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Alla [95]
3 years ago
5

Can somebody plz just answer this 1 question thx!!!

Chemistry
1 answer:
coldgirl [10]3 years ago
3 0

Answer:

cetacean i think thats the answer

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When 2.3 × 10^3 g of CaCO3 are heated, the actual yield of CaO is 1.09 × 10^3g. What is
iren [92.7K]

The percent yield : 4. 84.58%

<h3>Further explanation</h3>

Reaction

CaCO₃ ⇄ CaO+CO₂

mass CaCO₃ = 2.3 × 10³ g

mol CaCO₃ (MW=100.0869 g/mol) :

\tt mol=\dfrac{mass}{MW}\\\\mol=\dfrac{2.3\times 10^3}{100,0869}\\\\mol=22.98

From the equation, mol CaCO₃ : CaO = 1 : 1, so mol CaO=22.98

mass CaO(MW=56.0774 g/mol)⇒ (theoretical) :

\tt mass=mol\times MW\\\\mass=22.98\times 56,0774\\\\mass=1288.659~g

The percent yield :

\tt \%yield=\dfrac{actual}{theoretical}\times 100\%\\\\\%yield=\dfrac{1090}{1288.659}\times 100\%\\\\\5yield=84.58\%

5 0
3 years ago
Please help
Eddi Din [679]
CHA is not an example of a molecule because there is no element represented by “A”
6 0
3 years ago
(5.0 m3) ( 7.5 mmHg) = (P)(4.0m3)
anzhelika [568]

It looks like we are solving for a pressure.  All that is required is some algebraic manipulation to find our pressure in mmHg.

Given:

(5.0 m³)(7.5 mmHg) = (P)(4.0m³)

Multiply:

37.5 = 4.0P

Divide:

9.375 = P

P = 9.4 mmHg (remember sig figs)

<h3>Answer:</h3>

9.4 mmHg

7 0
3 years ago
Using the accepted value for the volume of 1 gram of water at the temperature of the room that you reported above, what is the a
inessss [21]

Write procedural steps that allow you to demonstrate the sun's role in the water cycle using common material - for each explain what you are modeling and how the materials you have chosen represent nature.

Using the accepted value for the volume of 1 gram of water at the temperature of the room that you reported above, what is the accepted value for the density of water

3 0
3 years ago
Question 1<br> 1 pts<br> How many mols of bromine are present in 35.7g of<br> Tin(IV) bromate?
sleet_krkn [62]

Answer:

n = 0.0814 mol

Explanation:

Given mass, m = 35.7g

The molar mass of Tin(IV) bromate, M = 438.33 g/mol

We need to find the number of moles of bromine. We know that,

No. of moles = given mass/molar mass

So,

n=\dfrac{35.7}{438.33}\\\\n=0.0814\ mol

So, there are 0.0814 moles of bromine in 35.7g of  Tin(IV) bromate.

3 0
3 years ago
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