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notsponge [240]
3 years ago
15

Question 2 of 5

Chemistry
1 answer:
Nataly_w [17]3 years ago
5 0

Answer:

C

Explanation:

The answer is C because machines such as tractors and sprinklers are counted as technology, and are better examples than pesticides.

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(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
NEEED HELP PLEASE IM VERY STRESSED
Arte-miy333 [17]

Answer:

C) It moves faster and collects more data

Explanation:

5 0
3 years ago
What would happen if the aspects that allow to maintain the characteristics of the states were altered?​
irinina [24]
An altered state of consciousness can be defined as any state of consciousness that deviates from normal waking consciousness, in terms of marked differences in our level of awareness, perceptions, memories, thinking, emotions, behaviours and sense of time, place and self-control.

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3 0
3 years ago
What is the [OH-] in a solution if the [H*] = 1.2 x 10-3 M?
spayn [35]

We know that [OH⁻] * [H⁺] = 10⁻¹⁴

plugging the value of [H⁺]

[OH⁻] * 1.2 * 10⁻³ = 10⁻¹⁴

[OH⁻] = 10⁻¹⁴ * (10³/1.2)

[OH⁻] = 833.3 * 10⁻¹⁴

[OH⁻] = 8.33 * 10⁻¹²

7 0
3 years ago
I will give the brainliest for who answer this questio but no guessing
xz_007 [3.2K]
IVA: nonmetal/other metals VIIA:halogens VIII: Nobel Gases
8 0
3 years ago
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