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liraira [26]
3 years ago
10

The equilibrium constant for the following reaction: H2(g) + Br2(g) ↔ 2HBr (g) is 1.10 x 103 at a certain temperature. Find the

equilibrium pressure of HBr if 7.40 atm of HBr is introduced into a sealed container at this temperature.
Chemistry
1 answer:
mihalych1998 [28]3 years ago
7 0

<u>Answer:</u> The partial pressure of HBr at equilibrium is 6.98 atm

<u>Explanation:</u>

We are given:

Initial partial pressure of HBr = 7.40 atm

As, initially HBr is present. So, the reaction is proceeding backwards.

For the given chemical equation:

                     H_2(g)+Br_2(g)\rightarrow 2HBr(g)

<u>Initial:</u>                                             7.40

<u>At eqllm:</u>          x            x          7.40-2x

The expression of K_p for above equation follows:

K_p=\frac{(p_{HBr})^2}{p_{H_2}\times p_{Br_2}}

We are given:

K_p=1.10\times 10^3

Putting values in above expression, we get:

1.10\times 10^3=\frac{(7.40-2x)^2}{x\times x}\\\\x=-0.237,0.210

Neglecting the negative value of 'x' because partial pressure cannot be negative.

So, equilibrium partial pressure of HBr = (7.40 - 2x) = [7.40 - 2(0.210)] = 6.98 atm

Hence, the partial pressure of HBr at equilibrium is 6.98 atm

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A sample of O2 gas occupies a volume of 571 mL at 26 ºC. If pressure remains constant, what would be the new volume if the tempe
Vlad1618 [11]

Answer: The new volume at different given temperatures are as follows.

(a) 109.81 mL

(b) 768.65 mL

(c) 18052.38 mL

Explanation:

Given: V_{1} = 571 mL,       T_{1} = 26^{o}C

(a) T_{2} = 5^{o}C

The new volume is calculated as follows.

\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\\frac{571 mL}{26^{o}C} = \frac{V_{2}}{5^{o}C}\\V_{2} = 109.81 mL

(b) T_{2} = 95^{o}F

Convert degree Fahrenheit into degree Cesius as follows.

(1^{o}F - 32) \times \frac{5}{9} = ^{o}C\\(95^{o}F - 32) \times \frac{5}{9} = 35^{o}C

The new volume is calculated as follows.

\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\\frac{571 mL}{26^{o}C} = \frac{V_{2}}{35^{o}C}\\V_{2} = 768.65 mL

(c) T_{2} = 1095 K = (1095 - 273)^{o}C = 822^{o}C

The new volume is calculated as follows.

\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\\frac{571 mL}{26^{o}C} = \frac{V_{2}}{822^{o}C}\\V_{2} = 18052.38 mL

8 0
2 years ago
According to the website of the National Aeronautics and Space Administration (NASA), the average temperature of the universe is
charle [14.2K]

Answer : The temperature in degree Celsius is, -270.45^oC

Explanation :

The conversion used for the temperature from Kelvin to degree Celsius is:

^oC=K-273.15

where,

K = temperature in Kelvin

^oC = temperature in centigrade

As we are given the temperature in Kelvin is, 2.7

Now we have to determine the temperature in Kelvin.

^oC=K-273.15

^oC=2.7-273.15

^oC=-270.45

Therefore, the temperature in degree Celsius is, -270.45^oC

3 0
3 years ago
determine mass of water formed when 12.5 L NH3(at298K and 1.50atm) is reacted with 18.9L of O2 (at 323K and 1.1atm)
sasho [114]

The  mass  of water formed  is


<u><em>calculation</em></u>

Use  the  ideal   gas  equation   to  calculate the  moles of  NH3  and O2

that  is  Pv= n RT

where;  P= pressure,  

V=  volume,

n = number  of  moles,

R=gas   constant  = 0.0821  l .atm/ mol.K

make n the formula of  the subject  by diving   both side  by  RT

n =  PV /RT

The   moles of NH3

n= (1.50 atm  x 12.5 L) /(  0.0821 L. atm /mol.k   x 298 K)  =0.766  moles

The  moles  of  O2

=(1.1 atm  x 18.9  L) /  (  0.0821 L. atm/ mol.k   x 323 K) = 0.784  moles


write the reaction  between  NH3  and  O2

4 NH3  + 5 O2  →4 No  +6H2O


from  equation above  0.766  moles of NH3  reacted to produce  

0.766 x 6/4 =1.149 moles of H2O


0.784  moles of O2   reacted to  produce  0.784  x 6/5=0.9408  moles  of H20


since  O2  is totally  consumed, O2  is the limiting  reagent  and therefore  the  moles of H2O  produced=  0.9408  moles


mass  of  H2O  = moles x molar mass

 from  periodic table the  molar mass  of H2O  =  (1 x2)+16= 18  g/mol

mass = 18 g/mol  x 0.9408  moles= 16.93  grams


3 0
3 years ago
Which characteristic of water helps it form droplets?
inessss [21]

Answer: C. High surface tension

Explanation:

Water has high specific heat as it require high heat to raise the temperature of 1 g of water through 1^0C.

Surface tension is the net downward force acting on the surface of liquids due to the cohesive nature of liquids.  

Water molecules are bonded by strong hydrogen bonding between the hydrogen atom and the electronegative oxygen atom making it polar. Thus water molecules present on the surface are strongly attracted by the molecules present below the surface and thus act as a stretched membrane.  

The surface acquires a minimum surface are and thus acquire a spherical shape.

3 0
3 years ago
What are the units of molality? A. mol/kg B. g/mol C. mol/L D. kg/mol
Serhud [2]

Answer:  A.mol/kg

The SI (international system) unit for molality is mol / kg, or solute moles per kg of peptides. A solution with a molality of 1 mol / kg is often described as "1 molal" or "1 m".

5 0
3 years ago
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