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Dafna1 [17]
3 years ago
6

Is this statement true or false?

Chemistry
1 answer:
gayaneshka [121]3 years ago
3 0
Many systems work together to control the internal environment of the body. This process is known as homeostasis. So the answer is False
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Conservation of mass was discussed in the background. describe how conservation of mass (actual, not theoretical) could be check
rosijanka [135]

Conservation of mass can be checked in an experiment . There are three steps to do it in a best way:

1. Weigh all the equipment and materials required in the experiment before the experiment.

2. Avoid spillage and evaporation during the experiment.

3. Weigh all the equipment and materials after the experiment.

If the mass is conserved then weight from step 1 is equal to weight from step 3.

7 0
3 years ago
(a) What is the total volume (in L) of gaseous products, measured at 350°C and 735 torr, when an automobile engine burns 100. g
Anarel [89]

Answer:

Part A

 The volume of the gaseous product  is  V = 787L

Part B

The volume of the the engine’s gaseous exhaust is  V_e = 2178 \ L

Explanation:

Part A

From the question we are told that

    The temperature is  T = 350^oC = 350 +273 =623K

     The pressure is  P = 735 \ torr = \frac{735}{760} =  0.967\ atm

     The of  C_8 H_{18} = 100.0g

The chemical equation for this combustion is

               2 C_8 H_{18}_{(l)} + 25O_2_{(l)} ----> 16CO_2_{(g)} + 18 H_2 O_{(g)}

 The number of moles of  C_8 H_{18} that reacted is mathematically represented as

               n = \frac{mass \ of \  C_8H_{18}  }{Molar \  mass \ of  C_8H_{18} }

The molar mass of  C_8 H_{18} is constant value which is

                  M = 114.23 \ g/mole  

So          n = \frac{100  }{114.23} }

             n = 0.8754 \ moles

The gaseous product in the reaction is CO_2_{(g)} and water vapour

Now from the reaction

    2 moles of C_8 H_{18}  will react with 25 moles of O_2 to give (16 + 18) moles of CO_2_{(g)} and  H_2 O_{(g)}

So

    1 mole of C_8 H_{18} will  react with 12.5 moles of  O_2 to give 17 moles of CO_2_{(g)} and  H_2 O_{(g)}

This implies that

    0.8754 moles of C_8 H_{18} will react with (12.5 * 0.8754 ) moles of O_2 to give  (17 * 0.8754) of CO_2_{(g)} and  H_2 O_{(g)}

So the no of moles of gaseous product is

         N_g = 17 * 0.8754

         N_g = 14.88 \ moles

From the ideal gas law

       PV = N_gRT

making V the subject

        V = \frac{N_gRT}{P}

Where R is the gas constant with a value R = 0.08206 \  L\cdot atm /K \cdot mole

Substituting values

          V = \frac{14.88* 0.08206 *623}{0.967}

          V = 787L

Part B

From the reaction the number of moles of oxygen that reacted is

         N_o = 0.8754 * 12.5

         N_o = 10.94 \ moles

The volume is

      V_o  = \frac{10.94 * 0.08206 *623}{0.967}

      V_o  = 579 \ L

No this volume is the 21% oxygen that reacted the 79% of air that did not react are the engine gaseous exhaust and this can be mathematically evaluated as

         V_e = V_o * \frac{0.79}{0.21}

Substituting values

       V_e = 579 * \frac{0.79}{0.21}

       V_e = 2178 \ L

3 0
3 years ago
Based on the data table, what evidence best suggests that this might be an object in freefall?
LuckyWell [14K]

Answer:

What datatable? Picture please!

Explanation:

5 0
3 years ago
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Does anyone know what Hydrogen's oxidation state would be if it was acting as an anion non-metal?
Ilia_Sergeevich [38]

Answer:

yeah,The oxidation state of an atom does not represent the "real" charge on that atom, or any other actual atomic property.Hydrogen has OS = +1, but adopts −1 when bonded as a hydride to metals or metalloids. Oxygen in compounds has OS = −2. This set of postulates covers .

Explanation:

3 0
3 years ago
Why is the r.a.m. Value used?
Andrew [12]

Answer:

The r.a.m. value is used to determine how many isotopes an element has.

Explanation:

5 0
3 years ago
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