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Dennis_Churaev [7]
3 years ago
8

Directions: Perform the Assessment in Module 7 because that will serve as your Performance Task for Module 5, 6 & 7. Kindly

read the rubric for you to be guided on how your output will be graded. Do not forget to submit your output to your teacher. Enjoy!
please answer nyo to nang maayos please ​

Chemistry
1 answer:
garik1379 [7]3 years ago
8 0

please put your question completely....

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Calculate the density of CO2 in g/cm3 at room temperature(25 degrees Celsuis) and pressure(1 atm) assuming it acts as an ideal g
Readme [11.4K]

Answer:

density=1.8x10^{-3}g/mL

Explanation:

Hello,

Considering the ideal equation of state:

PV=nRT

The moles are defined in terms of mass as follows:

n=\frac{m}{M}

Whereas M the gas' molar mass, thus:

PV=\frac{mRT}{M}

Now, since the density is defined as the quotient between the mass and the volume, we get:

P=\frac{m}{V} \frac{RT}{M}

Solving for m/V:

density= m/V=\frac{PM}{RT}

Thus, the result is given by:

density=\frac{(1atm)(44g/mol)}{[0.082atm*L/(mol*K)]*298.15K} \\density=1.8g/L=1.8x10^{-3}g/mL

Best regards.

8 0
3 years ago
What is the speed of sound in dry air at 20°C?
Lubov Fominskaja [6]
343 meters per second
4 0
3 years ago
Use the periodic table to identify the molar mass of each element below. Answer without doing any calculations.
Delicious77 [7]

Answer:

Beryllium (Be) : 9.01 g/mol

Silicon (Si) : 28.09 g/mol

Calcium (Ca) : 40.08 g/mol

Rhodium (Rh) : 102.91 g/mol

Explanation:

8 0
3 years ago
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What is the dependent variable in the experiment shown​
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8 0
3 years ago
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At 500 degree C, F_2 gas is stable and does not dissociate, but at 840 degree C, some dissociation occurs: F_2 (g) 2 F(g). A fla
bazaltina [42]

Answer:

2.73 is the equilibrium constant for the dissociation of F_2 gas at 840 degree Celsius.

Explanation:

F_2(g)\rightleftharpoons 2F(g)

Initial

0.600 atm    0

Equilibrium

(0.600 atm - p)        2p

Total pressure at equilibrium = P = 0.984 atm

P= 0.600 atm - p)+2p=0.984 atm

p = 0.384 atm

Partial pressure of the F_2 gas , p_{f_2}= (0.600 atm - 0.384 atm)=0.216 atm

Partial pressure of the F gas, p_{f} = 2(0.384 atm)=0.768 atm

K_p=\frac{(p_{F})^2}{p_{F_2}}

K_p=\frac{(0.768 atm)^2}{0.216 atm}=2.73

2.73 is the equilibrium constant for the dissociation of F_2 gas at 840 degree Celsius.

7 0
4 years ago
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