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The zero order reaction 2N2O→2N2+O2 has the reaction constant k is 6.28×10−3 molL s. If the initial concentration of N2O is 0.96

Talja [164]

Answer: The concentration of $N_2O$ in three significant figures will be 0.899 mol/L.

Explanation:

For the given reaction:

$2N_2O\rightarrow 2N_2+O_2$

The above reaction follows zero order kinetics. The rate law equation for zero order follows:

$k=\frac{1}{t}([A_o]-[A])$

where,

k = rate constant for the reaction = $6.28\times 10^{-3}\text{ mol }L^{-1}s^{-1}$

t = time taken = 10 sec

$[A_o]$ = initial concentration of the reactant = 0.962 mol/L

[A] = concentration of reactant after some time = ?

Putting values in above equation, we get:

$6.28\times 10^{-3}=\frac{1}{10}(0.962-[A])$

$[A]=0.899mol/L$

Hence, the concentration of $N_2O$ in three significant figures will be 0.899 mol/L.

5 0

3 years ago

If 20% of the nucleotides in a DNA molecule are adenine, what percentage of each of the other three bases would be found in this

liubo4ka [24]

What he said is correct!!

5 0

3 years ago

What is the molarity of a sodium carbonate solution containing 32.52 g Na2CO3 dissolved in enough water to make 822 ml of soluti

svet-max [94.6K]

The molarity of a Sodium carbonate solution : 0.373 M

<h3>Further explanation</h3>

Given

32.52 g Na₂CO₃

822 ml of solution = 0.822 L

Required

The molarity

Solution

Molarity shows the number of moles of solute in every 1 liter of solution.

$\large{\boxed {\bold {M ~ = ~ \frac {n} {V}}}$

mol of solute = mol of Na₂CO₃ :

= mass : MW Na₂CO₃

= 32.52 g : 106 g/mol

= 0.307

Molarity :

= n : V

= 0.307 mol : 0.822 L

= 0.373 M

8 0

3 years ago

Each elements on the periodic chart is made up of a certain kind of atom. argree or disagree reason why

Katena32 [7]

Agree cuz its one elemet

8 0

3 years ago

What is the concentration of NaCl in a solution if titration of 15.00 mL of the solution with 0.2503 M AgNO3 requires 20.22 mL o

Nina [5.8K]

Answer:

The concentration of NaCl = 0.3374 M

Explanation:

Given :

Molarity of AgNO₃ = 0.2503 M

Volume of AgNO₃ = 20.22 mL

The conversion of mL into L is shown below:

$1 mL= 10^{-3} L$

Thus, volume of the solution = 20.22×10⁻³ L

Molarity of a solution is the number of moles of solute present in 1 L of the solution.

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

The formula can be written for the calculation of moles as:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

Thus,

$Moles\ of\ AgNO_3 =Molarity \times {Volume\ of\ the\ solution}$

$Moles\ of\ AgNO_3 =0.2503 \times {20.22\times 10^{-3}}\ moles$

$Moles\ of\ AgNO_3 = 5.0611 \times 10^{-3} moles$

The chemical reaction taking place:

$AgNO_3_(aq) + NaCl_(aq) \rightarrow AgCl_(s) + NaNO_3_(aq)$

According to reaction stoichiometry:

1 mole of AgNO₃ reacts with 1 mole of NaCl

Thus,

5.0611×10⁻³ moles of AgNO₃ reacts with 5.0611×10⁻³ moles of NaCl

Thus, moles of NaCl required = 5.0611×10⁻³ moles

Volume of NaCl required = 15.00 mL

The conversion of mL into L is shown below:

$1 mL= 10^{-3} L$

Thus, volume of the solution = 15.00×10⁻³ L

Applying in the formula of molarity as:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$Molarity\ of\ NaCl=\frac{5.0611\times 10^{-3}}{15.00\times 10^{-3}}$

$Molarity\ of\ NaCl= 0.3374 M$

Thus, the concentration of NaCl = 0.3374 M

6 0

3 years ago