Answer:
177.3kg C₂₁H₄₄
Explanation:
Based on the chemical reaction:
C₂₁H₄₄ → 3C₂H₄ + C₁₅H₃₂
<em>Where 1 mole of C₂₁H₄₄ produce 3 moles of ethene, C₂H₄.</em>
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To solve this question we need to determine the moles of ethene in 50.4kg. 1/3 these moles are the moles of C₂₁H₄₄ that must be added:
<em>Moles Ethene -Molar mass: 28.05g/mol-</em>
50.4kg = 50400g * (1mol / 28.05g) = 1796.8 moles of ethene
<em>Moles C₂₁H₄₄:</em>
1796.8 moles of ethene * (1 mol C₂₁H₄₄ / 3 mol C₂H₄) = 589.93 moles C₂₁H₄₄
<em>Mass C₂₁H₄₄:</em>
589.93 moles C₂₁H₄₄ * (296g / mol) = 177283g =
<h3>177.3kg C₂₁H₄₄</h3>
The element itself is Yttrium, which is a transition metal and not that reactive. Since it has 39 protons it will also have 39 electrons.
Answer: 
Explanation: <u>Heats</u> <u>of</u> <u>formation</u> is the amount of heat necessary to create 1 mol of a compound from its molecular constituents. The basic conditions the substance is formed is at standard conditions: 1 atm and 25°C. Each compound has its own heat of formation per mol of compound (kJ/mol), but to an element is assigned a value of zero.
<u>Standard</u> <u>Enthalpy</u> <u>Change</u> is defined as the heat absorbed or released when a reaction takes place. It can be positive or negative, which means reaction is endothermic or exothermic, respectively.
Enthalpy change is calculated as the difference between the sum of heat formation of products and the sum of heat formation of the reactants:

For the reaction
2NH₃ + 3N₂O → 4N₂ + 3H₂O
2(-46.2) + 3(82.05) 4(0) + 3(-241.8)
![\Delta H^{0}=3(-241.8)-[ 2(-46.2)+3(82.05)]](https://tex.z-dn.net/?f=%5CDelta%20H%5E%7B0%7D%3D3%28-241.8%29-%5B%202%28-46.2%29%2B3%2882.05%29%5D)


<u>The standard enthalpy change for the reaction is </u>
<u> kJ</u>
The formula or chemical formula of a compound is same irrespective of source / mode of synthesis . Thus if a sample of compound has one carbon atom for every two atoms of oxygen (CO2), the formula will remains the same
So the answer is that for all other samples the compound X should hold this ration true.