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BigorU [14]
3 years ago
13

Magnesium oxide (MgO) forms when the metal burns in air. (a) If 1-25 9 of MgO contains 0.754 g of Mg, what is the mass ratio of

magnesium to oxide? (b) How many grams of Mg are in 534 g of MgO?
Chemistry
1 answer:
Vsevolod [243]3 years ago
5 0

Answer:

Explanation:

a )

1.25 g MgO contains .754 g of Mg .Rest will be O

so oxide = 1.25 - .754 = 0.496 g

ratio of magnesium to oxide = .754/.496 = 1.52

b) 1.25 g of MgO contains .754 g of Mg

534 g of MgO contains .754 x 534 / 1.25 g = 322.11 g

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The volume of  0. 250 mole sample of H_{2} gas occupy if it had a pressure of 1. 70 atm and a temperature of 35 °C is  3.71 L.

Calculation,

According to ideal gas equation which is known as ideal gas law,

PV =n RT

  • P is the pressure of the hydrogen gas  = 1.7 atm
  • Vis the volume of the hydrogen gas = ?
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By putting all the values of the given data like pressure temperature universal gas constant and number of moles in equation (i) we get ,

1.7 atm×V = 0.25 mole ×0.082 × 208 K

V = 0.25 mole ×0.082atm L /mole K × 308 K /1.7 atm

V = 3.71 L

So, volume of the sample of the hydrogen gas occupy is  3.71 L.

learn more about ideal gas equation

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The decomposition of N2O5 in solution in carbon tetrachloride proceeds via the reaction 2 N2O5(soln) → 4 NO2(soln) + O2(soln) Th
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<u>Answer:</u> The amount remained after 151 seconds are 0.041 moles

<u>Explanation:</u>

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Rate law expression for first order kinetics is given by the equation:

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k = rate constant  = 4.82\times 10^{-3}s^{-1}

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Putting values in above equation, we get:

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Hence, the amount remained after 151 seconds are 0.041 moles

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