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3 years ago

Balance equation of calcium hydroxide+carbon dioxide= calcium carbonate+water for grade 8 in step wise

Chemistry

1 answer:

larisa [96]3 years ago

8 0

Answer:

Ca(OH)2 +CO2========CaCO3 +H2O

Explanation:

since the number of elements in the right is equal to that of the elements in the left side shows it is balanced

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Alik [6]

I think it’s “number” and “type”

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3 years ago

An empty vial weighs 55.32 g. (a) If the vial weighs 185.56 g when filled with liquid mercury (d = 13.53 g/cm3). What i its volu

Tasya [4]

Answer:

a) Volume of vial= 9.626cm3

b) Mass of vial with water = 62.92 g

Explanation:

a) Mass of empty vial = 55.32 g

Mass of Vial + Hg = 185.56 g

Therefore,

$mass\ of\ Hg = 185.56-55.32 = 130.24 g$

Density of Hg = 13.53 g/cm3

$Volume\ of\ vial = Volume\ of\ Hg = \frac{Mass}{Density} \\\\= \frac{130.24g}{13.53g/cm3} = 9.626 cm3$

b) Volume of water = volume of vial = 9.626 cm3

Density of water = 0.997 g/cm3

$Mass\ of\ water = Density*volume = 0.997g/cm3*9.626cm3=9.60 g\\\\Total\ Mass\ of\ vial = Empty\ vial + mass\ of\ water\\= 53.32+9.60= 62.92g$

3 0

3 years ago

Select the letter of a category on which you should focus when proofreading.

oee [108]

B. Who, what, when, where, why, and how. Those are the most important and key componants in a story.

8 0

3 years ago

A certain metal forms a bromide containing percent by mass. what is the chemical formula of the compound?

77julia77 [94]

Answer:

w w w . g o o g l e . c o m

Explanation:

3 0

1 year ago

Consider the reaction CH4(g) 2O2(g)CO2(g) 2H2O(g) Using standard thermodynamic data at 298K, calculate the entropy change for th

zvonat [6]

Answer:

the entropy change for the surroundings when 1.62 moles of CH4(g) react at standard conditions is −8.343 J/K

Explanation:

The balanced chemical equation of the reaction in the question given is:

$CH_{4(g)} + 2O_{2(g)} \to CO_{2(g)} + 2 H_2O _{(g)}$

Using standard thermodynamic data at 298K.

The entropy of each compound above are listed as follows in a respective order.

Entropy of (CH4(g)) = 186.264 J/mol.K

Entropy of (O2(g)) = 205.138 J/mol.K

Entropy of (CO2(g)) = 213.74 J/mol.K

Entropy of (H2O(g)) = 188.825 J/mol.K

The change in Entropy (S) of the reaction is therefore calculated as follows:

$=1*S(CO2(g)) + 2*S(H2O(g)) - 1*S( CH4(g)) - 2*S(O2(g))$

$=1*(213.74) + 2*(188.825) - 1*(186.264) - 2*(205.138)$

= -5.15 J/mol.K

Given that :

the number of moles = 1.62 of CH4(g) react at standard conditions.

Then;

The change in entropy of the rxn $= 1.62 \ mol * -5.15 \ J/mol.K$

= −8.343 J/K

6 0

3 years ago