All categories

3 years ago

The table below shows a comparison of the different gas laws. Some cells have been left blank.

Chemistry

2 answers:

Nataly [62]3 years ago

7 0

The answer is C.

I hope I helped you

dlinn [17]3 years ago

3 0

Answer 1 :

Name : Boyle's law

Variables : Pressure, volume

Constants : Temperature, moles of gas

Equation : PV = k

Name : Charles’s law

Variables : Volume, temperature

Constants : Pressure, moles of gas

Equation : V = kT

Name : Gay-Lussac's law

Variables : Pressure, temperature

Constants : Volume, moles of gas

Equation : P = kT

Name : Combined gas law

Variables : Pressure, temperature, volume

Constants : Moles of gas

Equation : $\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

Explanation :

Boyle's Law : It is defined as the pressure is inversely proportional to the volume of the gas at constant temperature and number of moles.

$P\propto \frac{1}{V}$

Charles' Law : It is defined as the volume is directly proportional to the temperature of the gas at constant pressure and number of moles.

$V\propto T$

Gay-Lussac's Law : It is defined as the pressure is directly proportional to the temperature of the gas at constant volume and number of moles.

$P\propto T$

Avogadro's Law : It is defined as the volume is directly proportional to the number of moles of the gas at constant pressure and temperature.

$V\propto n$

Combined gas law : It is the combination of above four laws.

It is represented as,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

Answer 2 : The number of moles are assumed to be constant while using the combined gas law.

You might be interested in

5. Hurricane winds are strongest and rain is heaviest in the part of a hurricane called the<br>

Igoryamba

Answer:

In the center is the eye, with nearly clear skies, surrounded by the violent eyewall, with the strongest winds and very heavy rain.

Explanation:

3 0

3 years ago

Read 2 more answers

How many Helium atoms are in 12 grams of Helium?

horsena [70]

12gHe/1 × 1molHe/4.0026g × 6.02x10^23atomHe/1mol = 1.8 atoms

6 0

4 years ago

When the equation Sn + HNO₃ → SnO₂ + NO₂ + H₂O is balanced in acidic solution, what is the smallest whole-number coefficient for

balu736 [363]

Answer:

Explanation:

Hello,

In this case, for the given reaction we first assign the oxidation state for each species:

$Sn^0 + H^+N^{5+}O^{-2}_3 \rightarrow Sn^{4+}O_2 + N^{4+}O^{2-}_2 + H^+_2O^-$

Whereas the half reactions are:

$Sn^0+2H_2O \rightarrow Sn^{4+}O_2 +4H^++4e^-\\\\H^++H^+N^{5+}O^{-2}_3 +1e^-\rightarrow N^{4+}O^{2-}_2+H_2O$

Next, we exchange the transferred electrons:

$1\times(Sn^0+2H_2O \rightarrow Sn^{4+}O_2 +4H^++4e^-)\\\\4\times (H^++H^+N^{5+}O^{-2}_3 +1e^-\rightarrow N^{4+}O^{2-}_2+H_2O)\\\\\\Sn^0+2H_2O \rightarrow Sn^{4+}O_2 +4H^++4e^-\\\\4H^++4H^+N^{5+}O^{-2}_3 +4e^-\rightarrow 4N^{4+}O^{2-}_2+4H_2O$

Afterwards, we add them to obtain:

$Sn^0+2H_2O+4H^++4H^+N^{5+}O^{-2}_3 \rightarrow Sn^{4+}O_2 +4H^++4N^{4+}O^{2-}_2+4H_2O$

By adding and subtracting common terms we obtain:

$Sn^0+4H^+N^{5+}O^{-2}_3 \rightarrow Sn^{4+}O_2 +4N^{4+}O^{2-}_2+2H_2O$

Finally, by removing the oxidation states we have:

$Sn + 4HNO_3 \rightarrow SnO_2 + 4NO_2 + 2H_2O$

Therefore, the smallest whole-number coefficient for Sn is 1.

Regards.

6 0

3 years ago

Balance the following equation____H2 + ____O2 → ____H2O

Snezhnost [94]

Answer:

Explanation:

Since it has to be 2H2+02→2H20 for it to be balanced.

3 0

3 years ago

In a reaction between 6.0 g of oxygen gas, 4.0 g of hydrogen gas, and 5.0 g of solid sulfur at standard temperature and pressure

nikitadnepr [17]

Answer:

The limiting reagent is the O₂

Explanation:

We can think, this reaction

2O₂(g) + H₂(g) + S(s) → H₂SO₄

Mole of each = Mass / molar mass

6 g / 32 g/m = 0.187 mole O₂

4g / 2 g/m = 2 mole H₂

5g / 32.06 g/m = 0.156 mole S

Ratio between reactants is 2:1:1, 1:2:1, 1:1:2

For 2 mole of O₂, I need to react 1 mol of H₂ and 1 mol of S

0.187 mole of O₂, I need (the half)

0.093 mole of H₂ and 0.093 mole of S

For 1 mole of H₂, I need to react 2 mole of O₂ and 1 mol of S

2 mole of H₂, I need (the double of O₂ and the same for S)

4 mole of O₂ ; 2 mole of S

For 1 mol of S, I need to react 1 mol of H₂ and 2 mole of O₂

0.156 mole I need the same amount for H₂ and the double for O₂

0.156 mole of H₂ and 0.312 mole of O₂

In both cases, I can't make react, all the mass of oxygen, so this is the limiting reagent.

6 0

3 years ago