The answer is the first one: first ionization energy
Answer:
no of moles=mass in gm÷molar mass so let x be the mass in gm
4.5=x÷35.5×2
x=4.5×35.5×2 grams
x=319.5 gm
Explanation:
formula
8H⁺ + 5Fe²⁺ + MnO₄⁻ ⇒ Mn²⁺ + 5Fe³⁺ + 4H₂O
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According to the reaction, Fe</span>²⁺ and MnO4⁻<span> have following stoichiometric ratio:
n(</span>Fe²⁺) : n(MnO4⁻) = 5 : 1
n(MnO4⁻) = n(Fe²⁺) / 5
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So, for each mole of </span>Fe²⁺ it is needed 1/5 moles of MnO4⁻.<span>
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Answer:
396 g OF CO2 WILL BE PRODUCED BY 270 g OF GLUCOSE IN A RESPIRATION PROCESS.
Explanation:
To calculate the gram of CO2 produced by burning 270 g of gucose, we first write out the equation for the reaction and equate the two variables involved in the question;
C6H12O6 + 6O2 -------> 6CO2 + 6H2O
1 mole of C6H12O6 reacts to form 6 moles of CO2
Then, calculate the molar mass of the two variables;
Molar mass of glucose = ( 12 *6 + 1* 12 + 16* 6) g/mol = 180 g/mol
Molar mass of CO2 = (12 + 16 *2) g/mol = 44 g/mol
Next is to calculate the mass of glucose and CO2 involved in the reaction by multiplying the molar mass by the number of moles
1* 180 g of glucose yields 6 * 44 g of CO2
180 g of glucose = 264 g of CO2
If 270 g of glucose were to be used, how many grams of CO2 will be produced;
so therefore,
180 g of glucose = 264 g of CO2
270 g of glucose = x grams of CO2
x = 264 * 270 / 180
x = 71 280 / 180
x = 396 g of CO2.
In other words, 396 g of CO2 will be produced by respiration from 270 g of glucose.
