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Alinara [238K]
3 years ago
7

At 570. mm Hg and 25°C, a gas sample ne (in mm Hg) at a volume of 1250 mL and a temperature of 175°C and 25°C, a gas sample has

a volume of 2270 mL. What is the final pressure
Chemistry
1 answer:
maw [93]3 years ago
5 0

Answer:

final pressure ( P2) = 467.37 mm Hg

Explanation:

ideal gas:

  • PV = nRT

∴ P1 = 570 mm Hg * ( atm / 760 mm Hg ) = 0.75 atm

∴ T1 = 25 ° C = 298 K

∴ V1 = 1.250 L

∴ R = 0.082 atm L / K mol

⇒ n = P1*V1 / R*T1

⇒ n = (( 0.75 ) * ( 1.25 )) / (( 0.082 ) * ( 298 ))

⇒ n = 0.038 mol gas

∴ T2 = 175 °C ( 448 K )

∴ V2 = 2.270 L

⇒ P2 = nRT2 / V2

⇒ P2 = (( 0.038 ) * ( 0.082 ) * ( 448 )) / 2.270

⇒ P2 = 0.615 atm * ( 760 mm Hg / atm ) = 467.37 mm Hg

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Warm air _____.
svp [43]

Answer:

cools as it rises, then sinks back down

Explanation:

The movement of air is a convection current. Convection currents occur when warm air rises, cools down, and sinks due to gained density, replacing the warm air closer to the ground, creating a cycle.

5 0
3 years ago
Flourine is found to undergo 10% radioactivity decay in 366 minutes determine its halflife​
yuradex [85]

Answer:

\boxed{\text{2408 min}}

Explanation:

The integrated rate law for radioactive decay is

\ln\dfrac{N_{0}}{N_{t}} = kt

1. Calculate the decay constant

\begin{array}{rcl}\ln \dfrac{100}{90} & = & k \times 366\\\\1.054 & = & 366k\\\\k & = & \dfrac{1.054 }{366}\\\\k & = & 2.879 \times 10^{-4} \text{ min}^{-1}\\\end{array}\\\\

2. Calculate the half-life

t_{\frac{1}{2}} = \dfrac{\ln2}{k}\\\\t_{\frac{1}{2}} = \dfrac{\ln2}{2.879 \times 10^{-4} \text{ min}^{-1}} = \text{2408 min}\\\\\text{The half-life for decay is } \boxed{\textbf{2408 min}}

8 0
3 years ago
You wish to prepare an HC2H3O2 buffer with a pH of 4.24. If the pKa of is 4.74, what ratio of C2H3O2 /HC2H3O2 must you use?
LenaWriter [7]

To solve this problem, we can use the Henderson-Hasselbalch Equation which relates the pH to the measure of acidity pKa. The equation is given as:<span>

<span>pH = pKa + log ([base]/[acid])                ---> 1</span></span>

Where,

[base] = concentration of C2H3O2 in molarity or moles

<span>[acid] = concentration of  HC2H3O2 in molarity or moles</span>

 

For the sake of easy calculation, let us assume that:

[base] = 1

[acid] = x

<span>
Therefore using equation 1,
4.24 = 4.74 + log (1 / x) 

<span>log (1 / x) = - 0.5

1 / x = 0.6065 </span></span>

x = 1.65<span>

The required ratio of C2H3O2 /HC2H3O2 <span>is 1:1.65 or 3:5. </span></span>
3 0
3 years ago
Please help ASAP pleas help right answer please
Otrada [13]
It is -2 because the charge will be at zero and electrons lower the charge
5 0
3 years ago
A gas occupies 547 L at 331K. Find its volume at 376K. You must show all of your work to receive credit. Also, name the gas law
zysi [14]

621.4L

Explanation:

Given parameters:

Initial volume = 547L

Initial temperature = 331K

Final temperature = 376K

Unknown:

Final volume = ?

Solution:

The appropriate gas law to use is the Charles's law.

The Charles's law shows the relationship between the volume and temperature of a gas under constant pressure.

The law states that "The volume of a fixed of a gas varies directly as its absolute temperature if the pressure is constant".

Mathematically;

    \frac{V_{1} }{T_{1} }  = \frac{V_{2} }{T_{2} }

 V₁ is the initial volume

 T₁ is the initial temperature

 V₂ is the final volume

 T₂ is the final temperature

Since the unknown is the final volume, we make it the subject of the expression;

   V₂ = \frac{V_{1} }{T_{1} }  x T_{2}  = \frac{547}{331 }  x 376

 V₂ =  621.4L

learn more:

Boyle's law brainly.com/question/8928288

#learnwithBrainly

3 0
3 years ago
Read 2 more answers
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