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Flauer [41]
3 years ago
11

Charles's law states the relationship between the temperature and volume of a gas. Based on your observations so far, state Char

les's law in your own words. Write 1-2 complete sentences.
Chemistry
1 answer:
FromTheMoon [43]3 years ago
6 0

Answer:

Charles' and Gay-Lussac's Law states that at constant pressure, temperature and volume are directly proportional.

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BaCl₂(aq) + Na₂SO₄(aq) = BaSO₄(s) + 2NaCl(aq)

Ba²⁺(aq) + 2Cl⁻(aq) + 2Na⁺(aq) + SO₄²⁻(aq) = BaSO₄(s) + 2Na⁺(aq) + 2Cl⁻(aq)

Ba²⁺(aq) + SO₄²⁻(aq)= BaSO₄(s)



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Select the correct answer. if two half-lives have passed since a scientist collected a 1.00-gram sample of u-235, how much u-235
Ugo [173]

Answer:

0.25 g of U-235 isotope will left .

Formula used :

where,

N = amount of U-235 left after n-half lives = ?

= Initial amount of the U-235 = 1.00 g

n = number of half lives passed = 2

0.25 g of U-235 isotope will left .

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2 years ago
You want to prepare 500.0 mL of 1.000 M KNO3 at 20°C, but the lab (and water) temperature is 24°C at the time of preparation. Ho
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Explanation:

As per the given data, at a higher temperature, at 24^{o}C, the solution will occupy a larger volume than at 20^{o}C.

Since, density is mass divided by volume and it will decrease at higher temperature.

Also, concentration is number of moles divided by volume and it decreases at higher temperature.

At 20^{o}C, density of water=0.9982071 g/ml  

Therefore, \frac{concentration}{density} will be calculated as follows.

                 = \frac{C_{1}}{d_{1}}

                 = \frac{1.000 mol/L}{0.9982071 g/ml}

                 = 1.0017961 mol/g  

At 24^{o}C, density of water = 0.9972995 g/ml

Since, \frac{concentration}{density} = \frac{C_{2}}{d_{2}}

                           = \frac{C_{2}}{0.9972995}

Also,             \frac{C_{1}}{d_{1}} = \frac{C_{2}}{d_{2}}

so,                   1.0017961 mol/g = \frac{C_{2}}{0.9972995}

                      C_{2} = 1.0017961 \times 0.9972995

                                  = 0.9990907 mol/L

Therefore, in 500 ml, concentration of KNO_{3} present is calculated as follows.

             C_{2} = \frac{concentration}{volume}

               0.9990907 mol/L = \frac{concentration}{0.5 L}  

               concentration = 0.49954537 mol

Hence, mass (m'') = 0.49954537 mol \times 101.1032 g/mol = 50.5056 g       (as molar mass of KNO_{3} = 101.1032 g/mol).

Any object displaces air, so the apparent mass is somewhat reduced, which requires buoyancy correction.

Hence, using Buoyancy correction as follows,

                      m = m''' \times \frac{(1 - \frac{d_{air}}{d_{weights}})}{(1 - \frac{d_{air}}{d})}}

where,          d_{air} = density of air = 0.0012 g/ml

                     d_{weight} = density of callibration weights = 8.0g/ml                      

                     d = density of weighed object

Hence, the true mass will be calculated as follows.

           True mass(m) = 50.5056 \times \frac{(1 - \frac{0.0012}{8.0})}{(1 - (\frac{0.0012}{2.109})}

             true mass(m) = 50.5268 g

                                  = 50.53 g (approx)

Thus, we can conclude that 50.53 g apparent mass of KNO_{3} needs to be measured.

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3 years ago
Identify an alkene and carboxylic acid using primary observations
ehidna [41]

Answer:

The general formula for the carboxylic acids is C nH 2n+1COOH (where n is the number of carbon atoms in the molecule, minus 1).

Explanation:

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