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galina1969 [7]
3 years ago
12

Hydrogen iodide can decompose into hydrogen and iodine gases. 2 HI(g) H2(g) + 12(g) Kp for the reaction is 0.016. If 0.350 atm o

f HI(g) is sealed in a flask, what is the total pressure of the system when equilibrium is established? a. 0.385 atm b. 0.350 atm c. 0.279 atm d. 0.258 atm e. 0.412 atm
Chemistry
1 answer:
solniwko [45]3 years ago
4 0

Answer : The total pressure at equilibrium is 0.350 atm

Solution :  Given,

Initial pressure of HI = 0.350 bar

K_p = 0.016

The given equilibrium reaction is,

                             2HI(g)\rightleftharpoons H_2(g)+I_2(g)

Initially                  0.350           0       0

At equilibrium       (0.350-2x)    x       x

The total pressure at equilibrium = (0.350-2x)+x+x=0.350-2x+2x=0.350atm

Thus, the total pressure at equilibrium is 0.350 atm

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Name the two possible products in the precipitation reaction of copper (II) chloride and sodium phosphate. Use the charges on th
satela [25.4K]

Answer:

General equation for a double-displacement reaction:  

AB + CD --> AC + BD

• sodium chloride – NaCl copper sulfate – CuSO₄  

NaCl + CuSO₄ --> Na₂SO₄ + CuCl₂

The products formed are sodium sulfate and copper (II) chloride.

Copper (II) chloride forms a blue colored solution.

• sodium hydroxide – NaOH copper sulfate – CuSO₄  

NaOH + CuSO₄ --> Na₂SO₄ + Cu(OH)₂

The products formed are sodium sulfate and copper (II) hydroxide.

Copper (II) hydroxide forms a blue colored solution.

• sodium phosphate – Na₂HPO₂ copper sulfate – CuSO₄  

Na₂HPO₄ + CuSO₄ --> Na₂SO₄ + CuHPO₄

The products formed are sodium sulfate and copper (II) hydrogen phosphate.

Copper (II) hydrogen phosphate forms a blue colored solution.

• sodium chloride – NaCl silver nitrate – AgNO₃  

NaCl + AgNO₃--> AgCl + NaNO₃

The products formed are silver chloride and sodium nitrate.

Silver chloride forms a white precipitate.

• sodium hydroxide – NaOH silver nitrate – AgNO₃  

NaOH + AgNO₃   --> NaNO₃ + AgOH

The products formed are silver hydroxide and sodium nitrate.

Silver hydroxide forms a white precipitate.

• sodium phosphate – Na₂HPO₄ silver nitrate – AgNO₃

Na₂HPO₄ + AgNO₃  --> NaNO₃ +  Ag₂HPO₄

The products formed are sodium nitrate and silver hydrogen phosphate.

Silver hydrogen phosphate forms a colorless solution.

Explanation:

5 0
3 years ago
How many moles are in 2.98x10^23 particles?
Novosadov [1.4K]

Answer:

\boxed {\boxed {\sf 0.495 \ mol}}

Explanation:

We are given a number of particles and asked to convert to moles.

<h3>1. Convert Particles to Moles </h3>

1 mole of any substance contains the same number of particles (atoms, molecules, formula units) : 6.022 *10²³ or Avogadro's Number. For this question, the particles are not specified.

So, we know that 1 mole of this substance contains 6.022 *10²³ particles. Let's set up a ratio.

\frac { 1 \ mol }{6.022*10^{23 } \ particles}}

We are converting 2.98*10²³ particles to moles, so we multiply the ratio by that value.

2.98*10^{23} \ particles *\frac { 1 \ mol }{6.022*10^{23 } \ particles}}

The units of particles cancel.

2.98*10^{23}  *\frac { 1 \ mol }{6.022*10^{23 } }}

\frac { 2.98*10^{23}}{6.022*10^{23 } }}  \ mol

0.4948522086 \ mol

<h3>2. Round</h3>

The original measurement of particles (2.98*10²³) has 3 significant figures, so our answer must have the same.

For the number we found, 3 sig figs is the thousandth place.

The 8 in the ten-thousandth place (0.4948522086) tells us to round the 4 up to a 5 in the thousandth place.

0.495 \ mol

2.98*10²³ particles are equal to approximately <u>0.495 moles.</u>

3 0
3 years ago
Determine the boiling point of water at 620 mm Hg?
AfilCa [17]
For example, at sea level the atmospheric pressure is 760 mm Hg<span> (also expressed as 760 torr, 101325 Pa, 101.3 kPa, 1013.25 mbar or 14.696 psi) and pure </span>water<span> boils at 100°C. However, in Calgary (approx. 1050m above sea level) the atmospheric pressure is approximately 670 </span>mm Hg<span>, and </span>water<span> boils at about 96.6°C.</span>
7 0
3 years ago
Read 2 more answers
Need help asap
zzz [600]
Question 4 is C. 13 I believe
5 0
3 years ago
How many moles are in 18.3 grams of Carbon? ​
Vadim26 [7]

Answer:

1.5236414197340638

7 0
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