Answer:
179 L of CO2
Explanation:
Given the equation of the reaction;
C2H6(g) + 7/2 O2(g) -------> 2CO2(g) + 3H2O(g)
Now 1 mole of ethane yields 2 moles of CO2 from the balanced reaction equation
1 mole of a gas occupies 22.4 L volume so,
22.4 L of ethane yields 44.8 L of CO2
89.5 L of ethane yields 89.5 * 44.8/22.4 = 179 L of CO2
Answer:
125000
Explanation:
Because it is halved and halved again.
Answer:
Explanation:
The formula of the reaction:
KClO₂ → KCl + O₂
To assign oxidation numbers, we have to obey some rules:
- Elements in an uncombined state or one whose atoms combine with one another to form molecules have an oxidation number of zero.
- The charge on simple ions signifies their oxidation number.
- The algebraic sum of all the oxidation number of all atoms in a neutral compound is zero. For radicals with charges, their oxidation number is the charge.
The oxidation number of K in KClO₂:
K + (-1) + 2(-2) = 0
K-5 = 0
K = +5
The oxidation number of K in KCl:
K + (-1) = 0
K = +1
The oxidation number Cl in KClO₂ is -1
For Cl in KCl, the oxidation number is -1
For O in KClO₂, the oxidation number is (2 x -2) = -4
For O in O₂, the oxidation number is 0
K moves from an oxidation state of +5 to +1. This is a gain of electrons and K has undergone reduction. We then say K is reduced.
O moves from an oxidation state of -4 to 0. This is a loss of electrons and O has undergone oxidation. We say O is oxidized.
It should be 24g of carbon
Answer:
Oxygen and sulfur have similar chemical properties because both elements have six electrons in their outermost electron shells. Indeed, both oxygen and sulfur form molecules with two hydrogen atoms: water and hydrogen sulfide
Explanation:
