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ArbitrLikvidat [17]

2 years ago

10

A gas mixture contains helium (He ), argon (Ar) , neon (Ne ), Xenon ( Xe), and krypton (Kr) gases. The total pressure of the gas

mixture is 1252.5 mm Hg. The partial pressures of He, Ar , Ne, and Xe are 0.32, 0.21, 0.44, and 0.19 atm, respectively. What is the partial pressure exerted by Kr?

1 answer:

MA_775_DIABLO [31]2 years ago

5 0

Answer:

0.488atm = Pressure Kr

Explanation:

The total pressure in a mixture of gases could be defined as the sum of the partial pressures of a mixture. For the mixture of gases in the problem:

Total pressure = Pressure He + Pressure Ar + Pressure Ne + Pressure Xe + Pressure Kr

Converting the total pressure to atm:

1252.5mm Hg * (1atm / 760 mmHg) = 1.648 atm

Replacing:

1.648atm = 0.32atm + 0.21atm + 0.44atm + 0.19atm + Pressure Kr

<h3>0.488atm = Pressure Kr</h3>

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Shkiper50 [21]

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6 0

3 years ago

Read 2 more answers

In an experiment, 2.54 grams of copper completely reacts with sulfur, producing 3.18 grams of copper(I) sulfide.

kozerog [31]

Answer:

0.64 g of S

Solution:

The balance chemical equation is as follow,

2 Cu + S ----> Cu₂S

According to equation,

127 g (2 mole) Cu produces = 159 g (1 mole) of Cu₂S

So,

2.54 g Cu will produce = X g of Cu₂S

Solving for X,

X = (2.54 g * 159 g) / 127 g

X = 3.18 g of Cu₂S

Now, it is confirmed that the reaction is 100% ideal. Therefore,

As,

127 g (2 mole) Cu required = 32 g (1 mole) of S

So,

2.54 g Cu will require = X g of S

Solving for X,

X = (2.54 g * 32 g) / 127 g

X = 0.64 g of S

5 0

2 years ago

Read 2 more answers

PLEASE HELP this is physical science.

Andru [333]

Bone age : 22,920 years

<h3>Further explanation</h3>

Given

Nt = 2.5 g C-14

No = 40 g

half-life = 5730 years

Required

time of decay

Solution

General formulas used in decay:

$\large{\boxed{\bold{N_t=N_0(\dfrac{1}{2})^{t/t\frac{1}{2} }}}$

t = duration of decay

t 1/2 = half-life

N₀ = the number of initial radioactive atoms

Nt = the number of radioactive atoms left after decaying during T time

Input the value :

$\tt 2.5=40.\dfrac{1}{2}^{t/5730}\\\\\dfrac{2.5}{40}=\dfrac{1}{2}^{t/5730}\\\\(\dfrac{1}{2})^4=\dfrac{1}{2}^{t/5730}\\\\4=t/5730\rightarrow t=22920~years$

6 0

2 years ago

A gas has a volume of 590 mL at temperature of -55.0 C. What volume will the gas occupy at 30.0 C show your work

DENIUS [597]

Data:
$V_{initial} = 590\:mL$
$T_{initial} = -55.0^0C$
converting to Kelvin
TK = TC + 273
TK = -55.0 + 273 → TK = 218.0 → $T_{initial} = 218.0\:K$
$V_{final} = ? (in\:milliliters)$
$T_{final} = 30.0^0C$
TK = TC + 273
TK = 30.0 + 273 → TK = 303.0 → $T_{final} = 303.0\:K$

By the first Law of Charles and Gay-Lussac, we have:
$\frac{ V_{i} }{ T_{i} } = \frac{ V_{f} }{ T_{f} }$

Solving:
$\frac{ V_{i} }{ T_{i} } = \frac{ V_{f} }{ T_{f} }$
$\frac{ 590 }{ 218.0 } = \frac{ V_{f} }{ 303.0 }$
Product of extremes equals product of means:
$218.0* V_{f} = 590*303.0$
$218.0 V_{f} = 178770$
$V_{f} = \frac{178770}{218.0}$
$\boxed{\boxed{V_{f} \approx 820.04\:mL}}\end{array}}\qquad\quad\checkmark$

7 0

2 years ago

A sample of CH4 is confined in a water manometer. The temperature of the system is 30.0 °C and the atmospheric pressure is 98.70

kakasveta [241]

Explanation:

The given data is as follows.

$P_{atm}$ = 98.70 kPa = 98700 Pa,

T = $30^{o}C$ = (30 + 273) K = 303 K

height (h) = 30 mm = 0.03 m (as 1 m = 100 mm)

Density = 13.534 g/mL = $13.534 g/mL \times \frac{10^{6}cm^{3}}{1 m^{3}} \times \frac{1 kg}{1000 g}$

= 13534 $kg/m^{3}$

The relation between pressure and atmospheric pressure is as follows.

P = $P_{atm} + \rho gh$

Putting the given values into the above formula as follows.

P = $P_{atm} + \rho gh$

= $98700 Pa + 13534 \times 9.81 \times 0.03 m$

= 102683.05 Pa

= 102.68 kPa

thus, we can conclude that the pressure of the given methane gas is 102.68 kPa.

8 0

3 years ago