A 3.0 L flask containing helium at 145 mmHg is connected by a closed valve to a 2.0 L flask containing argon at 355 mmHg. When the valve is opened and the gases are allowed to mix equally in the two flasks, what is the total pressure (in mmHg) in the two connected flasks after mixing ?

Answers

Answer 1

Answer:

Answer:

Assuming that both helium and argon act like ideal gases, the total pressure after mixing would be approximately $229\; \rm mmHg$ .

Explanation:

By the ideal gas equation, $P\cdot V = n \cdot R \cdot T$ , where

$P$ is the pressure of the sample.
$V$ is the volume of the container.

$n$ is the number of moles of gas particles in the sample.
$R$ is the ideal gas constant.
$T$ is the temperature of the sample.

Rewrite to obtain:

$\displaystyle n = (P \cdot V)/(R\cdot T)$ , and
$\displaystyle P = (n \cdot R \cdot T)/(V)$ .

Assume that the two samples have the same temperature, $T$ . Also, assume that mixing the two gases did not affect the temperature.

Apply the equation $\displaystyle n = (P \cdot V)/(R\cdot T)$ to find the number of moles of gas particles in each container:

In the helium container, $V = 3.0\; \rm L$ and $P = \rm 145\; mmHg$ . Hence, $\displaystyle n_1 = (P\cdot V)/(R \cdot T) = \frac{(3.0\; \text{L}) \cdot (145\; \text{mmHg})}{R\cdot T}$ .
In the argon container, $V = 2.0\; \rm L$ and $P = 355\; \rm mmHg$ . Hence, $\displaystyle n_2 = (P\cdot V)/(R \cdot T) = \frac{(2.0\; \text{L}) \cdot (355\; \text{mmHg})}{R\cdot T}$ .

After mixing, $V = 2.0 + 3.0 = 5.0\; \rm L$ . Assuming that temperature $T$ stays the same.

$\displaystyle n_1 + n_2 = \frac{(3.0\; \text{L}) \cdot (145\; \text{mmHg})}{R\cdot T} + \frac{(2.0\; \text{L}) \cdot (355\; \text{mmHg})}{R\cdot T}$ .

Apply the equation $\displaystyle P = (n \cdot R \cdot T)/(V)$ to find the pressure after mixing.

$\begin{aligned}P &= \displaystyle \frac{\displaystyle \displaystyle \left(\frac{(3.0\; \text{L}) \cdot (145\; \text{mmHg})}{R\cdot T} + \frac{(2.0\; \text{L}) \cdot (355\; \text{mmHg})}{R\cdot T}\right) \cdot R \cdot T}{5.0\; \rm L} \n &= (3.0\; \rm L * 145\; \rm mmHg + 2.0\; \rm L * 355\; \rm mmHg)/(5.0\; \rm L) \n &\approx 229\; \rm mmHg\end{aligned}$ .

Answer 2

Answer:

Answer:

The total pressure is 229 atm

Explanation:

Step 1: Data given

Volume of helium flask = 3.0 L

Pressure helium flask = 145 mm Hg

Volume of argon flask = 2.0 L

Pressure argon flask = 355 mmHg

total volume = 5.0 L

Step 2: Partial pressure helium

pHe = 145 *(3/5) = 87.0 atm

Step 3: Calculate pressure argon

pAr = 355*(2/5) = 142.0 atm

Step 4: Calculate total pressure

Total pressure = 87.0 + 142.0 atm

Total pressure = 229 atm

The total pressure is 229 atm

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Conduct metric Titration of H_2(SO_4) and Ba(OH)_2 Write an equation (including states of matter) for the reaction between H_2(SO_4) and Ba(OH)_2 At the very start of the titration, before any titrant has been added to the beaker, what is present in the solution? What is the conducting species in this initial solution? Describe what happens as titrant is added to the beaker. Why does the conductivity of the solution decrease? What is the identity of the solid formed? What is the conducting species present in the beaker? What happens when the conductivity value reaches its minimum value (which is designated as the equivalence point for this type of titration)? What is the conducting species in the beaker? Describe what happens at additional titrant is added past the equivalence point. Why does the conductivity of the solution increase? What is the conducting species present in the beaker?

