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Inga [223]
3 years ago
13

The monoprotic acid from among the following is

Chemistry
2 answers:
oksian1 [2.3K]3 years ago
3 0

Monoprotic acid are acids having only one hydrogen atoms after dissociation into ions from its compound. The monoprotic acid from among the following is HCl. The answer is letter D. HCl → H+ + Cl-. Note that there is only one H+ ion upon dissociation.

Marianna [84]3 years ago
3 0

Answer: Option (d) is the correct answer.

Explanation:

An acid which gives only one hydrogen ion or H^{+} ions upon dissociation in an aqueous solution is known as a monoprotic acid.

Whereas when an acid gives two hydrogen or H^{+} ions upon dissociation in an aqueous solution is known as a diprotoc acid.

And when an acid gives three hydrogen or H^{+} ions upon dissociation in an aqueous solution is known as a triprotoc acid.

Therefore dissociation of the given acids in an aqueous solution will be as follows.

  H_{2}CO_{3} \rightarrow 2H^{+} + CO^{2-}_{3}

  H_{2}SO_{4} \rightarrow 2H^{+} + SO^{2-}_{4}

  H_{3}PO_{4} \rightarrow 3H^{+} + PO^{2-}_{4}

  HCl \rightarrow H^{+} + Cl^{-}

Hence, we can conclude that out of the given options, HCl is the monoprotic acid.

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Explanation:

if A=10.0ml let B be x

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Approximately how many ice cubes must melt to cool 650 milliliters of water from 29°C to 0°C? Assume that each ice cube contains
qwelly [4]

Answer : The number of ice cubes melt must be, 13

Explanation :

First we have to calculate the mass of water.

\text{Mass of water}=\text{Density of water}\times \text{Volume of water}

Density of water = 1.00 g/mL

Volume of water = 650 mL

\text{Mass of water}=1.00g/mL\times 650mL=650g

Now we have to calculate the heat released on cooling.

Heat released on cooling = m\times c\times (T_2-T_1)

where,

m = mass of water = 650 g

c = specific heat capacity of water = 4.18J/g^oC

T_2 = final temperature = 29^oC

T_2 = initial temperature = 0^oC

Now put all the given values in the above expression, we get:

Heat released on cooling = 650g\times 4.18J/g^oC\times (29-0)^oC

Heat released on cooling = 78793 J = 78.793 kJ   (1 J = 0.001 kJ)

As, 1 ice cube contains 1 mole of water.

The heat required for 1 ice cube to melt = 6.02 kJ

Now we have to calculate the number of ice cubes melted.

Number of ice cubes melted = \frac{\text{Total heat}}{\text{Heat for 1 ice cube}}

Number of ice cubes melted = \frac{78.793kJ}{6.02kJ}

Number of ice cubes melted = 13.1 ≈ 13

Therefore, the number of ice cubes melt must be, 13

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Since the half-life of 235U (7. 13 x 108 years) is less than that of 238U (4.51 x 109 years), the isotopic abundance of 235U has
Ymorist [56]

Answer:

\mathtt{ t_1-t_2= In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2} \ years}

Explanation:

Given that:

The Half-life of ^{235}U = 7.13 \times 10^8 \ years is less than that of ^{238} U = 4.51 \times 10^9 \ years

Although we are not given any value about the present weight of ^{235}U.

So, consider the present weight in the percentage of ^{235}U to be  y%

Then, the time elapsed to get the present weight of ^{235}U = t_1

Therefore;

N_1 = N_o e^{-\lambda \ t_1}

here;

N_1 = Number of radioactive atoms relating to the weight of y of ^{235}U

Thus:

In( \dfrac{N_1}{N_o}) = - \lambda t_1

In( \dfrac{N_o}{N_1}) =  \lambda t_1 --- (1)

However, Suppose the time elapsed from the initial stage to arrive at the weight of the percentage of ^{235}U to be = t_2

Then:

In( \dfrac{N_o}{N_2}) =  \lambda t_2  ---- (2)

here;

N_2 =  Number of radioactive atoms of ^{235}U relating to 3.0 a/o weight

Now, equating  equation (1) and (2) together, we have:

In( \dfrac{N_o}{N_1}) -In( \dfrac{N_o}{N_2}) =  \lambda( t_1-t_2)

replacing the half-life of ^{235}U = 7.13 \times 10^8 \ years

In( \dfrac{N_2}{N_1})  = \dfrac{In 2}{7.13 \times 10^9}( t_1-t_2)      ( since \lambda = \dfrac{In 2}{t_{1/2}} )

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\mathtt{In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2}= t_1-t_2}

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Thus, The time elapsed is  \mathtt{ t_1-t_2= In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2} \ years}

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