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3 years ago

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What is the vapor pressure of CS2CS2 in mmHgmmHg at 26.5 ∘C∘C? Carbon disulfide, CS2CS2, has PvapPvap = 100 mmHgmmHg at −−5.1 ∘C

∘C and ΔHvapΔHvap = 28.0 kJ/molkJ/mol .

1 answer:

Digiron [165]3 years ago

7 0

Answer: 26.5 mm Hg

Explanation:

The vapor pressure is determined by Clausius Clapeyron equation:

$ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}(\frac{1}{T_1}-\frac{1}{T_2})$

where,

$P_1$ = initial pressure at $26.5^oC$ = ?

$P_2$ = final pressure at $-5.1^oC$ = 100 mm Hg

= enthalpy of vaporisation = 28.0 kJ/mol =28000 J/mol

R = gas constant = 8.314 J/mole.K

$T_1$ = initial temperature = $26.5^oC=273+26.5=299.5K$

$T_2$ = final temperature = $-5.1^oC=273+(-5.1)=267.9K$

Now put all the given values in this formula, we get

$\log (\frac{P_1}{100})=\frac{28000}{2.303\times 8.314J/mole.K}[\frac{1}{299.5}-\frac{1}{267.9}]$

$\log (\frac{P_1}{100})=-0.576$

$\frac{P_1}{100}=0.265$

$P_1=26.5mmHg$

Thus the vapor pressure of $CS_2CS_2$ in mmHg at 26.5 ∘C is 26.5

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Question 2 (0 points)

bazaltina [42]

Answer:

The probability of spinning red with the spinner or rolling an odd number with the die is $\dfrac{1}{8}$

Explanation:

Given that,

Total color in spinner = 4

Let the area of four parts is equal in the spinner.

We need to calculate the probability of spinning red

Using formula of probability

$P(A) = \dfrac{Number\ of\ red\ color\ outcome}{Total\ number\ of\ colors\ in\ the\ spinner}$

Put the value into the formula

$P(A)=\dfrac{1}{4}$

We need to calculate the probability of odd number of the die

Using formula of probability

$P(B) = \dfrac{Number\ of\ odd\ digits\ outcome}{Total\ number\ of\ digits\ in\ the\ die}$

Put the value into the formula

$P(B)=\dfrac{3}{6}$

$P(B)=\dfrac{1}{2}$

We need to calculate the probability of spinning red with the spinner or rolling an odd number with the die

Using formula of probability of two events which is independent

$P(A\ and\ B)=P(A)\times P(B)$

Put the value into the formula

$P(A\ and\ B)=\dfrac{1}{4}\times\dfrac{1}{2}$

$P(A\ and\ B)=\dfrac{1}{8}$

Hence, The probability of spinning red with the spinner or rolling an odd number with the die is $\dfrac{1}{8}$

5 0

3 years ago

Ammonia and oxygen react to produce nitric oxide and water. Which of the following chemical equations describes this reaction an

aksik [14]

Answer:

Option B. 4NH3(g) + 5O2(g) —> 4NO(g) + 6H2O(g)

Explanation:

The reaction between ammonia, NH3 and oxygen, O2, will produce nitric oxide and water as shown below:

NH3(g) + O2(g) —> NO(g) + H2O(g)

Now, let us balance the equation.

This is illustrated below:

NH3(g) + O2(g) —> NO(g) + H2O(g)

There are 3 atoms of H on the left side and 2 atoms on the right side. It can be balance by putting 4 in front of NH3 and 6 in front of H2O as shown below:

4NH3(g) + O2(g) —> NO(g) + 6H2O(g)

There are 4 atoms of N on the left side and 1 atom on the right side. It can be balance by putting 4 in front of NO as shown below:

4NH3(g) + O2(g) —> 4NO(g) + 6H2O(g)

Now, the are 2 atoms of O on the left side and a total of 10 atoms on the right side. It can be balance by putting 5 in front of O2 as shown below:

4NH3(g) + 5O2(g) —> 4NO(g) + 6H2O(g)

Now, we can see that the equation is balanced.

Therefore, option B gives the balanced chemical equation for the reaction.

5 0

3 years ago

A method used by the U.S. Environmental Protection Agency (EPA) for determining the concentration of ozone in air is to pass the

baherus [9]

Answer: 1. $9.08\times 10^{-6}$ moles

2. 90 mg

Explanation:

$O_3(g)+2NaI(aq)+H_2O(l) \rightarrow O_2(g)+I_2(s)+2NaOH(aq)$

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus $4.54 \times 10^{-6}$ moles of ozone is removed by = $\frac{2}{1}\times 4.54 \times 10^{-6}=9.08\times 10^{-6}$ moles of sodium iodide.

Thus $9.08\times 10^{-6}$ moles of sodium iodide are needed to remove $4.54\times 10^{-6}$ moles of $O_3$

2. $\text{Number of moles of ozone}=\frac{0.01331g}{48g/mol}=0.0003moles$

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus 0.0003 moles of ozone is removed by = $\frac{2}{1}\times 0.0003=0.0006$ moles of sodium iodide.

Mass of sodium iodide= $moles\times {\text {molar mass}}=0.0006\times 150g/mol=0.09g=90mg$ (1g=1000mg)

Thus 90 mg of sodium iodide are needed to remove 13.31 mg of $O_3$ .

3 0

3 years ago

The gauge pressure inside a vessel is ‐40kPa, at an elevation of 5000m. a) What is the absolute pressure? b) At this elevation,

hoa [83]

Answer:

a) Pabs = 48960 KPa

b) T = 433.332 °C

Explanation:

Pabs = Pgauge + d*g* h

∴ d = 1000 Kg/m³

∴ g = 9.8 m/s²

∴ h = 5000 m

∴ P gauge = - 40 KPa * ( 1000 Pa / KPa ) = - 40000 Pa; Pa≡Kg/m*s²

⇒ Pabs = - 40000 Kg/ms² + ( 1000 Kg/m³ * 9.8 m/s² * 5000 m )

⇒ Pabs = 48960000 Pa = 48960 KPa

a) at that height and pressure, we find the temperature at which the water boils by means of an almost-exponential graph which has the following equation:

P(T) = 0.61094 exp ( 17.625*T / ( T + 243.04 ))......P (KPa) ∧ T (°C)....from literature

∴ P = 48960 KPa

⇒ ( 48960 KPa / 0.61094 ) = exp ( 17.625T / (T+ 243.04))

⇒ 80138.803 = exp ( 17.625T / ( T + 243.04))

⇒ Ln ( 80138.803) = 17.625T / ( T + 243.04))

⇒ 11.292 * ( T + 243.04 ) = 17.625T

⇒ 11.292T + 2744.289 = 17.625T

⇒ 2744.289 = 17.625T - 11.292T

⇒ 2744.289 = 6.333T

⇒ T = 433.332 °C

3 0

3 years ago

Help in chem please!!!!!!

IgorLugansk [536]

This is a one-step unit analysis problem. Since we are staying in moles, grams of our compound, and thus molar mass, is not needed.

1 mole is equal to 6.022x10²³ particles as given, so:

$1.5x10^{24} particles FeO_{2} (\frac{1mol}{6.022x10^{23}particles } )=2.49 molFeO_{2}$

<h3>Answer:</h3>

2.49 mol

Let me know if you have any questions.

3 0

3 years ago