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2 years ago

4

A reaction was experimentally determined to follow the rate law, Rate = k[A] where k = 0.15 s-1. Starting with [A]o = 0.225M, ho

w many seconds will it take for [A]t = 0.0350M?

1 answer:

arlik [135]2 years ago

3 0

Answer:

The time the reaction takes is 12.4s

Explanation:

A reaction that follows the rate law:

Rate =k[A] is order 1 and follows the equation:

Ln[A] = -kt + ln[A]₀

<em>Where [A] is concentration of the reaction after time t: 0.0350M</em>

<em>k is rate constant = 0.15s⁻¹</em>

<em>t is time in seconds</em>

<em>[A]⁰ is initial concentration = 0.225M</em>

<em />

Ln[0.0350] = -0.15s⁻¹*t + ln[0.225]

-1.86075 = -0.15s⁻¹*t

12.4s = t

<h3>The time the reaction takes is 12.4s</h3>

<em />

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vredina [299]

Answer:

volume in L = 0.25 L

Explanation:

Given data:

Mass of Cu(NO₃)₂ = 2.43 g

Volume of KI = ?

Solution:

Balanced chemical equation:

2Cu(NO₃)₂ + 4KI → 2CuI + I₂ + 4KNO₃

Moles of Cu(NO₃)₂:

Number of moles = mass/ molar mass

Number of moles = 2.43 g/ 187.56 g/mol

Number of moles = 0.013 mol

Now we will compare the moles of Cu(NO₃)₂ with KI.

Cu(NO₃)₂ : KI

2 : 4

0.013 : 4 × 0.013=0.052 mol

Volume of KI:

<em>Molarity = moles of solute / volume in L</em>

volume in L = moles of solute /Molarity

volume in L = 0.052 mol / 0.209 mol/L

volume in L = 0.25 L

6 0

2 years ago

You have a mixture that contains 0.380 moles of Ne(g), 0.250 moles of He(g), and 0.500 moles CH4(g) at 400 K and 7.25 atm. What

boyakko [2]

I think it is the pacific ocean

5 0

2 years ago

Suppose you wanted to make a buffer of exactly ph 7.00 using kh2po4 and na2hpo4. if the final solution was 0.10 m in kh2po4, wha

OleMash [197]

Answer:- 0.138 M

Solution:- The buffer pH is calculated using Handerson equation:

$pH=pKa+log(\frac{base}{acid})$

$KH_2PO_4$ acts as a weak acid and $Na_2HPO_4$ as a base which is pretty conjugate base of the weak acid we have.

The acid hase two protons(hydrogen) where as the base has only one proton. So, we could write the equation as:

$H_2PO_4^-\rightleftharpoons H^++HPO_4^-^2$

Phosphoric acid gives protons in three steps. So, the above equation is the second step as the acid has only two protons and the base has one proton.

So, we will use the second pKa value. The acid concentration is given as 0.10 M and we are asked to calculate the concentration of the base to make a buffer of exactly pH 7.00.

Let's plug in the values in the equation:

$7.00=6.86+log(\frac{base}{0.10})$

$7.00-6.86=log(\frac{base}{0.10})$

$0.14=log(\frac{base}{0.10})$

Taking antilog:

$10^0^.^1^4=\frac{base}{0.10}$

$1.38=\frac{base}{0.10}$

On cross multiply:

[base] = 1.38(0.10)

[base] = 0.138

So, the concentration of the base that is $Na_2HPO_4$ required to make the buffer is 0.138M.

5 0

2 years ago

The volume of distilled water that should be added to 10.0 mL of 6.00 M HCl(aq) in order to prepare a 0.500 M HCl(aq) solution i

bija089 [108]

Answer:

110ml

Explanation:

<em>Using the dilution equation, C1V1 = C2V2</em>

<em>Where C1 is the initial concentration of solution</em>

<em>C2 is final concentration of solution</em>

<em>V1 is intital volume of solution</em>

<em>V2 is final volume of solution.</em>

From the question , C1=6M, C2=0.5M, V1=10ml, V2=?

$V2 =\frac{C1V1}{C2}$

$V2 =\frac{10*6}{0.5}$

$V2 =120ml$

volume of water added = final volume -initial volume

= 120-10

=110ml

3 0

2 years ago

What would be the composition and ph of an ideal buffer prepared from lactic acid (ch3chohco2h), where the hydrogen atom highlig

Mashutka [201]

Answer:

$P_H =2.86$

$c=1.4\times 10^{-4}$

Explanation:

first write the equilibrium equaion ,

$C_3H_6O_3$ ⇄ $C_3H_5O_3^{-} +H^{+}$

assuming degree of dissociation $\alpha$ =1/10;

and initial concentraion of $C_3H_6O_3$ =c;

At equlibrium ;

concentration of $C_3H_6O_3 = c-c\alpha$

$[C_3H_5O_3^{-} ]= c\alpha$

$[H^{+}] = c\alpha$

$K_a = \frac{c\alpha \times c\alpha}{c-c\alpha}$

$\alpha$ is very small so $1-\alpha$ can be neglected

and equation is;

$K_a = {c\alpha \times \alpha}$

$[H^{+}] = c\alpha$ = $\frac{K_a}{\alpha}$

$P_H =- log[H^{+} ]$

$P_H =-logK_a + log\alpha$

$K_a =1.38\times10^{-4}$

$\alpha = \frac{1}{10}$

$P_H= 3.86-1$

$P_H =2.86$

composiion ;

$c=\frac{1}{\alpha} \times [H^{+}]$

$[H^{+}] =antilog(-P_H)$

$[H^{+} ] =0.0014$

$c=0.0014\times \frac{1}{10}$

$c=1.4\times 10^{-4}$

6 0

2 years ago