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poizon [28]
3 years ago
5

A(g) + 2B(g) → C(g) + D(g)

Chemistry
1 answer:
WITCHER [35]3 years ago
6 0

Answer:

0.169

Explanation:

Let's consider the following reaction.

A(g) + 2B(g) ⇄ C(g) + D(g)

We can find the pressures at equilibrium using an ICE chart.

       A(g) + 2 B(g) ⇄ C(g) + D(g)

I       1.00     1.00        0        0

C       -x        -2x        +x       +x

E    1.00-x  1.00-2x     x         x

The pressure at equilibrium of C is 0.211 atm, so x = 0.211.

The pressures at equilibrium are:

pA = 1.00-x = 1.00-0.211 = 0.789 atm

pB = 1.00-2x = 1.00-2(0.211) = 0.578 atm

pC = x = 0.211 atm

pD = x = 0.211 atm

The pressure equilibrium constant (Kp) is:

Kp = pC × pD / pA × pB²

Kp = 0.211 × 0.211 / 0.789 × 0.578²

Kp = 0.169

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The untrue statement is that they high melting points.  

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When using science to investigate physical phenomena, which characteristic of the event must exist?
Maru [420]
 "Observable" is the one characteristic of event that must exist when <span>using science to investigate physical phenomena. </span>
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3 years ago
When making an observation you should provide a general description of the subject rather than going into too much detail true o
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3 years ago
A solution containing 60 grams of nano3 completely dissolved in 50. Grams of water at 50°c is classified as being
Igoryamba

Answer:

<em>A solution containing 60 grams of nano3 completely dissolved in 50. Grams of water at 50°c is classified as being</em> <u>supersaturaded</u>

Explanation:

This question is about solubility.

Regarding solubility, the solutions may be classified as:

  • Unsaturated: the concentration is below the maximum concentration permited at the given temperature.

  • Saturated: the concentration is the maximum permitted at the given temperature, under normal conditions.

  • Supersaturated: the concentration has overcome the maximum permitted at the given temperature. This is possible only under special conditions and is a very unstable state.

Each substance has its own, unique solubility properties. So, in order to tell the state of the solution you need to compare with either solubility tables, or solubility curves; or run you own experiments.

  • In internet you can find the solubility curve of NaNO₃ showing the solubility for a wide range of temperatures.

  • In such curve the solubility of NaNO₃ at 50°C is about 115 g of NaNO₃ per 100 g  of water.

  • Hence, do the proportion to determine the amount of solute that can be dissolved in 50 grams of water at 50°CÑ

       115 g NaNO₃ / 100 g H₂O = x / 50 g H₂O  ⇒ x =  57.5 g NaNO₃

  • <u>Conclusion</u>: 50 grams of water can contain 57.5 g of NaNO₃ dissolved; so, <em>a solution containing 60 g of NaNO₃ completely dissolved in 50 grams of water is supersaturated.</em>

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3 0
3 years ago
The density of concentrated ammonia, which is 28.0% w/w nh3, is 0.899 g/ml. what volume of this reagent should be diluted to 1.0
vlada-n [284]

Answer: 2.4 ml

Solution :

Molar mass of NH_3 = 17 g/mole

Given,: 28% w/w of NH_3 solution means 28 g of ammonia in 100 g of solution.

Mass of solution = 100 g

Now we have to calculate the volume of solution.

Volume=\frac{Mass}{Density}=\frac{100g}{0.899g/ml}=111.2ml  

Molarity : It is defined as the number of moles of solute present in one liter of solution.

Molarity=\frac{n\times 1000}{V_s}

where,

n = moles of solute NH_3=\frac{\text {given mass}}{\text {molar mass}}=\frac{28}{17}=1.65moles

V_s = volume of solution in liter = 0.11 L

Now put all the given values in the formula of molarity, we get

Molarity=\frac{1.65moles}{0.11L}=15mole/L

Using molarity equation:

M_1V_1=M_2V_2

15\times V_1=0.036\times 1.0\times 10^{3}

V_1=2.4ml

6 0
3 years ago
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