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Maksim231197 [3]

2 years ago

11

Pacing thresholds of temporary epicardial electrodes: Variation with electrode type, time, and epicardial position

1 answer:

Rudiy272 years ago

7 0

Pacing thresholds of temporary epicardial electrodes: Variation with electrode type of anxiety (low vs. high) and stress.

Everyone experiences anxiety occasionally, but persistent anxiety can reduce your quality of life. Though likely best known for altering behavior, worry can have negative effects on our electrodes health. Anxiety speeds up our heartbeat and breathing, concentrating blood flow to the parts of our brains that need it. You are getting ready for a electrodes situation by having this extremely bodily reaction. Test performance may be impacted by anxiety. According to studies, pupils with low levels of test anxiety perform better on multiple-choice question (MCQ) exams than pupils with high levels of anxiety. Studies have indicated that female students have greater levels of test anxiety than male students.

Learn more about anxiety here:

brainly.com/question/4913240

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A scientist and his friend walked into a bar. The Scientist said: "May I please get some H2O". the waitress asked his friend wha

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Phosphorus trichloride gas and chlorine gas react to form phosphorus pentachloride gas: PCl3(g)+Cl2(g)⇌PCl5(g). A 7.5-L gas vess

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Answer:

The equilibrium constant in terms of concentration that is, $K_c=3.6243\times 10^{3}$ .

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$PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)$

The relation of $K_c\& K_p$ is given by:

$K_p=K_c(RT)^{\Delta n_g}$

$K_p$ = Equilibrium constant in terms of partial pressure.=98.1

$K_c$ = Equilibrium constant in terms of concentration =?

T = temperature at which the equilibrium reaction is taking place.

R = universal gas constant

$\Delta n_g$ = Difference between gaseous moles on product side and reactant side= $n_{g,p}-n_{g.r}=1-2=-1$

$98.1=K_c(RT)^{-1}$

$98.1 =\frac{K_c}{RT}$

$K_c=98.1\times 0.0821 L atm/mol K\times 450 K=3,624.30=3.6243\times 10^{3}$

The equilibrium constant in terms of concentration that is, $K_c=3.6243\times 10^{3}$ .

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