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marusya05 [52]
3 years ago
10

What is the temperature of 0.750 mol of a gas stored in a 6,850 mL cylinder at 2.21 atm? Use PV=nRT and R=0.0821 (L•atm)/(mol•K)

Chemistry
2 answers:
nadya68 [22]3 years ago
3 0
Using the formula PV=nRT
we know:
n= 0.750mol
R= 0.0821 (L•atm)/(mol•K)
V= 6850 ml = 6.85 L
P= 2.21 atm
and we need to find T
all you have to do is divide PV by nR and you will find T, like this:
t =  \frac{p \times v}{n \times r} \\ t =  \frac{2.21atm \times 6.85l}{0.750mol \times 0.0821 \frac{atm \times l}{mol \times k} }  \\ t = 246k
The temperature of the gas is D)246 K
SIZIF [17.4K]3 years ago
3 0

Answer:

d.) 246 K

Explanation:

this is the answer on edge, i just took quiz

hope this helps!!!

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Give the systematic names for the formulas or give the formulas for the names:
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Lead (II) acetate trihydrate is the systematic name for the formula Pb(C₂H₃O₂)₂ . 3H₂O.

<h3>What is Molecular Formula ?</h3>

The chemical formula that gives total number of atoms of each element in one molecule of a compound is called Molecular Formula.

<h3>What is Oxidation State ?</h3>

Oxidation state is also known as oxidation number. It is defined as the atom is equal to the total number of electrons which have been removed from the element in order to form chemical bond with other atom.

Now find the oxidation state of Pb in Pb(C₂H₃O₂)₂ .3H₂O

Assume the oxidation state of Pb in Pb(C₂H₃O₂)₂ .3H₂O be x

x + 2 × (-1) + 3 × 0 = 0

x - 2 + 0 = 0

x = 2

Oxidation state of Pb is +2 or (II)

Thus from the above conclusion we can say that The systematic name for the formula Pb(C₂H₃O₂)₂ .3H₂O is Lead (II) acetate trihydrate.

Learn more about the Molecular Formula here: brainly.com/question/15960587

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Explanation:

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Calculate delta Gº for the following system:
Llana [10]

Answer:

ΔG = - 442.5  KJ/mol    

Explanation:

Data Given

delta H = -472 kJ/mol

delta S = -108 J/mol K

So,

delta S = -0.108 J/mol K

delta Gº = ?

Solution:

The answer will be calculated by the following equation for the Gibbs free energy

                    G = H - TS

Where

G = Gibbs free energy

H = enthalpy of a system (heat

T = temperature

S = entropy

So the change in the Gibbs free energy at constant temperature can be written as

                ΔG = ΔH - TΔS . . . . . . (1)

Where

ΔG = Change in Gibb’s free energy

ΔH = Change in enthalpy of a system

ΔS = Change in entropy

if system have standard temperature then

T = 273.15 K

Now,

put values in equation 1

                  ΔG = (-472 kJ/mol) - 273.15 K (-0.108 KJ/mol K)

                 ΔG = (-472 kJ/mol) - (-29.5 KJ/mol)

                 ΔG = -472 kJ/mol + 29.5 KJ/mol

                 ΔG = - 442.5  KJ/mol          

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2 years ago
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