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Reptile [31]
3 years ago
6

subtract the mass of the filter paper (0.27 g) from the mass of the paper and copper (0.98 g). Record the difference in the data

table as the amount of copper.
Chemistry
1 answer:
vova2212 [387]3 years ago
7 0
When measuring copper that is collected on a filter paper, the mass measurement obtained will include both the mass of the filter paper and copper.
therefore we have to find the mass of the empty filter paper before copper is on the paper.
the mass of the filter paper alone is 0.27 g
then the mass of copper and filter paper is 0.98 g
therefore to find the mass of copper alone we have to take the difference between the 2 masses
mass of copper = (mass of copper + filter paper ) - mass of filter paper
                          = 0.98 g - 0.27 g 
mass of copper = 0.71 g
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Answer:

7.186

Explanation:

The mean is the average of some given data from a sample point.

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From the given pH of the solutions, we can form a table:

      x                                     f                                          fx

    7.15                                  1                                         7.15

    7.16                                  1                                         7.16

    7.18                                  1                                         7.18

    7.19                                  1                                         7.19

    7.20                                 3                                        21.6

    7.21                                  1                                         7.21

Now ∑fx = 7.15 +7.16 +7.18 + 7.19 + 7.20 + 7.21 = 57.49

         ∑f = 1 + 1 + 1 + 1 + 3 + 1 = 8

The mean = \frac{57.49}{8} = 7.18625 = 7.186

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3 years ago
Explain why the larger the hydrocarbon molecule is the more
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6 0
3 years ago
Read 2 more answers
Deterioration of buildings, bridges, and other structures through the rusting of iron costs millions of dollars a day. The actua
GrogVix [38]

The value of ∆H when 0.250kg of iron rusts is -1.846 × 10³kJ.

The rust forms when 4.85X10³ kJ of heat is released is 888.916 g.

<h3>Chemical reaction:</h3>

4 Fe + 3O2 ------ 2Fe2O3

∆H = -1.65×10³kJ

A) Given,

mass of iron = 0.250kg = 250 g

<h3>Calculation of number of moles</h3>

moles = given mass/ molar mass

= 250/ 55.85 g/mol.

= 4.476 mol

As we know that,

For the rusting of 4 moles of Fe, ∆H = -1.65×10³kJ

For the rusting of 4.476 moles of Fe ∆H required can be calculated as

-1.65×10³kJ × 4.476 mol/ 4mol

∆H required = -1.846 × 10³kJ

Now,

when 2 mol of Fe2O3 formed, ∆H = - 1.65×10³kJ

It can be said that,

-1.65×10³kJ energy released when 2 mol of Fe2O3 formed

So, -4.6 × 10³kJ energy released when 2 mol of Fe2O3 formed

= 2 × -4.6 × 10³kJ / -1.65×10³kJ

= 5.57 mol of Fe2O3 formed

Now,

mass of Fe2O3 formed = 5.57 mol × 159.59 g/mol

= 888.916 g

Thus, we calculated that the rust forms when 4.85X10³ kJ of heat is released is 888.916 g. and the value of ∆H when 0.250kg of iron rusts is -1.846 × 10³kJ.

learn more about ∆H:

brainly.com/question/24170335

#SPJ4

DISCLAIMER:

The given question is incomplete. Below is the complete question

QUESTION:

Deterioration of buildings, bridges, and other structures through the rusting of iron costs millions of dollars a day. The actual process requires water, but a simplified equation is 4Fe(s) + 3O₂(g) → 2Fe₂O₃(s) ΔH = -1.65×10³kJ

a) What is the ∆H when 0.250kg iron rusts.

(b) How much rust forms when 4.85X10³ kJ of heat is released?

7 0
1 year ago
In an isothermal gas chromatography experiment using an ECD detector, 1.69 nmols of nchlorohexane, C6H13Cl, was added as an inte
prohojiy [21]

Answer:

The amount of Chlorodecane in the unknown is 0.105nmols

Explanation:

a) Since the GC is in an isothermal state, Chlorohexane C6H13Cl (1.69 nmols) because of its lower boiling point will elute first and Chlorodecane C12H21Cl will elute second.

The area of the first peak corresponding to Chlorohexane is 32434 units.

The area of the second peak corresponding to chlorodecane is 2022 units.

Since the response factor of the compound is not given in question and considering the response factor is same for both the compounds, the answer will be as follow:

1.69 nmols of Chlorohexane gives 32434 units

How much of chlorodecane gives 2022 units

By cross multiplication;

Moles of Chlorodecane = 2022*1.69/32434

                                         =0.105nmols

4 0
3 years ago
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