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Artyom0805 [142]
3 years ago
8

Question 4 of 5

Chemistry
2 answers:
Leokris [45]3 years ago
3 0
Should be rabbit and owl
Marysya12 [62]3 years ago
3 0

Answer:

Rabbit and Mouse

Explanation:

The rabbit and mouse are the prey of foxes.

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Studying the decay of radioactive isotopes in dead organisms helps scientists to identify fossilized remains. The ratio of C-12
satela [25.4K]

Answer:

Only Organism Z is alive, and Organism X has been dead longer than Organism Y.

Explanation:

After the death of an organism, there is radioactive disintegration of C-14 allotrope of carbon, which increases the ratio of C-12 to C-14 in a dead organism as compared to a living organism.

7 0
3 years ago
The equilibrium constant Kp for the reaction I2(g) + Br2(g) ⇀↽ 2 IBr(g) + 11.7 kJ is 280 at 150◦C. Suppose that a quantity of IB
mixas84 [53]

Answer:

\large \boxed{\text{0.0120 atm }}

Explanation:

The balanced equation is

I₂(g) + Br₂(g) ⇌ 2IBr(g)

Data:

      Kp = 280

p(IBr) = 0.200 atm

1. Set up an ICE table.

Let p = the initial pressure of IBr. Then

\begin{array}{ccccccc}\rm \text{I}_{2}& + & \text{Br}_{2} & \, \rightleftharpoons \, & \text{2IBr} &  &  \\0 & & 0 & & p & & \\+x &   & +x  & & -2x &   &\\x &   & x} &   & 280 & & \\\end{array}

2. Calculate p(I₂)

\begin{array}{rcl}K_{\text{p}}&=&\dfrac{p_{\text{IBr}}^{2}} {p_{\text{I}_2}^{2}}\\\\280&=&{\dfrac{0.200^{2}}{x^{2}}&&\\\\280x^{2} & = &0.0400\\x^{2} & = &\dfrac{0.0400}{280 }\\\\& = & 1.429 \times 10^{-4}\\x & = & \textbf{0.0120 atm}\\\end{array}\\\text{The partial pressure of iodine is $\large \boxed{\textbf{0.0120 atm }}$}}

Check:

\begin{array}{rcl}{\dfrac{0.200^{2}}{0.0120^{2}}}&=&280\\\\280& =& 280\\\end{array}

8 0
3 years ago
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