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tester [92]
3 years ago
5

How does reproduction leads to natural selection in a population

Chemistry
1 answer:
Advocard [28]3 years ago
6 0

Natural selection requires variation between individuals. Mutations and reproduction increase genetic variation in a population. Natural selection occurs when environmental pressures favor certain traits that are passed on to offspring.

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Is cornstarch able to disintegrate in fire or an open flame?
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Answer:

Yes

Explanation:

I just quickly looked it up. Cornstarch is extremely flammable.

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What happens when you add acetic acid vinegar and sodium bicarbonate baking soda nahco​
Sophie [7]

Explanation:

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10) How many grams are there in 1.00 x 10^24 molecules of BC13?
lana66690 [7]

194.5 g of BCl₃ is present in 1 × 10²⁴ molecules of BCl₃.

Explanation:

In order to convert the given number of molecules of BCl₃ to grams, first we have to convert the molecules to moles.

It is known that 1 moles of any element has 6.022×10²³ molecules.

Then 1 molecule will have \frac{1}{6.022*10^{23} } moles.

So 1*10^{24} molecules = \frac{1*10^{24} }{6.022*10^{23} } =1.66 moles

Thus, 1.66 moles are included in BCl₃.

Then in order to convert it from moles to grams, we have to multiply it with the molecular mass of the compound.

As it is known as 1 mole contains molecular mass of the compound.

As the molecular mass of BCl₃ will be

Molecular mass of BCl_{3}= Mass of Boron + (3*Mass of chlorine)

Mass of boron is 10.811 g and the mass of chlorine is 35.453 g.

Molar mass of BCl₃ = 10.811+(3×35.453)=117.17 g.

grams of BCl_{3}=Molar mass of BCl_{3}*Number of moles

grams of BCl_{3}=117.17*1.66=194.5 g

So, 194.5 g of BCl₃ is present in 1 × 10²⁴ molecules of BCl₃.

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3 years ago
Which of the following [H+] values describes the most basic (alkaline) solution?
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4 years ago
Never mind jhvjycdtrsesetdfyguhbjnk
OlgaM077 [116]

Complete Question

methanol can be synthesized in the gas phase by the reaction of gas phase carbon monoxide with gas phase hydrogen, a 10.0 L reaction flask contains carbon monoxide gas at 0.461 bar and 22.0 degrees Celsius. 200 mL of hydrogen gas at 7.10 bar and 271 K is introduced. Assuming the reaction goes to completion (100% yield)

what are the partial pressures of each gas at the end of the reaction, once the temperature has returned to 22.0 degrees C express final answer in units of bar

Answer:

The partial  pressure of  methanol is  P_{CH_3OH_{(g)}} =0.077 \  bar

The partial  pressure of carbon monoxide is  P_{CO} = 0.382 \ bar

The partial  pressure at  hydrogen is  P_H =  O \  bar

Explanation:

From the question we are told that

  The volume of the  flask is  V_f = 10.0 \  L

   The initial pressure of carbon monoxide gas is  P_{CO} = 0.461 \ bar

   The initial  temperature of carbon monoxide gas is T_{CO} = 22.0^oC

   The volume of the hydrogen gas is  V_h  =  200 mL = 200 *10^{-3} \  L

    The initial  pressure of the hydrogen is P_H  =  7.10 \  bar

    The initial temperature of the hydrogen  is  T_H = 271 \  K

The reaction of  carbon monoxide and  hydrogen is  represented as

         CO_{(g)} + 2H_2_{(g)} \rightarrow CH_3OH_{(g)}

Generally from the ideal gas equation the initial number of moles of carbon monoxide is  

        n_1  =  \frac{P_{CO} *  V_f }{RT_{CO}}

Here R is the gas constant with value  R  = 0.0821 \ L \cdot atm \cdot mol^{-1} \cdot K

=>     n_1  =  \frac{0.461  *  10 }{0.0821 * (22 + 273)}

=>     n_1  = 0.19

Generally from the ideal gas equation the initial number of moles of Hydrogen  is  

       n_2  =  \frac{P_{H} *  V_H }{RT_{H}}

      n_2  =  \frac{ 7.10 *  0.2 }{0.0821 * 271 }

=> n_2  =  0.064

Generally from the chemical equation of the reaction we see that

        2 moles of hydrogen gas reacts with 1 mole of CO

=>      0.064 moles of  hydrogen gas will react with  x  mole of  CO

So

          x = \frac{0.064}{2}

=>       x = 0.032 \ moles \ of  \  CO

Generally from the chemical equation of the reaction we see that

        2 moles of hydrogen gas reacts with 1 mole of CH_3OH_{(g)}

=>      0.064 moles of  hydrogen gas will react with  z  mole of  CH_3OH_{(g)}

So

          z = \frac{0.064}{2}

=>       z = 0.032 \ moles \ of  \ CH_3OH_{(g)}

From this calculation we see that the limiting reactant is hydrogen

Hence the remaining CO after the reaction is  

          n_k = n_1 - x

=>       n_k = 0.19  - 0.032

=>       n_k = 0.156

So at the end of the reaction , the partial pressure for  CO is mathematically represented as

      P_{CO} = \frac{n_k  *  R *  T_{CO}}{V}

=>    P_{CO} = \frac{0.158   *  0.0821 *  295}{10}

=>    P_{CO} = 0.382 \ bar

Generally the partial pressure of  hydrogen is  0 bar because hydrogen was completely consumed given that it was the limiting reactant

Generally the partial  pressure of the methanol is mathematically represented as

         P_{CH_3OH_{(g)}} = \frac{z  *  R *  T_{CO}}{V_f}

Here  T_{CO} is used because it is given the question that the   temperature  returned to 22.0 degrees C

So

      P_{CH_3OH_{(g)}} = \frac{0.03 * 0.0821 *  295}{10}

     P_{CH_3OH_{(g)}} =0.077 \  bar

6 0
3 years ago
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