A rock sample has a mass of 125 grams and a volume of 50 mL. What is the density of this rock sample?

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

(04.02 LC) Based on the cell theory, which of the following is true?

prohojiy [21]

Answer:

c

Explanation:

there are billions of cells in living things. We are made out of millions of cells

4 0

3 years ago

Conversion factors are useful in solving problems in which a given measurement must be expressed in

spin [16.1K]

Conversion factors are useful in solving problems in which a given measurement must be expressed in some other units of measure.

Conversion factors are helpful in solving issues where one unit of measurement needs to be expressed in another. In general, a measurement's numerical value changes when it is multiplied by a conversion factor, while the quantity being measured's real size stays the same.
A conversion factor is a number that is used to multiply or divide one set of units into another. If a conversion is required, it must be done using the correct conversion factor to get an identical value. For instance, 12 inches equals one foot when converting between inches and feet.

To learn more about Conversion factors please visit -brainly.com/question/23510660
#SPJ1

4 0

2 years ago

Which of the graphs below might represent a mixture of pure water and ice exposed to a room temperature of 3°C?

Y_Kistochka [10]

Answer:

C. Graph C

Explanation:

We have a mixture of water and ice.

At 0 °C they are at equilibrium.

water-to-ice rate = ice-to-water rate

Next, we lower the temperature to -3 °C — just slightly below freezing.

The water will slowly turn to ice.

The water-to-ice rate will be slightly faster than the ice-to-water rate.

The purple bar will be slightly higher than the blue bar.

Graph C best represents the relative rates

A. is wrong. The ice-to-water rate is faster, so the water is melting. The temperature is slightly above freezing (say, 3 °C).

B. is wrong. The two rates are equal, so the temperature is 0 °C.

D. is wrong. The water-to-ice rate (freezing) is much greater than the ice-to-water rate, so the temperature is well below freezing( say, -10 °C).

6 0

4 years ago

A 2.300×10−2 m solution of nacl in water is at 20.0∘c. the sample was created by dissolving a sample of nacl in water and then b

lilavasa [31]

The given question is incomplete. The complete question is as follows.

A 2.300×10−2 m solution of nacl in water is at 20.0∘c. the sample was created by dissolving a sample of nacl in water and then bringing the volume up to 1.000 l. it was determined that the volume of water needed to do this was 999.4 ml . the density of water at 20.0∘c is 0.9982 g/ml.

Calculate the molality of the salt solution.

Express your answer to four significant figures and include the appropriate units.

Explanation:

Molality is defined as the number of moles present in kg of a solvent.

Mathematically, Molality = $\frac{\text{moles of solute}}{\text{mass of solvent}}$

Also,

Mole of solute = Molarity of solute x Volume of solution

= (0.0230 M) x (1.000 L) = 0.0230 mol of solute

Therefore, mass of solvent will be as follows.

$999.4 mL \times (\frac{0.9983 g}{1 mL})$

= 997.7 g

= 0.9977 kg (as 1 kg = 1000 g)

Therefore, we will calculate the molality as follows.

Molality = $\frac{0.0230 mol}{0.9977 kg}$

= 0.02306 mol/kg

thus, we can conclude that molality of the given solution is 0.02306 mol/kg.

8 0

3 years ago