Answers

Answer:

a) H₂SO₄ + Ba(OH)₂ ⇄ BaSO₄(s) + 2 H₂O(l)

b) H₂SO₄, H⁺, HSO₄⁻, SO₄²⁻. H₂O, H⁺, OH⁻.

c) H⁺, HSO₄⁻, SO₄²⁻

d) As the titration takes place, reaction [1] proceeds to the right. The conductivity of the solution decreases because the amount of H⁺, HSO₄⁻, SO₄²⁻ decreases. The formed solid is barium sulfate BaSO₄. Since BaSO₄ is very insoluble, the main responsible for conductivity are still H⁺, HSO₄⁻ and SO₄²⁻,

e) At the equivalence point equivalent amounts of H₂SO₄ and Ba(OH)₂ react. The conducting species are Ba²⁺, SO₄²⁻, H⁺ and OH⁻.

f) After the equivalence point there is an excess of Ba(OH)₂. The ions Ba²⁺ and OH⁻ are responsible for the increase in the conductivity, being the major conducting species.

Explanation:

a) Write an equation (including states of matter) for the reaction between H₂SO₄ and Ba(OH)₂.

The balanced equation is:

H₂SO₄ + Ba(OH)₂ ⇄ BaSO₄(s) + 2 H₂O(l) [1]

b) At the very start of the titration, before any titrant has been added to the beaker, what is present in the solution?

In the beginning there is H₂SO₄ and the ions that come from its dissociation reactions: H⁺, HSO₄⁻, SO₄²⁻. There is also H₂O and a very small amount of H⁺ and OH⁻ coming from its ionization.

H₂SO₄(aq) ⇄ H⁺(aq) + HSO₄⁻(aq)

HSO₄⁻(aq) ⇄ H⁺(aq) + SO₄²⁻(aq)

H₂O(l) ⇄ H⁺(aq) + OH⁻(aq)

c) What is the conducting species in this initial solution?

The main responsible for conductivity are the ions coming from H₂SO₄: H⁺, HSO₄⁻, SO₄²⁻.

d) Describe what happens as titrant is added to the beaker. Why does the conductivity of the solution decrease? What is the identity of the solid formed? What is the conducting species present in the beaker?

As the titration takes place, reaction [1] proceeds to the right. The conductivity of the solution decreases because the amount of H⁺, HSO₄⁻, SO₄²⁻ decreases. The formed solid is barium sulfate BaSO₄. Since BaSO₄ is very insoluble, the main responsible for conductivity are still H⁺, HSO₄⁻ and SO₄²⁻,

e) What happens when the conductivity value reaches its minimum value (which is designated as the equivalence point for this type of titration)? What is the conducting species in the beaker?

At the equivalence point equivalent amounts of H₂SO₄ and Ba(OH)₂ react. Only BaSO₄ and H₂O are present, and since they are weak electrolytes, there is a small amount of ions to conduct electricity. The conducting species are Ba²⁺ and SO₄²⁻ coming from BaSO₄ and H⁺ and OH⁻ coming from H₂O.

f) Describe what happens at additional titrant is added past the equivalence point. Why does the conductivity of the solution increase? What is the conducting species present in the beaker?

After the equivalence point there is an excess of Ba(OH)₂. The ions Ba²⁺ and OH⁻ are responsible for the increase in the conductivity, being the major conducting species.

Final answer:

The chemical reaction between H2SO4 and Ba(OH)2 forms BaSO4 and water, reducing conductivity by reducing the number of free ions. Beyond the equivalence point, the conductivity increases due to the dissociated ions from the excess Ba(OH)2 in the solution.

Explanation:

Chemical Reaction and Metric Titration

Firstly, the equation representing the reaction between sulfuric acid (H2SO4) and barium hydroxide (Ba(OH)2) is:

Ba(OH)2 (aq) + H2SO4 (aq) → BaSO4 (s) + 2H2O (l)

In the beginning, the solution only contains H2SO4 with its dissociated ions serving as the conducting species. As titrant (Ba(OH)2) is added, they react to form BaSO4, a solid precipitate reducing the number of free ions in the solution, thus decreasing conductivity. At the equivalence point, all H2SO4 has reacted, and conductivity reaches its minimum as there are lesser free ions for conduction. If additional titrant is added past the equivalence point, conductivity increases due to excess Ba(OH)2's dissociated ions that increase ion concentration in solution.

