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Monica [59]
3 years ago
12

Four gases were combined in a gas cylinder with these partial pressures: 3.5 atm N2, 2.8 atm O2, 0.25 atm Ar, and 0.15 atm He.

Chemistry
2 answers:
antiseptic1488 [7]3 years ago
7 0

Answer:

P(total) =  6.7 atm

Explanation:

Given data:

Partial pressure of N₂ = 3.5 atm

Partial pressure of O₂ = 2.8 atm

Partial pressure of Ar = 0.25 atm

Partial pressure of He = 0.15 atm

Total pressure = ?

Solution:

According to the Dalton law of partial pressure,

"The total pressure inside the gas cylinder having mixture of gases is equal to the sum of partial pressures of individual gas present in it"

Mathematical expression:

P(total) = P₁ + P₂ + P₃ + .........Pₙ

Now we will determine the total pressure of given gases.

P(total) = P₁ + P₂ + P₃ + P₄

P(total) =P(N₂) + P(O₂) + P(Ar) + P(He)

P(total) = 3.5 atm + 2.8 atm + 0.25 atm + 0.15 atm

P(total) =  6.7 atm

sp2606 [1]3 years ago
5 0

Answer:

6.7 atm

Explanation:

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m(t) = m_{0} e^{-kt}
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k = the decay constant
t = time, years.

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e^{-30k} =  \frac{1}{2} \\\ -30k = ln(0.5) \\ k =  \frac{ln(0.5)}{-30} =0.0231

After 60 years, the mass remaining is
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4 0
3 years ago
how many grams of oxygen gas are needed to produce 10.0 grams of carbon dioxide according to the balanced equation of CH4
Alinara [238K]
Molar mass of :

O2 = 16 * 2 = 32 g/mol

CO2 = 12 + 16 * 2 = 44 g/mol

<span>Balanced chemical equation :
</span>
1 CH4 + 2 O2 = 1 CO2 + 2 H2O
               ↓              ↓
             2 moles     1 mole

2* 32 g O2 ----------> 1* 44 g CO2
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44 x = 2 * 32*10.0

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3 years ago
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Which physical property do plasmas and gases have in common?
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A 100.0 g sample of aluminum released 1680 calories when cooled from 100.0c to 20.0 c what is the specific heat of the metal
vovangra [49]

Answer:

The specific heat of aluminium is 0.8792 J/g °C  or 0.21 Cal/g °C

Explanation:

Step 1 : Write formule of specific heat

Q=mcΔT

with Q = heat transfer (J)

with m = mass of the substance

with c = specific heat ⇒ depends on material and phase ( J/g °C)

with ΔT = Change in temperature

For this case :

Q = 1680 Calories = 7033.824 J ( 1 calorie = 4.1868 J)

m = 100.0g

c= has to be determined

ΔT = 100 - 20 = 80°C

<u>Step 2:  Calculating specific heat</u>

⇒ via the formule Q=mcΔT

7033.824 J = 100g * c * 80

7033.824 = 8000 *c

c = 7033.824 /8000

c = 0,879228 J/g °C

or 0.21 Cal / g°C

The specific heat of aluminium is 0.8792 J/g °C  or 0.21 Cal/g °C

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