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HACTEHA [7]
3 years ago
7

How many mols are in 15 of mg (0H) 2

Chemistry
1 answer:
SpyIntel [72]3 years ago
5 0
The answer is 0.0171468704904. We assume you are converting between moles Mg(OH)2 and gram. This compound is also known as Magnesium Hydroxide. 1 mole<span> is equal to </span>1 moles<span> Mg(OH)2, or 58.31968 grams.</span>
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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
Please help, I really don’t understand this!!!
kap26 [50]

<u>Analysing the Question:</u>

We are given the balanced equation:

C₆H₁₂O₆ + 6O₂→ 6CO₂ + 6H₂O

from this equation, we can say that: <em>for every 1 mole of Glucose, we need 6 moles of Oxygen</em>

<u>Moles of Glucose used in the reaction:</u>

Molar mass of Glucose = 180 grams / mol

Given mass of Glucose = 1 gram

Mole of Glucose = Given mass / Molar mass

Moles of Glucose = 1 / 180 moles

<u>Mass of Oxygen required:</u>

We know that for every mole of Glucose, we need 6 moles of Oxygen

So, for 1/180 moles of Glucose, we need 6 / 180 = 1 / 30 moles of Oxygen

Mass of 1 / 30 moles of Oxygen:

Mass = Molar mass * number of moles

Mass of Oxygen = 32 * 1/30

Mass of Oxygen  = 32 / 30

Mass of Oxygen = 1.06 grams

5 0
3 years ago
quizlet a basement is successfully sealed so that more radon may not enter the space, but 5.7×107 radon atoms are already trappe
12345 [234]

After 25 days, it remains radon 5.9x10^5 atoms.

Half-life is the time required for a quantity (in this example number of radioactive radon) to reduce to half its initial value.

N(Ra) = 5.7×10^7; initial number of radon atoms

t1/2(Ra) = 3.8 days; the half-life of the radon is 3.8 days

n = 25 days / 3.8 days

n = 6.58; number of half-lifes of radon

N1(Ra) = N(Ra) x (1/2)^n

N1(Ra) = 5.7×10^7 x (1/2)^6.58

N1(Ra) = 5.9x10^5; number of radon atoms after 25 days

The half-life is independent of initial concentration (size of the sample).

More about half-life: brainly.com/question/1160651

#SPJ4

6 0
2 years ago
Bae wya .......... please bae I can't find you please show up ​
Elan Coil [88]

Answer:

Your babe is in your school

Explanation:

go ask babes out

7 0
3 years ago
Read 2 more answers
In science, Bob learns that the energy of a wave is directly proportional to the square of the waves amplitude. If the energy of
Illusion [34]
Jamuuj ExpertThe energy of a wave is directly proportional to the square of the waves amplitude. Therefore, E = A² where A is the amplitude. This therefore means when the amplitude of a wave is doubled the energy will be quadrupled, when the amplitude is tripled the energy increases by a nine fold and so on.Thus, in this case if the energy is 4J, then the amplitude will be  √4 = 2 .
4 0
3 years ago
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