΔH° 1 - 3*ΔH° 2

-2NH3 - 1/2O2 ---> - N2H4 - H2O, - ΔH°3 = 143 kJ

4N2H4 + H2O ---> 4N2 + 9H2 + 1/2O2, ΔH° 1 - 3*ΔH° 2 - ΔH°3

H2 + 1/2O2 ---> H2O, ΔH° 4 = - 286 kJ

9*(H2 + 1/2O2 ---> H2O, ΔH° 4 = - 286 kJ)

9H2+ 9/2 O2 ---> 9H2O, 9*ΔH° 4 = 9*(- 286) kJ

4N2H4 + H2O ---> 4N2 + 9H2 + 1/2O2, ΔH° 1 - 3*ΔH° 2 - ΔH°3

9H2+ 9/2 O2 ---> 9H2O, 9*ΔH° 4 = 9*(- 286) kJ

4N2H4 +4O2 --->4N2+8H2O, ΔH° 1 - 3*ΔH° 2 - ΔH°3 + 9*ΔH° 4

1/4*(4N2H4 +4O2 --->4N2+8H2O, ΔH° 1 - 3*ΔH° 2 - ΔH°3 + 9*ΔH° 4

N2H4 +O2 --->N2+2H2O, 1/4( ΔH° 1 - 3*ΔH° 2 - ΔH°3 + 9*ΔH° 4)

N2H4 +O2 --->N2+2,

1/4( ΔH° 1 - 3*ΔH° 2 - ΔH°3 + 9*ΔH° 4)=

=1/4(-1010kJ - 3*(-317kJ) - (-143kJ) + 9*(-286kJ))= - 622.5 kJ

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3 years ago

What is the solution with a POH value of two considered

AfilCa [17]

Answer:

it is considered a base

Explanation:

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Which of the following is NOT a part of Dalton's atomic theory

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Answer:

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What volume of 3.00 MM HClHCl in liters is needed to react completely (with nothing left over) with 0.750 LL of 0.500 MM Na2CO3N

AlladinOne [14]

Answer: The volume of HCl needed is 0.250 L

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

For sodium carbonate:

Molarity of sodium carbonate solution = 0.500 M

Volume of solution = 0.750 L

Putting values in above equation, we get:

$0.500M=\frac{\text{Moles of sodium carbonate}}{0.750}\\\\\text{Moles of sodium carbonate}=(0.500mol/L\times 0.750L)=0.375mol$

The chemical equation for the reaction of sodium carbonate and HCl follows:

$Na_2CO_3+2HCl\rightarrow 2NaCl+H_2CO_3$

By Stoichiometry of the reaction:

1 mole of sodium carbonate reacts with 2 moles of HCl

So, 0.375 moles of sodium carbonate will react with = $\frac{2}{1}\times 0.375=0.750mol$ of HCl

Now, calculating the volume of HCl by using equation 1:

Moles of HCl = 0.750 moles

Molarity of HCl = 3.00 M

Putting values in equation 1, we get:

$3.00M=\frac{0.750mol}{\text{Volume of solution}}\\\\\text{Volume of solution}=\frac{0.750mol}{3.00mol/L}=0.250L$

Hence, the volume of HCl needed is 0.250 L

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Which type of bonding involves the sharing of electrons, in which the electrons are confined close to the more electronegative e

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Answer:

metallic bonding

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