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Kazeer [188]
3 years ago
15

How many grams of carbon dioxide will form if 5.5 g of C3H8 burns in 15 g of O2?

Chemistry
1 answer:
mr Goodwill [35]3 years ago
5 0
C3H8+3O2--->3CO2+8H
Therefore for every 1:3 there are 3 Carbon dioxides that form. That means find the limiting reactant from the two reactants.
5.5g(1mole C3H8/44.03g of C3H8)=0.1249 moled of C3H8 and if for every one C3H8 we can form three CO2. We can assume 0.3747 miles of CO2 will be produced.
15g of O2(1 mole O2/32g of O2)=0.4685moles O2 and if for every three O2 we can produce three CO2 we may assume a 1:1 ratio.
This means C3H8 will be your limiting reactant. Therefore 0.3747 moles of CO2 will be produced.
0.3747 moles of CO2(48.01 g of CO2/1 mole of CO2)= 17.99 grams of CO2
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What are the two types of numbers in experimental calculations?
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8 0
3 years ago
It takes 167 s for an unknown gas to effuse through a porous wall and 99 s for the same volume of n2 gas to effuse at the same t
irakobra [83]

Answer : The molar mass of the unknown gas will be 79.7 g/mol


Explanation : To solve this question we can use graham's law;


Now we can use nitrogen as the gas number 2, which travels faster than gas 1;

So, 167 / 99 = 1.687 So the nitrogen gas is 1.687 times faster that the unknown gas 1

We can compare the rates of both the gases;


So here, Rate of gas 2 / Rate of gas 1 = \sqrt{(molar mass 1 / molar mass 2)}

Now, 1.687 = square root [\sqrt{(molar mass 1) / (28.01 g/mol N_{2})} ]


When we square both the sides we get;


2.845 = (molar mass 1) / (28.01 g/mol N2)


On rearranging, we get,


2.845 X (28.01 g/mol N2) = Molar mass 1

So the molar mass of unknown gas will be = 79.7 g/mol

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3 years ago
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drek231 [11]
<h3>Solution-:</h3>

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#<em>o</em><em>f</em><em>f</em><em>i</em><em>c</em><em>a</em><em>i</em><em>l</em><em> </em><em>Nazo</em>

<em>ll </em><em>Radhe</em><em> Radhe</em><em> ll</em>

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2 years ago
Using VSEPR theory which of the following would be the correct shape for nitrogen trifluoride?
RUDIKE [14]

Answer:

trigonal pyramidal

Explanation:

In NF3, the nitrogen atom is sp3 hybridized. Now we must remember that according to the VSEPR theory, the number of electron pairs in the valence shell of the central atom in a molecule determines its shape.

Here, the nitrogen atom is the central atom and its outermost shell is surrounded by four electron pairs - one lone pair and three bond pairs. This means that it has a tetrahedral electron pair geometry.

However, due to the lone pair, the three fluorine atoms are arranged in a  trigonal pyramidal geometry. Hence the correct shape of the molecule is trigonal pyramidal.

6 0
2 years ago
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