Answer: 8.28g Na
Explanation: use ideal gas law
PV= nRT
Solve for moles of Cl2
n= PV/ RT
Substitute:
= 1 atm x 4.0 L / 0.08205 L.atm/ mol. K x 273 K
= 0.18 moles Cl2
Do stoichiometry to solve for m of Na
2 Na + Cl2 => 2 NaCl2
=0.18 moles Cl2 x 2 mol Na/ 1 mol Cl2 x 23g Na / 1 mol Na
= 8.28 g Na.
The rate constant : k = 9.2 x 10⁻³ s⁻¹
The half life : t1/2 = 75.3 s
<h3>Further explanation</h3>
Given
Reaction 45% complete in 65 s
Required
The rate constant and the half life
Solution
For first order ln[A]=−kt+ln[A]o
45% complete, 55% remains
A = 0.55
Ao = 1
Input the value :
ln A = -kt + ln Ao
ln 0.55 = -k.65 + ln 1
-0.598=-k.65
k = 9.2 x 10⁻³ s⁻¹
The half life :
t1/2 = (ln 2) / k
t1/2 = 0.693 : 9.2 x 10⁻³
t1/2 = 75.3 s
The maximum mass of B₄C that can be formed from 2.00 moles of boron (III) oxide is 55.25 grams.
<h3>What is the stoichiometry?</h3>
Stoichiometry of the reaction gives idea about the relative amount of moles of reactants and products present in the given chemical reaction.
Given chemical reaction is:
2B₂O₃ + 7C → B₄C + 6CO
From the stoichiometry of the reaction, it is clear that:
2 moles of B₂O₃ = produces 1 mole of B₄C
Now mass of B₄C will be calculated by using the below equation:
W = (n)(M), where
- n = moles = 1 mole
- M = molar mass = 55.25 g/mole
W = (1)(55.25) = 55.25 g
Hence required mass of B₄C is 55.25 grams.
To know more about stoichiometry, visit the below link:
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Answer:
Explanation:
Hello,
In this case, the enthalpy of combustion is understood as the energy released when one mole of fuel, in this case octene, is burned in the presence of oxygen and is computed with the enthalpies of formation of the fuel, carbon dioxide and water as shown below (oxygen is circumvented as it is a pure element):
Thus, since we already know the enthalpy of combustion of the fuel, for carbon and water we have -393.5 and -241.8 kJ/mol respectively, thereby, the enthalpy of combustion turns out:
Best regards.
Answer:
the answer is Fructose
Explanation:
the reason is because when it brakes down it forms a sort of fructose
