Answer:
-250.3kJ
Explanation:
Based in the reactions and using -<em>Hess's law-</em>:
(1) P₄(s) + 6 Cl₂(g) → 4PCl₃(g) ΔH₁ = -4439kJ
(2) 4PCl₅(g) → P₄(s) + 10Cl₂ ΔH₂ = 3438kJ
The sum of (1) + (2) is:
4PCl₅(g) → 4PCl₃(g) + 4 Cl₂ ΔH = -4439kJ + 3438kJ = -1001kJ
Dividing this reaction in 4:
PCl₅(g) → PCl₃(g) + Cl₂ ΔH = -1001kJ / 4 = <em>-250.3kJ</em>
When pancakes are being cooked they get stuck to each other <span><span>piece,</span> in the pancake an get more tangled up</span>
It's unable to identify a decrease in LOS linked to corticosteroid exposure during hospitalization for ocular cellulitis in this database search. After two days of hospitalization, operational episodes and the prescription of corticosteroids were related to admission to the PICU.
Within two days of admission, 1347 (24%) of the 5462 children who were included in the research received a corticosteroid prescription. In analyses that controlled for age, the existence of meningitis, abscess, or visual problems, as well as the surgical episode and PICU admission within 2 days, corticosteroid prescription was not linked with LOS (e = 1.01, 95% confidence interval [CI]: 0.97-1.06). Among patients with a primary diagnosis of orbital cellulitis, corticosteroid exposure was linked to surgical events after two days of hospitalization (odds ratio = 2.05, 95% CI: 1.29-3.27) and 30-day readmission (odds ratio = 2.40, 95% CI: 1.52-3.78). Prospective, randomized control trials are required prior to the widespread usage of corticosteroids.
Learn more about orbital cellulitis here:-
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A) Head to tail joining of monomers. :) (confirmed correct answer, I took the test)
<h3><u>Answer;</u></h3>
NH3/NH4+
<h3><u>Explanation;</u></h3>
From the equation;
NH3(aq)+HNO3(aq)→NH4+(aq)+NO3−(aq)
NH3 is the base; while NH4+ is the conjugate acid
HNO3 is the acid; while NO3- is the conjugate base
- The conjugate base of a Brønsted-Lowry acid is species that is formed after an acid donates a proton while the conjugate acid of a Brønsted-Lowry base is the species formed after a base accepts a proton.