Which equation would be used to calculate the rate constant from initial<br> concentrations?

2 SO3 (g) + Heat <-----> 2 SO2 (g) + O2 (g)

Step2247 [10]

Answer:

The concentration of SO₂ will decreases

Explanation:

As you can see in the reaction

2 moles of gas ⇆ 3 moles of gas

Based on Le Châtelier's principle, a change doing in a system will produce that the system reacts in order to counteract the change made.

If the pressure is increased, the system will shift to the left in order to produce less moles of gas and decrease, thus, the pressure.

As the system shift to the left, the concentration of SO₂ will decreases

7 0

2 years ago

In an acid-base titration experiment, 50 mL of a 0.05 M solution of acetic acid (Ka= 1.75 x 10-5 ) was titrated with a 0.05M sol

Viktor [21]

Answer:

(3) 5.36

Explanation:

Since this is a titration of a weak acid before reaching equivalence point, we will have effectively a buffer solution. Then we can use the Henderson-Hasselbalch equation to answer this question.

The reaction is:

HAc + NaOH ⇒ NaAc + H₂O

V NaOH = 40 mL x 1 L/1000 mL = 0.040 L

mol NaOH reacted with HAc = 0.040 L x 0.05 mol/L = 0.002 mol

mol HAC originally present = 0.050 L x 0.05 mol/L = 0.0025 mol

mol HAc left after reaction = 0.0025 - 0.002 = 0.0005

Now that we have calculated the quantities of the weak acid and its conjugate base in the buffer, we just plug the values into the equation

pH = pKa + log ((Ac⁻)/(HAc))

(Notice we do not have to calculate the molarities of Ac⁻ and HAc because the volumes cancel in the quotient)

pH = -log (1.75 x 10⁻⁵) + log (0.002/0.0005) = 5.36

THe answer is 5.36

5 0

2 years ago

Read 2 more answers

Wendy made the following diagram shown to represent the solar eclipse.

Pavlova-9 [17]

The moon should be between the sun and Earth

7 0

3 years ago

Read 2 more answers

What is the mass of the solid NH4Cl formed when 75.5 g of NH3 is mixed with an equal mass of HCl? What is the volume of the gas

Gekata [30.6K]

Answer : The volume of the gas remaining is 56.5 liters.

The gas is hydrochloric acid and the formula of the gas is HCl.

The mass of $NH_4Cl$ produced is, 110.7 grams.

Explanation :

The balanced chemical reaction will be:

$NH_3+HCl\rightarrow NH_4Cl$

First we have to calculate the moles of $NH_3$ and HCl

$\text{Moles of }NH_3=\frac{\text{Mass of }NH_3}{\text{Molar mass of }NH_3}$

Molar mass of NH_3 = 17 g/mole

$\text{Moles of }NH_3=\frac{75.5g}{17g/mole}=4.44mole$

and,

$\text{Moles of }HCl=\frac{\text{Mass of }HCl}{\text{Molar mass of }HCl}$

Molar mass of HCl = 36.5 g/mole

$\text{Moles of }HCl=\frac{75.5g}{36.5g/mole}=2.07mole$

Now we have to calculate the limiting and excess reagent.

From the balanced reaction we conclude that

As, 1 mole of HCl react with 1 mole of $NH_3$

So, 2.07 mole of HCl react with 2.07 mole of $NH_3$

From this we conclude that, $NH_3$ is an excess reagent because the given moles are greater than the required moles and HCl is a limiting reagent and it limits the formation of product.

The remaining moles of HCl gas = 4.44 - 2.07 = 2.37 moles

Now we have to calculate the volume of the gas remaining.

Using ideal gas equation :

PV = nRT

where,

P = Pressure of gas = 752 mmHg = 0.989 atm (1 atm = 760 mmHg)

V = Volume of gas = ?

n = number of moles of gas = 2.37 moles

R = Gas constant = 0.0821 L.atm/mol.K

T = Temperature of gas = $14.0^oC=273+14.0=287K$

Putting values in above equation, we get:

$0.989atm\times V=2.37mole\times (0.0821L.atm/mol.K)\times 287K$

V = 56.5 L

Now we have to calculate the moles of $NH_4Cl$

As, 1 mole of HCl react with 1 mole of $NH_4Cl$

So, 2.07 mole of HCl react with 2.07 mole of $NH_4Cl$

Now we have to calculate the mass of $NH_4Cl$

$\text{ Mass of }NH_4Cl=\text{ Moles of }NH_4Cl\times \text{ Molar mass of }NH_4Cl$

Molar mass of $NH_4Cl$ = 53.5 g/mole

$\text{ Mass of }NH_4Cl=(2.07moles)\times (53.5g/mole)=110.7g$

Thus, the volume of the gas remaining is 56.5 liters.

The gas is hydrochloric acid and the formula of the gas is HCl.

The mass of $NH_4Cl$ produced is, 110.7 grams.

3 0

2 years ago

El etanol combustiona con el oxigeno produciendo dióxido de carbono gaseoso y agua: En un proceso se hacen reaccionar 450.2 g de

Olenka [21]

Answer:

uh oh stinky uh oh stinky

6 0

2 years ago

Which equation would be used to calculate the rate constant from initial concentrations?