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suter [353]
3 years ago
14

Help with all of this fill in the blank if it’s right you will be marked as brain list

Chemistry
1 answer:
seraphim [82]3 years ago
6 0

Answer:

your bad dude

Explanation:

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What is the mass of 3.35 mol Hg(IO3)2? 1,700 g 1,840 g 1,960 g 2,110 g
vichka [17]

Answer:- 1840 g.

Solution:- We have been given with 3.35 moles of  and asked to calculate it's mass.

To convert the moles to grams we multiply the moles by the molar mass of the compound. Molar mass of the compound is the sum of atomic masses of all the atoms present in it.

molar mass of  = atomic mass of Hg + 2(atomic mass of I) + 6(atomic mass of O)

= 200.59+2(126.90)+6(16.00)

= 200.59+253.80+96.00

= 550.39 gram per mol

Let's multiply the given moles by the molar mass:

3.35mol(\frac{550.39g}{1mol})

= 1843.8 g

Since, there are three sig figs in the given moles of compound, we need to round the calculated my to three sig figs also. So, on rounding off to three sig figs the mass becomes 1840 g.



5 0
3 years ago
Briefly explain whether the following procedural errors
cluponka [151]
Can you please tell me what the procedures are

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3 years ago
Which is a characteristic of a decomposition reaction? Substance P replaces X in the compound XY. Substance XYZ breaks down into
Law Incorporation [45]

Substance P replaces X in the compound XY

this is the characteristic of decomposition reaction

6 0
3 years ago
A mixture of 15.0 g of the anesthetic halothane (C2HBrClF3 197.4 g/mol) and 22.6 g of oxygen gas has a total pressure of 862 tor
AlexFokin [52]

Answer : The partial pressure of C_2HBrClF_3 and O_2 are, 84 torr and 778 torr respectively.

Explanation : Given,

Mass of C_2HBrClF_3 = 15.0 g

Mass of O_2 = 22.6 g

Molar mass of C_2HBrClF_3 = 197.4 g/mole

Molar mass of O_2 = 32 g/mole

First we have to calculate the moles of C_2HBrClF_3 and O_2.

\text{Moles of }C_2HBrClF_3=\frac{\text{Mass of }C_2HBrClF_3}{\text{Molar mass of }C_2HBrClF_3}=\frac{15.0g}{197.4g/mole}=0.0759mole

and,

\text{Moles of }O_2=\frac{\text{Mass of }O_2}{\text{Molar mass of }O_2}=\frac{22.6g}{32g/mole}=0.706mole

Now we have to calculate the mole fraction of C_2HBrClF_3 and O_2.

\text{Mole fraction of }C_2HBrClF_3=\frac{\text{Moles of }C_2HBrClF_3}{\text{Moles of }C_2HBrClF_3+\text{Moles of }O_2}=\frac{0.0759}{0.0759+0.706}=0.0971

and,

\text{Mole fraction of }O_2=\frac{\text{Moles of }O_2}{\text{Moles of }C_2HBrClF_3+\text{Moles of }O_2}=\frac{0.706}{0.0759+0.706}=0.903

Now we have to partial pressure of C_2HBrClF_3 and O_2.

According to the Raoult's law,

p^o=X\times p_T

where,

p^o = partial pressure of gas

p_T = total pressure of gas

X = mole fraction of gas

p_{C_2HBrClF_3}=X_{C_2HBrClF_3}\times p_T

p_{C_2HBrClF_3}=0.0971\times 862torr=84torr

and,

p_{O_2}=X_{O_2}\times p_T

p_{O_2}=0.903\times 862torr=778torr

Therefore, the partial pressure of C_2HBrClF_3 and O_2 are, 84 torr and 778 torr respectively.

6 0
3 years ago
What characteristics do atoms of carbon-12, carbon-13, and carbon-14 have in common?
Sophie [7]
Their proton and electron numbers are same
7 0
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