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hammer [34]
3 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]3 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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uranmaximum [27]

Answer:

Wavelength, \lambda=3.3\times 10^{-18}\ m

Explanation:

We have,

Velocity of the object, v=5\times 10^5\ m/s

Mass of the object, m=4\times 10^{-22}\ kg

It is required to find the wavelength of the object. The relation between wavelength and velocity of the object is given by :

\lambda=\dfrac{h}{mv}

h is Planck's constant

\lambda=\dfrac{6.6\times 10^{-34}}{4\times 10^{-22}\times 5\times 10^5}\\\\\lambda=3.3\times 10^{-18}\ m

So, the wavelength of the object is 3.3\times 10^{-18}\ m.      

5 0
3 years ago
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Answer:

Explanation:

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6 0
3 years ago
2 HI(g) ⇄ H2(g) + I2(g) Kc = 0.0156 at 400ºC 0.550 moles of HI are placed in a 2.00 L container and the system is allowed to rea
Alex_Xolod [135]

<u>Answer:</u> The concentration of hydrogen gas at equilibrium is 0.0275 M

<u>Explanation:</u>

Molarity is calculated by using the equation:

\text{Molarity}=\frac{\text{Number of moles}}{\text{Volume of solution (in L)}}

Moles of HI = 0.550 moles

Volume of container = 2.00 L

\text{Initial concentration of HI}=\frac{0.550}{2}=0.275M

For the given chemical equation:

                          2HI(g)\rightleftharpoons H_2(g)+I_2(g)

<u>Initial:</u>                  0.275

<u>At eqllm:</u>           0.275-2x      x         x

The expression of K_c for above equation follows:

K_c=\frac{[H_2][I_2]}{[HI]^2}

We are given:

K_c=0.0156

Putting values in above expression, we get:

0.0156=\frac{x\times x}{(0.275-2x)^2}\\\\x=-0.0458,0.0275

Neglecting the negative value of 'x' because concentration cannot be negative

So, equilibrium concentration of hydrogen gas = x = 0.0275 M

Hence, the concentration of hydrogen gas at equilibrium is 0.0275 M

4 0
3 years ago
2ch4(g) c2h2(g) 3h2(g) describe what is happening within the system when it is at equilibrium in terms of concentrations, reacti
SashulF [63]

Balanced chemical reaction: 2CH₄(g) ⇄ C₂H₂(g) + 3H₂(g).

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2) At equilibrium, both the forward and reverse reactions are still occurring.

3) Reaction rates of the forward and backward reactions are equal and  there are no changes in the concentrations of the reactants and products.

4 0
3 years ago
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A. Fission creates new elements from which electricity can be generated.

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3 years ago
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