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Wewaii [24]
3 years ago
15

Complete and balance the molecular equation for the reaction of aqueous sodium sulfate, Na2SO4, and aqueous barium nitrate, Ba(N

O3)2. Include physical states.molecular equation:Na_{2}SO_{4}(aq) + Ba(NO_{3})_{2}(aq) ->Na2SO4(aq)+Ba(NO3)2(aq)⟶Enter the balanced net ionic equation for this reaction. Include physical states.net ionic equation:
Chemistry
1 answer:
Rina8888 [55]3 years ago
7 0

Answer:

1. The balanced molecular equation is given below:

Na2SO4(aq) + Ba(NO3)2(aq) —> BaSO4(s) + 2NaNO3(aq)

2. The net ionic equation is given below:

SO4^2-(aq) + Ba^2+(aq) —> BaSO4(s)

Explanation:

1. The balanced molecular equation

Na2SO4(aq) + Ba(NO3)2(aq) —> BaSO4(s) + NaNO3(aq)

The above equation can be balance as follow:

Na2SO4(aq) + Ba(NO3)2(aq) —> BaSO4(s) + NaNO3(aq)

There are 2 atoms of Na on the left side and 1 atom on the right side. It can be balance by putting 2 in front of NaNO3 as shown below:

Na2SO4(aq) + Ba(NO3)2(aq) —> BaSO4(s) + 2NaNO3(aq)

Now, the equation is balanced.

2. The bal net ionic equation.

This can be obtained as follow:

Na2SO4(aq) + Ba(NO3)2(aq) —>

In solution, Na2SO4 and Ba(NO3)2 will dissociate as follow:

Na2SO4(aq) —> 2Na^+(aq) + SO4^2-(aq)

Ba(NO3)2(aq) —> Ba^2+(aq) + 2NO3^-(aq)

Na2SO4(aq) + Ba(NO3)2(aq) —>

2Na^+(aq) + SO4^2-(aq) + Ba^2+(aq) + 2NO3^-(aq) —> BaSO4(s) + 2Na^+(aq) + 2NO3^-(aq)

Cancel the spectator ions i.e Na^+ and NO3^- to obtain the net ionic equation.

SO4^2-(aq) + Ba^2+(aq) —> BaSO4(s)

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14. An enzyme-catalyzed reaction was carried out with the substrate concentration initially a thousand times greater than the Km
emmainna [20.7K]

Answer:

27 min

Explanation:

The kinetics of an enzyme-catalyzed reaction can be determined by the equation of Michaelis-Menten:

v = \frac{vmax[S]}{Km + [S]}

Where v is the velocity in the equilibrium, vmax is the maximum velocity of the reaction (which is directed proportionally of the amount of the enzyme), Km is the equilibrium constant and [S] is the concentration of the substrate.

So, initially, the velocity of the formation of the substrate is 12μmol/9min = 1.33 μmol/min

If Km is a thousand times smaller then [S], then

v = vmax[S]/[S]

v = vmax

vmax = 1.33 μmol/min

For the new experiment, with one-third of the enzyme, the maximum velocity must be one third too, so:

vmax = 1.33/3 = 0.443 μmol/min

Km will still be much smaller then [S], so

v = vmax

v = 0.443 μmol/min

For 12 μmol formed:

0.443 = 12/t

t = 12/0.443

t = 27 min

7 0
3 years ago
DDT, an insecticide harmful to fish, birds, and humans, is produced by the following reaction. 2 C6H5Cl + C2HOCl3 → C14H9Cl5 + H
Elena-2011 [213]

Answer:

1,070.41 grams of DDT will be formed .Explanation:

1) 2 C_6H_5Cl + C_2HOCl_3\rightarrow C_{14}H_9Cl_5 + H_2O

Moles of chlorobenzene = \frac{1197 g}{112.5 g/mol}=10.64 mol

According to reaction, 2 moles of chloro benzene reacts with 1 mole of chloral . Then 10.64 moles of chloro benzene will react with :

\frac{1}{2}\times 10.64 mol=5.32 mole of chloral

2) Moles of chloral = \frac{456 g}{147.5 g/mol}=3.0915 mol

According to reaction, 1 moles of chloral reacts with 2 mole of chlorobenzene . Then 3.0915 moles of chloral will react with :

\frac{2}{1}\times 3.0915 mol=6.1830 mol of chloro benzene

As we can see that chloral is in limiting amount and chloro benzene is in excessive amount. So, amount of DDT will depend upon amount of chloral.

According to reaction, 1 mole chloral gives 1 mole DDT.Then 3.0195 moles of chloral will give :

\frac{1}{1}\times 3.0195 mol=3.0195 mol

Mass of 3.0195 moles of DTT :

3.0195 mol × 354.5 g/mol = 1,070.41 g

1,070.41 grams of DDT will be formed .

3 0
3 years ago
The formula for the nitrate ion: NO3 NO2 NO3+​
tekilochka [14]

Answer:

NO3-

Explanation:

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3 0
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Explanation:

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3 years ago
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Twenty five grams of Iron 3 oxide react with an excess of carbon monoxide to form 15 g of Fe. Carbon dioxide is the other produc
densk [106]
<h3>Answer:</h3>

Theoretical mass = 17.42 g

Percent yield of Fe = 86.11%

<h3>Explanation:</h3>

The equation for the reaction between iron (iii) oxide and carbon monoxide is given by;

Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g)

We are required to calculate the theoretical yield and the percentage yield of Iron.

Step 1: Moles of iron (iii) oxide

Moles are given by dividing the mass of the compound by the molar mass.

Molar mass of Iron(iii) oxide = 159.69 g/mol

Moles of Iron(III) oxide = 25 g ÷ 159.69 g/mol

                                     = 0.156 moles

Step 2: Moles of Iron produced

From the equation 1 mole of Iron(iii) oxide reacts to produce 2 moles of Fe.

Therefore, the mole ratio of Fe₂O₃ to Fe is 1 : 2.

Thus, moles of Fe = Moles of Fe₂O₃ × 2

                              = 0.156 moles × 2

                              = 0.312 moles

Step 3: Theoretical mass of iron produced

To calculate the mass of iron we multiply the number of moles of iron with the relative atomic mass.

Relative atomic mass = 55.845

Mass of iron = 0.312 moles × 55.845

                    = 17.42 g

Step 4: Percent yield of iron

% yield = (Actual mass ÷ Theoretical mass)×100

            = (15 g ÷ 17.42 g) × 100 %

            = 86.11%

7 0
3 years ago
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