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A car travels for 0.5 hours and goes 25 miles. What is the car's speed? (Remember sad + t)

Answers

Answer:

s = 22.352 m/s

Explanation:

Given data:

Time taken = 0.5 hours

Distance cover = 25 miles

Car speed = ?

Solution:

First of all we will convert the units.

1 hour = 3600 sec

0.5 hr ×3600 sec / 1hr = 1800 sec

miles to meter:

25 mi × 1609 m/ 1 mi = 40233.6 m

Formula:

s = d/t

s = speed

d = distance

t = time

Now we will put the values in formula.

s = 40233.6 m / 1800 sec

s = 22.352 m/s

Final answer:

The speed of the car, calculated by dividing the distance travelled (25 miles) by the time taken (0.5 hours), is 50 miles per hour.

Explanation:

The subject of the question falls under Mathematics, specifically a section of it named rate, time, and distance problems. The problem is asking us to calculate the speed of a car which can be obtained by dividing the distance travelled by the time taken. Given that the distance travelled by the car is 25 miles and the time taken is 0.5 hours, we can calculate the speed as follows.

Denote the Speed as S, the distance as D, and the time as T.
Since Speed = Distance/Time, we substitute our values into this equation
Therefore, Speed = 25 miles / 0.5 hour = 50 miles/hour

So, the speed of the car is 50 miles per hour.

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Please helpplease please..help

Answers

Answer:

vague symptoms are characteristic of an acute toxin, because of the of the lack of well defined consistency that these symptoms have in relation to the course of the disease progress.

Iridium-192 is used in medicine to treat prostate cancer. Iridium-192 has two modes of radioactive decay: 96% of the time it decays by beta emission and 4% of the time it decays by electron capture. What are the daughter nuclides of these two decay processes?

Answers

Answer:

The daughter nuclides of these two decay processes are $^(192)_(78)Pt$ and $^(192)_(76)Os$ .

Explanation:

The beta emission is represented by:

A = (Z + 1) + (n - 1) = is invariant

n: neutron

p: proton

$^(A)_(Z)X \rightarrow ^(A)_(Z+1)Y$

Hence, the daughter nuclide of the beta emission of Ir-192 is:

$^(192)_(77)Ir \rightarrow ^(192)_(78)Pt$

Now, electron capture is represented by:

A = (Z - 1) + (n + 1) = is invariant

$^(A)_(Z)X \rightarrow ^(A)_(Z-1)Y$

Then, the daughter nuclide of the electron capture of Ir-192 is:

$^(192)_(77)Ir \rightarrow ^(192)_(76)Os$

Therefore, the daughter nuclides of these two decay processes are $^(192)_(78)Pt$ and $^(192)_(76)Os$ .

I hope it helps you!

What is the mass of ammonium nitrate in 250 mL of a 75% by mass solution (density = 1.725 g/mL)?

Answers

Explanation:

Density is the mass present in per unit volume.

Mathematically, Density = $(mass)/(volume)$

Therefore, first calculate the mass of solution as follows.

Density = $(mass)/(volume)$

1.725 g/mL = $(mass)/(250 mL)$

mass = 431.25 g

Now, calculate mass of ammonium nitrate as follows.

Percentage by mass = $\frac{\text{mass of ammonium nitrate}}{\text{mass of solution}} * 100$

75 = $\frac{\text{mass of ammonium nitrate}}{431.25g} * 100$

Mass of ammonium nitrate = 323.43 g

Thus, we can conclude that mass of ammonium nitrate is 323.43 g.

1) We need to find mass of the solution

m=D*V
D= 1.725 g/mL
V= 250 mL
m=1.725 g/mL*250 mL= 431.25 g

2) 75% = 0.75
0.75*431.25 ≈ 323 g of NH4NO3

The rate constant for the second-order reaction 2NOBr(g) ¡ 2NO(g) 1 Br2(g) is 0.80/M ? s at 108C. (a) Starting with a concentration of 0.086 M, calculate the concentration of NOBr after 22 s. (b) Calculate the half-lives when [NOBr]0 5 0.072 M and [NOBr]0 5 0.054 M.