Why are the first bubbles of oxygen prduced not collected?

Which gas law (choices are Charles' Law, Gay-Lussac's Law, or Boyle's Law) explains each scenario:________.

posledela

Answer:

A. Boyle's Law

B. Charles' Law

C. Gay-Lussac's Law

Explanation:

An air bag inflates due to the decomposition of sodium azide or NaN₃ to completely fill the bag with nitrogen gas which is an example of Boyle's law, which states that the pressure of a given mass of gas is inversely proportional to its volume, hence due to the estricted volume of the airbag, the pressure of the nitrogen gas in the bag increses protecting the occupants of a cr from injuries in a crash

Helium balloon decrease in sice in a freezer is an example of Charlles law which states that the volume of a given mass of gas is nverslely proportionl to its temperature at constant pressure

A can of spray paint will explode if tossed into a fire is an example of Gay-Lussac's Law which states that the pressure of a given mass of gas is directly proportional to its temperature hence the increased pressure causes the can ti explode

4 0

3 years ago

Which would result in radioactive decay? Check all that apply

Gekata [30.6K]

6+2=115 and its good it took test

4 0

3 years ago

Read 2 more answers

When 0.42 g of a compound containing C, H, and O is burned completely, the products are 1.03 g CO2 and 0.14 g H2O. The molecular

Luda [366]

Answer: The empirical and molecular formula for the given organic compound is $C_2H_2O_4$ and $C_6H_4O_2$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=1.03g$

Mass of $H_2O=0.14g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 1.03 g of carbon dioxide, $\frac{12}{44}\times 1.03=0.28g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 0.14 g of water, $\frac{2}{18}\times 0.14=0.016g$ of hydrogen will be contained.

Mass of oxygen in the compound = (0.42) - (0.28 + 0.016) = 0.124 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.28g}{12g/mole}=0.023moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.016g}{1g/mole}=0.016moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.124g}{16g/mole}=0.00775moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.00775 moles.

For Carbon = $\frac{0.023}{0.00775}=2.96\approx 3$

For Hydrogen = $\frac{0.016}{0.00775}=2.06\approx 2$

For Oxygen = $\frac{0.00775}{0.00775}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 3 : 2 : 1

Hence, the empirical formula for the given compound is $C_3H_{2}O_1=C_3H_2O$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

$n=\frac{\text{molecular mass}}{\text{empirical mass}}$

We are given:

Mass of molecular formula = 108.10 g/mol

Mass of empirical formula = 54 g/mol

Putting values in above equation, we get:

$n=\frac{108.10g/mol}{54g/mol}=2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(3\times 2)}H_{(2\times 2)}O_{(1\times 2)}=C_6H_4O_2$

Thus, the empirical and molecular formula for the given organic compound is $C_3H_2O$ and $C_6H_4O_2$

6 0

3 years ago

Will mark as Brainliest.

SVETLANKA909090 [29]

If I done the math correctly it is 3729J because you multiply 16.5 g by the 2260 J/g and get 3729 J

4 0

4 years ago

As the temperature of a gas in a closed container increases, what is true of the particles in the substance? (2 points)

Misha Larkins [42]

The rate of movement increases, as they get faster with more energy.

8 0

3 years ago

Read 2 more answers

Why are the first bubbles of oxygen prduced not collected?​

Why are the first bubbles of oxygen prduced not collected?