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hammer [34]
2 years ago
6

KHP is a monoprotic acid which provides one H+ ion. How would your results be affected if a diprotic acid (such as sulfuric acid

, H2SO4) was used which provides two H+ ions? Calculate the molarity of the NaOH solution if 17.50 mL of N a O H was titrated against 10.50 mL of 0.75 M H2SO4.
Chemistry
1 answer:
kari74 [83]2 years ago
7 0

Explanation:

A.

In a diprotic acid, 2 moles of H+ ions is released. Therefore, number of moles of H+ in a diprotic acid = 2 × number of moles of H+ of monoprotic acid.

B.

Equation of the reaction

2NaOH + H2SO4 --> Na2SO4 + 2H2O

Number of moles of H2SO4 = molar concentration × volume

= 0.75 × 0.0105

= 0.007875 moles.

By stoichiometry, since 1 mole of H2SO4 reacts with 2 moles of NaOH. Therefore, number of moles of NaOH = 2 × 0.007875

= 0.01575 moles.

Molar concentration of NaOH = number of moles ÷ volume

= 0.01575 ÷ 0.0175

= 0.9 M of NaOH.

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Answer:Molarity tells us the number of moles of solute in exactly one liter of a solution. (Note that molarity is spelled with an "r" and is represented by a capital M.) We need two pieces of information to calculate the molarity of a solute in a solution: The moles of solute present in the solution

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3 years ago
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frozen [14]

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3 years ago
Ocean water contains 3.5 nacl by mass. what mass of ocean water in grams contains 45.8 g of nacl
Romashka-Z-Leto [24]

Mass percentage of sodium chloride(NaCl) in ocean waters = 3.5 %

That means 3.5 g sodium chloride(NaCl) is present for every 100 g of ocean water.

The given mass of sodium chloride(NaCl) is 45.8 g

Calculating the mass of ocean waters that would contain 45.8 g sodium chloride(NaCl):

45.8 g NaCl *\frac{100g ocean water}{3.5g NaCl}

                     = 1309 g ocean water

Therefore, 45.8 g sodium chloride is present in 1309 g ocean water.

3 0
3 years ago
Which item does not depend on rare earth elements?
Alisiya [41]

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2 years ago
A gas mixture contains 3.00 atm of H2 and 1.00 atm of O2 in a 1.00 L vessel at 400K. If the mixture burns to form water while th
sleet_krkn [62]

Answer:

p_{H_2O}=2.00atm

Explanation:

Hello!

In this case, according to the following chemical reaction:

2H_2+O_2\rightarrow 2H_2O

It means that we need to compute the moles of hydrogen and oxygen that are reacting, via the ideal gas equation as we know the volume, pressure and temperature:

n_{H_2}=\frac{3.00atm*1.00L}{0.08206\frac{atm*L}{mol*K}*400K}=0.0914molH_2 \\\\n_{O_2}=\frac{1.00atm*1.00L}{0.08206\frac{atm*L}{mol*K}*400K}=0.0305molH_2

Thus, the yielded moles of water are computed by firstly identifying the limiting reactant:

n_{H_2O}^{by\ H_2} = 0.0914molH_2*\frac{2molH_2O}{2molH_2} =0.0914molH_2O\\\\n_{H_2O}^{by\ O_2} = 0.0305molO_2*\frac{2molH_2O}{1molO_2} =0.0609molH_2O

Thus, the fewest moles of water are 0.0609 mol so the limiting reactant is oxygen; in such a way, by using the ideal gas equation once again, we compute the pressure of water:

p_{H_2O}=\frac{0.0609molH_2O*0.08206\frac{atm*L}{mol*K}*400K}{1.00L}\\\\ p_{H_2O}=2.00atm

Best regards!

7 0
3 years ago
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