Answer:
Option (B) 3.
Explanation:
In the model represented above, the two extreme represent carbon atoms since no other group are attached to it. The joint at the middle also represent carbon atom.
Thus, we can write a more simplify illustration for the model above as
C—C—C
From the above illustration, we can see that the model contains 3 carbon atom.
Answer:- 1.62 moles
Solution:- At constant temperature and pressure, volume is directly proportional to the moles of the gas.
from given data, 
= 5.17 L, 
= 1.05 moles
= 8.00 L, 
= ?
Let's plug in the values in the formula:
On cross multiply:
= 1.62 moles
So, now the toy contains 1.62 moles of the air.
In a chemical reaction, 0. 3 moles of H2O result in the production of 0.2 moles of CO2. Chemical reactions are the means through which one group of chemical compounds are changed into another.
Chemical reactions are typically defined as changes that only affect the positions of electrons in the formation and breakage of chemical bonds between atoms, with no change to the nuclei (i.e., no change to the elements present). These types of changes are often included in the term chemical reactions.
Number of moles Definition We utilize this enormous quantity to measure atoms. Additionally, it equals the 6.022* 10 23 atoms that make up 12 grammes of carbon-12, or atoms.
C2H6 + 7/2O2 = 2CO2 + 3H2O,
where moles(CO2)=(2*0,3)/3=0.2 mol,
and n(CO2)=(2*0,3)/3=n(H2O);
n(CO2)=n(CO2)=n(H2O).
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Innovated Understanding Curious Logical
Answer:
2.1056L or 2105.6mL
Explanation:
We'll begin by calculating the number of mole in 10g of Na2CO3. This can be obtained as follow:
Molar mass of Na2CO3 = (23x2) + 12 + (16x3) = 106g/mol
Mass of Na2CO3 = 10g
Mole of Na2CO3 =.?
Mole = mass /molar mass
Mole of Na2CO3 = 10/106
Mole of Na2CO3 = 0.094 mole
Next, we shall determine the number of mole CO2 produced by the reaction of 0.094 mole of Na2CO3. This is illustrated below:
Na2CO3 + 2HCl —> 2NaCl + H2O + CO2
From the balanced equation above,
1 mole of Na2CO3 reacted to produce 1 mole of CO2.
Therefore, 0.094 mole of Na2CO3 will also react to 0.094 mole of CO2.
Next, we shall determine the volume occupied by 0.094 mole of CO2 at STP. This is illustrated below:
1 mole of a gas occupy 22.4L at STP. This implies that 1 mole CO2 occupies 22.4L at STP.
Now, if 1 mole of CO2 occupy 22.4L at STP, then, 0.094 mole of CO2 will occupy = 0.094 x 22.4 = 2.1056L
Therefore, the volume of CO2 produced is 2.1056L or 2105.6mL
