Answer:
D.
Explanation:
Adding more gas particles to a set volume will increase the number of collisions, thus increasing collision force and pressure.
Answer:
using a more concentrated potassium hydroxide
Explanation:
<em>The option that would likely increase the rate of reaction would be to use a more concentrated potassium hydroxide.</em>
<u>The concentration of reactants is one of the factors that affect the rate of reaction. The more the concentration of the reactants, the faster the rate of reaction. </u>
Granted that there are enough of the other reactants, increasing the concentration of one of the reactants will lead to an increased rate of reaction.
Hence, using a more concentrated potassium hydroxide which happens to be one of the reactants would likely increase the rate of reaction.
Models in science help you get the idea of what something looks like that's why your teacher may ask you to draw a diagram to help you remember what the object looks like.
Hope this helps.
We know that, M1V1 = M2V2
(Initial) (Final)
where, M1 and M2 are initial and final concentration of soution respectively.
V1 and V2 = initial and final volume of solution respectively
Given: M1 = 12 m, V1 = 35 ml and V2 = 1.2 l = 1200 ml
∴ M2 = M1V1/V2 = (12 × 35)/ 1200 = 0.35 m
Final concentration of solution is 0.35 m
Answer:
-2.86x10³ kJ
Explanation:
The enthalpy of a reaction (ΔH) is defined as the heat produced or consumed by a reaction. In the reaction:
2 C₂H₆(g) + 7 O₂(g) → 4 CO₂(g) + 6 H₂O(g)
The ΔH is the heat envolved in the reaction per 2 moles of C₂H₆. 1.43x10³ kJ are involved when 1 mole reacts. Thus, when 2 moles react, involved heat is:
1.43x10³ kJ ₓ 2 = <em>2.86x10³ kJ</em>. As the reaction is a combustion reaction (Produce CO₂ and H₂O), the heat involved in the reaction is <em>PRODUCED, </em>that means ΔH is negative, <em>-2.86x10³ kJ</em>
