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3 years ago

Mercury(II) oxide (HgO) decomposes to form mercury (Hg) and oxygen (O2). The balanced chemical equation is shown below.

Chemistry

2 answers:

Andreas93 [3]3 years ago

8 0

Answer:

15.63 moles of HgO are needed to produce 250.0 g of $O_{2}$

Explanation:

According to given balanced equation, 1 mole of $O_{2}$ is produced from 2 moles of HgO.

Number of moles $=\frac{mass}{molar mass}$

So 250 g of $O_{2}$ $=\frac{250}{32}moles O_{2}$

So moles of HgO needed $=2\times \frac{250}{32}moles HgO$

$=15.63 moles HgO$

____ [38]3 years ago

5 0

The answer to the question is C. 15.63

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A gas has a volume of 590ml at a tempature of -55.0 Celsius. What volume will the gas occupy at 30.0 Celsius? Show your work

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3 years ago

I really need someone to explain how to find the average atomic mass.... the gizmos was really confusing because they also had w

evablogger [386]

Answer:

Option C. 52.057

Explanation:

The following data were obtained from the question:

Isotope >> Mass number > Percentage

A (Cr-50) > 50 >>>>>>>>>> 4.3

B (Cr-52) > 52 >>>>>>>>>> 83.8

C (Cr-53) > 53 >>>>>>>>>> 9.5

D (Cr-54) > 54 >>>>>>>>>> 2.4

Average atomic mass =?

The average atomic mass of chromium, Cr can be obtained as follow:

Average atomic mass = [(Mass of A × A%) /100] + [(Mass of B × B%) /100] + [(Mass of C × C%) /100] + [(Mass of D × D%) /100]

Atomic mass of Cr = [50×4.3)/100] + [52×83.8)/100] + [53×9.5)/100] + [54×2.4)/100]

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Therefore, the atomic mass of chromium, Cr is 52.057

5 0

3 years ago

The tabulated data were collected for this reaction:

erastovalidia [21]

Answer:

ai) Rate law, $Rate = k [CH_3 Cl] [Cl_2]^{0.5}$

aii) Rate constant, k = 1.25

b) Overall order of reaction = 1.5

Explanation:

Equation of Reaction:

$CH_{3} Cl (g) + 3 Cl_2 (g) \rightarrow CCl_4 (g) + 3 HCl (g)$

If $A + B \rightarrow C + D$ , the rate of backward reaction is given by:

$Rate = k [A]^{a} [B]^{b}\\k = \frac{Rate}{ [A]^{a} [B]^{b}}\\k = \frac{Rate}{ [CH_3 Cl]^{a} [Cl_2]^{b}}$

k is constant for all the stages

Using the information provided in lines 1 and 2 of the table:

$0.014 / [0.05]^a [0.05]^b = 00.029/ [0.100]^a [0.05]^b\\0.014 / [0.05]^a [0.05]^b = 00.029/ [2*0.05]^a [0.05]^b\\0.014 / = 0.029/ 2^a\\2^a = 2.07\\a = 1$

Using the information provided in lines 3 and 4 of the table and insering the value of a:

$0.041 / [0.100]^a [0.100]^b = 0.115 / [0.200]^a [0.200]^b\\0.041 / [0.100]^a [0.100]^b = 0.115 / [2 * 0.100]^a [2 * 0.100]^b\\$

$0.041 = 0.115 / [2 ]^a [2]^b\\ \[[2 ]^a [2]^b = 0.115/0.041\\ \[[2 ]^a [2]^b = 2.80\\\[[2 ]^1 [2]^b = 2.80\\\[[2]^b = 1.40\\b = \frac{ln 1.4}{ln 2} \\b = 0.5$