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snow_tiger [21]
3 years ago
6

A 8.2 L sample of gas has a pressure of 0.8 atm at a temperature of 259 K. If the temperature increases to 301 K, causing the vo

lume to increase to 11.5 L, what is the new pressure? Round your answer to the nearest tenth
Chemistry
2 answers:
Rina8888 [55]3 years ago
8 0

The answer is 0.7 atm


pashok25 [27]3 years ago
8 0

Answer:

0.7 atm

Explanation:

For the Clapeyron's equation, we can calculate the change in the physical properties that happens when there is a state change:

P1*V1/T1 = P2*V2/T2

Where P is the pressure, V is the volume, T is the temperature, 1 is the initial state, and 2 the final state. So:

0.8*8.2/259 = P2*11.5/301

0.02533 = 0.038206*P2

P2 = 0.7 atm

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What are the electrons for calcium
IRINA_888 [86]
The traditional calcium atom has twenty protons and twenty electrons making it neutral.


The calcium in the pic is a calcium ion so the number of protons and electrons are not equivalent.


Since it's 2+ that means the ion is positively charged and for that to happen electrons are away.

So 20-2=18

There are 18 electrons
5 0
3 years ago
Bond energies do not account for the energy associated with the formation of aqueous solutions. Explain what energy is not accou
Tems11 [23]

Answer:

Why do we all not know the answer to this on the practical

Explanation:

5 0
2 years ago
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
How many molecules of XeF6 are formed from 12.9 L of F2 (at 298 K and 2.6 atm) according to 11) the following reaction? Assume t
ddd [48]

Answer:

#Molecules XeF₆ = 2.75 x 10²³ molecules XeF₆.

Explanation:

Given … Excess Xe + 12.9L F₂ @298K & 2.6Atm => ? molecules XeF₆

1. Convert 12.9L 298K & 2.6Atm to STP conditions so 22.4L/mole can be used to determine moles of F₂ used.

=> V(F₂ @ STP) = 12.6L(273K/298K)(2.6Atm/1.0Atm) = 30.7L F₂ @ STP

2. Calculate moles of F₂ used

=> moles F₂ = 30.7L/22.4L/mole = 1.372 mole F₂ used

3. Calculate moles of XeF₆ produced from reaction ratios …

Xe + 3F₂ => XeF₆ => moles of XeF₆ = ⅓(moles F₂) = ⅓(1.372) moles XeF₆ = 0.4572 mole XeF₆

4. Calculate number molecules XeF₆ by multiplying by Avogadro’s Number  (6.02 x 10²³ molecules/mole)

=> #Molecules XeF₆ = 0.4572mole(6.02 x 10²³ molecules/mole)

                                  = 2.75 x 10²³ molecules XeF₆.

8 0
2 years ago
How many moles of carbon are in 3.5 L?
SashulF [63]

Answer:

0.16mole

Explanation:

To solve this problem, we are going to assume that the number of moles of carbon to be determined is that at STP, standard temperature and pressure.

The number of moles of a substance at STP is given as;

 Number of moles  = \frac{volume}{22.4}  

Given volume  = 3.5L

Now, insert the parameters;

     Number of moles  = \frac{3.5}{22.4}   = 0.16mole

8 0
3 years ago
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