7 days ago

A 0.5376 g sample of an unknown compound is found to contain 0.3044 g of carbonate. Could this compound be calcium carbonate?

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The atomic weight of iodine is less than the atomic weight of tellurium. However, Mendeleev listed iodine after tellurium in his

eduard [2782]

Answer:

The basis for the doubt stemmed from the way iodine chemically reacted and its other properties, suggesting that it should be grouped with chlorine and bromine, while the much denser tellurium ought to be classified in the preceding group

This doubt was confirmed when subsequent researchers discovered that tellurium's atomic number is 52 and iodine's is 53

Explanation:

5 0

3 months ago

A large balloon is initially filled to a volume of 25.0 L at 353 K and a pressure of 2575 mm Hg. What volume of gas will the bal

Tems11 [2777]

Answer:

A balloon will hold a gas volume of 45.0 L at a pressure of 1.35 atm and a temperature of 253 K.

Explanation:

Applying the Ideal Gas Law for the same gas amount, we have:

$\frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}$

Parameters are given as follows:

V₁ = 25.0 L

V₂ =?

P₁ = 2575 mm Hg

To convert pressure to atmospheres: P (atm) = P (mm Hg) / 760

P₁ = 2575 / 760 atm = 3.39 atm

P₂ = 1.35 atm

T₁ = 353 K

T₂ = 253 K

Using the earlier formula, we derive:

$\frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}$

$\frac{{3.39}\times {25.0}}{353}=\frac{{1.35}\times {V_2}}{253}$

$\frac{1.35V_2}{253}=\frac{3.39\times \:25}{353}$

By solving for V₂, we find:

V₂ = 45.0 L

A balloon will hold a gas volume of 45.0 L at a pressure of 1.35 atm and a temperature of 253 K.

4 0

3 months ago