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2 years ago

10

Answer the question PLS 100 points!!!

1 answer:

cluponka [151]2 years ago

5 0

1atm=760mm Hg

Now

1.36atm
1.36(760)
1033.6mm Hg

Done!

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When a variable is determined by a factor outside of the function or model being evaluated, it is said to be A. statistically i

inna [77]

Answer:

The correct answer is B. exogenous

Explanation:

Let us try to describe exogenous and endogenous variables an exogenous variable value is influenced only by factors outside a model or system and is forced onto the model, while a change in an exogenous variable is known as an exogenous change. Also an endogenous variable is one whose value is influenced only by the system or model under study.

4 0

4 years ago

If the pH changes from 5 to 6, the H+ would concentration would?

Strike441 [17]

Decrease by factor 10

5 0

3 years ago

If a radioactive isotope has a half-life of 26.5 days, how many days does it take for a sample of the isotope to decrease by 35.

dangina [55]

Answer:

16.5 days

Explanation:

Given that:

Half life = 26.5 days

$t_{1/2}=\frac {ln\ 2}{k}$

Where, k is rate constant

So,

$k=\frac {ln\ 2}{t_{1/2}}$

$k=\frac {ln\ 2}{26.5}\ days^{-1}$

The rate constant, k = 0.02616 days⁻¹

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration

Given:

35.0 % is decomposed which means that 0.35 of $[A_0]$ is decomposed. So,

$\frac {[A_t]}{[A_0]}$ = 1 - 0.35 = 0.65

t = 7.8 min

$\frac {[A_t]}{[A_0]}=e^{-k\times t}$

$0.65=e^{-0.02616\times t}$

<u>t = 16.5 days.</u>

7 0

3 years ago

From these two reactions at 298 K, V2O3(s) + 3CO(g) → 2V(s) + 3CO2(g); ΔH° = 369.8 kJ; ΔS° = 8.3 J/K V2O5(s) + 2CO(g) → V2O3(s)

gregori [183]

Answer:

ΔG° = -133,1 kJ

Explanation:

For the reactions:

<em>(1) </em>V₂O₃(s) + 3CO(g) → 2V(s) + 3CO₂(g); ΔH° = 369,8 kJ; ΔS° = 8,3 J/K

<em>(2) </em>V₂O₅(s) + 2CO(g) → V₂O₃(s) + 2CO₂(g); ΔH° = –234,2 kJ; ΔS° = 0,2 J/K

By Hess's law it is possible to obtain the ΔH° and ΔS° of:

2V(s) + 5CO₂(g) → V₂O₅(s) + 5CO(g)

Substracting -(1)-(2), that means:

ΔH° = -369,8 kJ - (-234,2 kJ) = <em>-135,6 kJ</em>

ΔS° = - 8,3 J/K - 0,2 J/K =<em> -8,5 J/K</em>

Using: ΔG° = ΔH° - TΔS° at 298K

ΔG° = -135,6 kJ - 298K×-8,5x10⁻³kJ/K

<em>ΔG° = -133,1 kJ</em>

I hope it helps!

7 0

4 years ago

his chemical equation represents the burning of methane, but the equation is incomplete. What is the missing coefficient in both

Aleksandr [31]

It’s Supposed to be CH4 + 2O2 > CO2 + 2H2O

7 0

3 years ago