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koban [17]
3 years ago
5

calculate the number of moles of sulfuric acid that is contained in 250 mL of 8.500 M sulfuric acid solution

Chemistry
1 answer:
san4es73 [151]3 years ago
5 0

Answer : The moles of H_2SO_4 are, 2.125 mole.

Explanation : Given,

Molarity of H_2SO_4 = 8.500 M

Volume of solution = 250 mL  = 0.250 L    (1 L = 1000 mL)

Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.

Formula used :

\text{Molarity}=\frac{\text{Moles of }H_2SO_4}{\text{Volume of solution (in L)}}

Now put all the given values in this formula, we get:

8.500M=\frac{\text{Moles of }H_2SO_4}{0.250L}

\text{Moles of }H_2SO_4=2.125mol

Therefore, the moles of H_2SO_4 are, 2.125 mole.

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Answer:

T_2=335.42K=62.27^oC

Explanation:

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In this case, by using the general gas law, that allows us to understand the pressure-volume-temperature relationship as shown below:

\frac{P_1V_1}{T_1} =\frac{P_2V_2}{T_2}

Thus, solving for the temperature at the end (considering absolute units of Kelvin), we obtain:

T_2=\frac{P_2V_2T_1}{P_1V_1}=\frac{1.8L*0.75atm*(25+273.15)K}{1.2L*1.0atm} \\\\T_2=335.42K=62.27^oC

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How are Kelvin scale and celcius scale similar <br>​
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Answer: The Kelvin scale is related to the Celsius scale. The difference between the freezing and boiling points of water is 100 degrees in each, so that the kelvin has the same magnitude as the degree Celsius.

Explanation:

Celsius is, or relates to, the Celsius temperature scale (previously known as the centigrade scale). The degree Celsius (symbol: °C) can refer to a specific temperature on the Celsius scale as well as serve as a unit increment to indicate a temperature interval(a difference between two temperatures or an uncertainty). “Celsius” is named after the Swedish astronomer Anders Celsius (1701-1744), who developed a similar temperature scale two years before his death.

K = °C + 273.15

°C = K − 273.15

Until 1954, 0 °C on the Celsius scale was defined as the melting point of ice and 100 °C was defined as the boiling point of water under a pressure of one standard atmosphere; this close equivalence is taught in schools today. However, the unit “degree Celsius” and the Celsius scale are currently, by international agreement, defined by two different points: absolute zero, and the triple point of specially prepared water. This definition also precisely relates the Celsius scale to the Kelvin scale, which is the SI base unit of temperature (symbol: K). Absolute zero—the temperature at which nothing could be colder and no heat energy remains in a substance—is defined as being precisely 0 K and −273.15 °C. The triple point of water is defined as being precisely 273.16 K and 0.01 °C.

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A volcano violently erupts near a small town and wildlife preserve. Which effect might occur? Some local species develop new ada
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A weather balloon is filled with helium that occupies a volume of 5.57 104 L at 0.995 atm and 32.0°C. After it is released, it r
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6.52 × 10⁴ L. (3 sig. fig.)

<h3>Explanation</h3>

Helium is a noble gas. The interaction between two helium molecules is rather weak, which makes the gas rather "ideal."

Consider the ideal gas law:

P\cdot V = n\cdot R\cdot T,

where

  • P is the pressure of the gas,
  • V is the volume of the gas,
  • n is the number of gas particles in the gas,
  • R is the ideal gas constant, and
  • T is the absolute temperature of the gas in degrees Kelvins.

The question is asking for the final volume V of the gas. Rearrange the ideal gas equation for volume:

V = \dfrac{n \cdot R \cdot T}{P}.

Both the temperature of the gas, T, and the pressure on the gas changed in this process. To find the new volume of the gas, change one variable at a time.

Start with the absolute temperature of the gas:

  • T_0 = (32.0 + 273.15) \;\text{K} = 305.15\;\text{K},
  • T_1 = (-14.5 + 273.15) \;\text{K} = 258.65\;\text{K}.

The volume of the gas is proportional to its temperature if both n and P stay constant.

  • n won't change unless the balloon leaks, and
  • consider P to be constant, for calculations that include T.

V_1 = V_0 \cdot \dfrac{T_1}{T_2} = 5.57\times 10^{4}\;\text{L}\times \dfrac{258.65\;\textbf{K}}{305.15\;\textbf{K}} = 4.72122\times 10^{4}\;\text{L}.

Now, keep the temperature at T_1 =258.65\;\text{K} and change the pressure on the gas:

  • P_1 = 0.995\;\text{atm},
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The volume of the gas is proportional to the reciprocal of its absolute temperature \dfrac{1}{T} if both n and T stays constant. In other words,

V_2 = V_1 \cdot\dfrac{\dfrac{1}{P_2}}{\dfrac{1}{P_1}} = V_1\cdot\dfrac{P_1}{P_2} = 4.72122\times 10^{4}\;\text{L}\times\dfrac{0.995\;\text{atm}}{0.720\;\text{atm}}=6.52\times 10^{4}\;\text{L}

(3 sig. fig. as in the question.).

See if you get the same result if you hold T constant, change P, and then move on to change T.

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