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Alik [6]
2 years ago
11

A container holds a mixture of 3.01 x 1023 molecules of H₂ gas. 1.20 x 1024 molecules of CO₂ gas, and 1.50 x 10

Chemistry
1 answer:
Rom4ik [11]2 years ago
5 0

Answer:

.4998 atm    ( rounded to .5 atm   ) <===== not one of the answers listed

Explanation:

The partial pressure exerted by O2   is   equal to the fraction that is O2 times the pressure

1.5 / ( 1.5 + 1.2 + .301 )   * 1 atm = .4998 atm  

<u>( 1) Are you sure it is   ONE atm pressure in the container?</u>

<u>(2) IS the first answer supposed to be   5 x 10^-1  atm    perhaps?</u>

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Suppose that 25,0 mL of a gas at 725 mmHg and 298K is converted to
posledela

The new volume : 21.85 ml

<h3>Further explanation</h3>

Given

V1=25,0 ml

P1=725 mmHg

T1=298K is converted to

T2=273'K

P2=760 mmHg atm

Required

V2

Solution

Combined gas law :

\tt \dfrac{P_1.V_1}{T_1}=\dfrac{P_2.V_2}{T_2}

Input the value :

V2=(P1.V1.T2)/(P2.T1)

V2=(725 x 25 ml x 273)/(760 x 298)

V2=21.85 ml

5 0
3 years ago
Caleb is an inventor who has designed a new kind of fishing net. He is looking for a strong, flexible material that can be used
UkoKoshka [18]

Answer:

Its B Nylon

Explanation:

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5 0
3 years ago
Read 2 more answers
What mass of barium sulfate (233 g/mol) is produced when 125 mL of a 0.150 M solution of barium chloride is mixed with 125 mL of
Rzqust [24]

Answer:

4.37 g of barium sulphate

Explanation:

The reaction equation is;

3BaCl2(aq) + Fe2(SO4)3(aq) ---->3 BaSO4(s) + 2FeCl3(aq)

From the question, the number of moles of both barium chloride and FeSO4 = 125/1000 L × 0.150 M = 0.01875 moles

To find the limiting reactant;

3 moles of barium chloride yields 3 moles of barium sulphate

0.01875 moles of barium chloride yields 3 × 0.01875 moles/3 = 0.01875 moles of barium sulphate

1 mole of iron III sulphate yields 3 moles of barium sulphate

0.01875 molesof iron III sulphate yields 0.01875 moles ×3/1 = 0.05625 moles of barium sulphate

Hence,barium chloride is the limiting reactant

Amount of barium sulphate produced = 0.01875 moles × 233 g/mol = 4.37 g of barium sulphate

4 0
3 years ago
How many bananas are equal to 7.50 moles of bananas?​
Veseljchak [2.6K]

Answer:

4.52×10^24

Explanation:

N = n × Na

where; N = no. of bananas

n = no. of moles

Na = Avogadro's constant

Which is 6.02×10^23

N = 7.5 × 6.02×10^23

N =4.515×10^24

7 0
3 years ago
N2 + 3H2 + 2NH3 .
HACTEHA [7]

Answer:

7

Explanation:

7 0
3 years ago
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