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3 years ago

11

find the weight of the same object on a planet where the gravitational attraction has been reduced to 1/10 of the earth pull sho

w all work

1 answer:

nataly862011 [7]3 years ago

6 0

Answer:

9.9Xkgms-¹

Explanation:

Let mass of object be xkg

Gravitational pull is 10m/s-1/10=99/10=9.9

9.9xkgms-¹

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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:

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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:

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

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