Molar mass KI = 166.0028 g/mol and Pbl₂ = 461.0089 g/mol

2 Kl + Pb(NO₃)₂ ------> 2 KNO₃ + Pbl₂

2 x 166.0028 g ---------> 461.0089 g
? g KI -------------> 1.2 g

Mass KI = 1.2 x 2 x 166.0028 / 461.0089

Mass KI = 398.40672 / 461.0089

<span>Mass KI = </span>0.86 g

Answer B

<span>hope this helps!</span>

6 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

Which of the following graphs correctly represents the relationship between the temperature and the volume of an ideal gas that

Anon25 [30]

Answer:

The graph of the relationship of temperature one volume is a graphical representation of Charles law.

Explanation:

The graph shows the relationship between volume vs temperature plotted at constant pressure for a fixed amount of gas. As can be observed from the graph, the volume increases with an increase in the temperature, and vice versa. Thus, volume is directly proportional to temperature at a constant pressure, which is the statement of Charles's law.

Volume is plotted on the y- axis, and temperature is on x-axis. The graph is a straight line with a positive slope passing the origin. The equation of the line is V = kT, which is the equation of Charles's law. The slope of the line is k. As temperature approaches zero kelvin, volume also approaches zero.

Real gases do not obey Charles's law at low temperatures. As temperature approaches absolute zero (0 K), the real gases start deviating significantly from Charles's law.

4 0

3 years ago