What would be the original temperature of a gas that has a volume of 2.0 l and a pressure of 2.0 atm and an unknown temperature that the volume increased to 3.5 l in its pressure decreased to 1.0 atm if the final temperature is measured to be 11°c

= 51.57 °C

Explanation:

The combined gas equation shows that P₁V₁/T₁=P₂V₂/T₂ where P represents pressure, V represents volume and T absolute temperature

From the information provided in the question,

Since we are finding the initial or original temperature, we can make T₁ the subject of the formula.

T₁=P₁V₁T₂/P₂V₂

P₁=2.0 ATM

P₂= 1.0 ATM

V₁=2.0 L

V₂= 3.5 L

T₁=?

T₂= (11+273) K=284 K

Using these values in the formula:

T₁= (2.0 ATM × 2.0 L× 284 K)/(1.0 ATM × 3.5 L)

=324.57 K

324.57-273= 51.57 °C

a sequence is a function whose domain is the set of natural numbers.

natural numbers are all the negative and positive integers

a. the law of proportion states the combination of 2 atoms in a specific ratio to produce multiple compounds such as the given statement of question ‘1st compound contains 88.8 % cu ’ indicates 88.8 g of cu is present along with 11.2 g oxygen in 100 g of compound whereas the statement ‘2st compound contains 79.9 % cu ’ indicates 79.9 g of cu is present along with 20.1 g oxygen in 100 g of compound

1st compound contains 88.8 g cu

for cu

m = 63 g /mol

atomic % = 88.8 /63 = 1.42

for o

m = 16 g /mol

atomic % = 11.2 /16 = 0.7

b. the emperical formula of 1st compound is

cu1.4o0.7

cu2o