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Verdich [7]
3 years ago
13

How many total electron will fit in the 1st energy level

Chemistry
2 answers:
Vlad1618 [11]3 years ago
7 0

2, only 2 electrons fit in the 1st energy level or ring

nevsk [136]3 years ago
4 0
2 elections will fill the first energy level
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The symbol for gold is<br> a. go.<br> b. g.<br> c. au.<br> d. au.
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What is the median number of seeds that germinate at 32
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consider the reaction between calcium oxide and carbon dioxide: cao ( s ) + co 2 ( g ) → caco 3 ( s ) a chemist allows 14.4 g of
MakcuM [25]

Answer:

CaO is the limiting reagent

Theoritical yield = 25.71 g

% Yield = 75.44%

Explanation:

1 mole = Molar mass of the substance

Molar Mass of CaO = 56 g/mol

Molar Mass of CaCO3 = 100 g/mol

Molar mass of CO2 = 44 g/mol

The balanced Equation is :

CaO + CO_{2}\rightarrow CaCO_{3}

1 mole of CaO reacts with = 1 mole of CO2

56 g of CaO reacts with = 44 g of CO2

1 g of CaO reacts with =

\frac{44}{56}

= 0.785 g of CO2

So,

<u>14.4 g of CaO</u><u> </u>must react with = (14.4 x 0.785) g of CO2

= 11.31 g of CO2

<u>Needed = 11.31 g</u>

<u>Available CO2  = 13.8 g</u><u> </u>(given)

So CO2 is in excess , hence<u> CaO is the limiting reagent and product will produce from 14.4 g of CaO</u>

CaO + CO_{2}\rightarrow CaCO_{3}

1 mole of CaO will produce 1 mole pf CaCO3

56 g of CaO produce = 100 g of CaCO3

1 g  of CaO produce =

\frac{100}{56}

= 1.785 g of CaCO3

14.4 g of CaO will produce = (1.785 x 14.4) g of CaCO3

= 25.71 g of CaCO3

Theoritical Yield of CaCO3 = 25.71 g

Actual yield = 19.4 g

Percent Yield =

\frac{Actual\ yield}{Theoritical\ yield}\times 100

\frac{19.4}{25.71}\times 100

= 75.44 %

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3 years ago
To compare how different types of soil absorbed water, Christian set up an experiment. He poked equal numbers and sizes of holes
lilavasa [31]

Answer:

The empty cup

Explanation:

Got it right on the test

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