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3 years ago

Help me please thanks

Chemistry

1 answer:

Evgesh-ka [11]3 years ago

6 0

Then answer would be D. Answer D is correct because you would need to use a better solvent to see the ink separate on the chromatography paper. Hope that helps. :)

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What is the symbol for an atom containing 20 protons and 22 neutrons?

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Answer:

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Explanation:

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3 years ago

The decomposition of hydrogen peroxide was studied, and the following data were obtained at a particular temperature: Time (s) (

harina [27]

Answer:

Part a: The rate of the equation for 1st order reaction is given as $Rate=k[H_2O_2]$

Part b: The integrated Rate Law is given as $[H_2O_2]=[H_2O_2]_0 e^{-kt}$

Part c: The value of rate constant is $7.8592 \times 10^{-4} s^{-1}$

Part d: Concentration after 4000 s is 0.043 M.

Explanation:

By plotting the relation between the natural log of concentration of $H_2O_2$ , the graph forms a straight line as indicated in the figure attached. This indicates that the reaction is of 1st order.

Part a

Rate Law

The rate of the equation for 1st order reaction is given as

$Rate=k[H_2O_2]$

Part b

Integrated Rate Law

The integrated Rate Law is given as

$[H_2O_2]=[H_2O_2]_0 e^{-kt}$

Part c

Value of the Rate Constant

Value of the rate constant is given by using the relation between 1st two observations i.e.

t1=0, M1=1.00

t2=120 s , M2=0.91

So k is calculated as

$-k(t_2-t_1)=ln{\frac{M_2}{M_1}}\\-k(120-0)=ln{\frac{0.91}{1.00}}\\k=\frac{-0.09431}{-120}\\k=7.8592 \times 10^{-4} s^{-1}$

The value of rate constant is $7.8592 \times 10^{-4} s^{-1}$

Part d

Concentration after 4000 s is given as

$-k(t_2-t_1)=ln{\frac{M_2}{1.0}}\\-7.8592 \times 10^{-4}(4000-0)=ln{\frac{M_2}{1.00}}\\-3.1437=ln{\frac{M_2}{1.00}}\\M_2=e^{-3.1437}\\M_2=0.043 M$

Concentration after 4000 s is 0.043 M.

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3 years ago

The prefix milli- means 103 10-2 10-3 102

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3 years ago

How many grams of HCl(aq) are required to react completely with 1.25g of Zn(s) to form ZnCl2(aq) and H2

faltersainse [42]

1.39 g HCl

Explanation:

The balanced chemical equation for this reaction is given by

Zn(s) + 2HCl(aq) ---> ZnCl2(aq) + H2(g)

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1.25 g Zn × (1 mol Zn/65.38 g Zn) = 0.0191 mol Zn

Use the molar ratio to find the # of moles of HCl needed to react completely with the given amount of Zn:

0.0191 mol Zn × (2 mol HCl/1 mol Zn) = 0.0382 mol HCl

Convert this amount to grams:

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3 years ago