Answer:
Frozen Meats – Frozen raw meats that are properly vacuum sealed can be stored in the freezer from 1-3 years depending on the type of meat. However, raw meat that is not vacuum sealed will only last 1-12 months depending on the meat.
Explanation:
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Answer:
AsF3:C2CI6
4:3
1.3618 moles: 1.02135 moles(1.3618÷4×3)
C2CI6 is the limting reagent
So the number of moles for AsCI3 is 0.817 moles( number of moles of the limting reagant) ÷3 ×4 (according to ratio by balancing chemical equation)=1.09 moles(3 s.f.)
or
Balanced equation
4AsF3 + 3C2Cl6 → 4AsCl3 + 3C2Cl2F4
Use stoichiometry to calculate the moles of AsCl3 that can be produced by each reactant.
Multiply the moles of each reactant by the mole ratio between it and AsCl3 in the balanced equation, so that the moles of the reactant cancel, leaving moles of AsCl3.
Explanation:
More positive ions shows acidic whereas more negative ions indicates basic solution.
<h3>Which charge show acidic solution?</h3>
That side which has more positive charges is considered as acidic solution while on the other hand, that region where negative charges are present in large number as compared to positive charges is considered as basic or alkaline solution.
So we can conclude that more positive ions shows acidic whereas more negative ions indicates basic solution.
Learn more about charge here: brainly.com/question/25923373
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Answer:
ΔH₁₂ = -867.2 Kj
Explanation:
Find enthalpy for 3H₂ + O₃ => 3H₂O given ...
2H₂ + O₂ => 2H₂O ΔH₁ = -483.6 Kj
3O₂ => 2O₃ ΔH₂ = + 284.6 Kj
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3(2H₂ + O₂ => 2H₂O) => 6H₂ + 3O₂ => 6H₂O (multiply by 3 to cancel O₂)
6H₂ + 3O₂ => 6H₂O ΔH₁ = 3(-483.6 Kj) = -1450.6Kj
2O₃ => 3O₂ ΔH₂ = -284.6Kj (reverse rxn to cancel O₂)
_______________________________
6H₂ + 2O₃ => 6H₂O ΔH₁₂ = -1735.2 Kj (Net Reaction - not reduced)
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divide by 2 => target equation (Net Reaction - reduced)
3H₂ + O₃ => 3H₂O ΔH₁₂ = (-1735.2/2) Kj = -867.2 Kj
The normality that would be calculated will be to high because the change in volume will be greater than the actual change in volume. if the buret tip is not filled when reading the initial volume, the actual volume should be lesser with that reading. so if you will you the higher reading the change in volume or the volume you use in titration will be higher than the actual