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viktelen [127]
3 years ago
11

A sample of 4.50 g of diatomic oxygen (O2) occupies a volume of 15.0 L at a temperature of 203 K and a pressure of 5.0 atm. What

will be the volume of the same sample if the pressure increases to 15.0 atm and the temperature is raised to 303 K?
Chemistry
1 answer:
AleksandrR [38]3 years ago
3 0

Answer:

V = 0.23 Lt

Explanation:

Data: O₂ 4.50 g = 0.14 mol (considering a molecular mass of 32 g for O₂)

V₁ = 15.0 Lt , T₁ = 203 K , P₁ = 5.0 atm

V₂ = ? , T₂ = 303 K , P₂ = 15.0 atm

If we consider that this gas beahves as an ideal gas, then:

PV = nRT

V = nRT/P

V₂ = nRT₂/P₂ , where R gas constant = 0.082 Lt*atm/K*mol

V₂ = (0.14x0.082x303)/15.0 ⇒ V₂ = 0.23 Lt

Which is consistent, considering that the gas is subjected to a high pressure (from 5 atm to 15 atm), its volume should decrease (from 15 Lt to 0.23 Lt)

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Deutirium is the name of H-2 isotope
3 0
3 years ago
titration 25.0 ml of 3.4 x 10^ -3 M of Ba(OH) Neutralize 16.6 mL of HCL solution. What is the molarity of HCL solution?
marusya05 [52]
What we  are give: Concentration of base (CB) = 3.4 × 10^{-3}
Then convert all volume in ml to L.
Volume of base (VB) 25.0ml = 0.025L
Volume of acid (VA) 16.6ml = 0.0166L
Now that we have everything we use the formula CAVA=CBVB.
Make 'CA' the subject then solve. 
CA=\frac{CBVB}{VA}

8 0
3 years ago
PLEASE HELP!
Anon25 [30]

Answer:

C. Mole

Explanation:

Chemical formula deals with mole number

4 0
3 years ago
How many molecules are equal to 10.35 g of pyridine?
Lera25 [3.4K]

Answer:

0.78× 10 ²³ molecules

Explanation:

Given data:

Mass of pyridine = 10.35 g

Number of molecules = ?

Solution:

Number of moles of pyridine:

Number of moles = mass/molar mass

Number of moles = 10.35 g/ 79.1 g/mol

Number of moles = 0.13 mol

Number of molecules;

1 mole contain 6.022× 10 ²³ molecules

0.13 mol × 6.022× 10 ²³ molecules / 1mol

0.78× 10 ²³ molecules

7 0
3 years ago
The concentration of Pb²⁺ in a sample of wastewater is to be determined by using gravimetric analysis. To a 100.0-mL sample of t
svetoff [14.1K]

Answer:

0.005404 M

Explanation:

Pb^{2+}(aq) + Na_{2}CO_{3}(aq) ---> PbCO_{3}(s) + 2Na^{+}(aq)

Since you added an excess of sodium carbonate you warrantied that all the Pb^{+2} in the sample reacted with it. So we can say that the insoluble lead (II) carbonate PbCO_{3} contains all the Pb^{+2} ions in the original sample.  

The moles of PbCO_{3} are:

moles-of-PbCO_{3}=\frac{mass-of-PbCO_{3}}{Molecular-weight-of- PbCO_{3}}=\frac{0.1443g}{267\frac{g}{mol}}=0.00054 mol

One mol of Pb^{+2} is required to form one mol of PbCO_{3}. So, the stoichiometric relationship between them is 1:1.

Pb^{+2} +CO_{3}^{-2} - - - > PbCO_{3}

Knowing this, 0.00054 is also the number of moles of Pb^{+2} in the original sample.  

So, the concentration of Pb^{+2} in the original sample is:

M = \frac{mol-of-Pb^{+2}}{volume-wastewater-(liters)}=\frac{0.00054}{0.1L}=0.005404 M

8 0
3 years ago
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