The first order reaction is when the rate of the reaction is directly proportional to the concentrations of one of the reactants.
if the reactant whose concentration determines the rate - N
rate equation is
rate = k [N]¹
where k - rate constant
the differential equation is as follows;
[N] = [N₀] e^ (-kt)
where N - concentration after t minutes
N₀ - concentration when t = 0 minutes
k -rate constant
t - time taken in minutes
substituting the values given following equation is obtained
0.088 m = 0.13 e^ (-0.33 *t)
e^ (-0.33 *t) = 0.088/ 0.13
-0.33t = ln 0.676
- 0.33t = -0.39
t = 1.18 minutes
it takes 1.18 minutes for reactant concentration to decrease from 0.13 m to 0.088 m
Answer:
56.5J
Explanation:
<em>To </em><em>find</em><em> </em><em>the </em><em>heat </em><em>energy</em><em> </em><em>required</em><em> </em><em>use </em><em>the </em><em>formula</em><em> </em><em>for </em><em>the </em><em>specific</em><em> </em><em>heat </em><em>capacity</em><em> </em><em>which </em><em>is </em>
<em>c=</em><em>quantity</em><em> of</em><em> </em><em>heat/</em><em>mass×</em><em>c</em><em>h</em><em>a</em><em>n</em><em>g</em><em>e</em><em> </em><em>in </em><em>temperature</em>
<em>in </em><em>this </em><em>question</em><em> </em><em>c </em><em>is </em><em>0</em><em>.</em><em>2</em><em>2</em><em>6</em><em>j</em><em>/</em><em>g,</em><em>the </em><em>mass </em><em>is </em><em>5</em><em>0</em><em>g</em><em> </em><em>and </em><em>the </em><em>change </em><em>in </em><em>temperature</em><em> </em><em>is </em><em>3</em><em>0</em><em>-</em><em>2</em><em>5</em><em>=</em><em>5</em>
<em>therefore</em>
<em>0</em><em>.</em><em>2</em><em>2</em><em>6</em><em>=</em><em>Q/</em><em>5</em><em>0</em><em>×</em><em>5</em>
<em>Q=</em><em>0</em><em>.</em><em>2</em><em>2</em><em>6</em><em>×</em><em>2</em><em>5</em><em>0</em>
<em> </em><em> </em><em> </em><em>=</em><em>56.5J</em>
<em>I </em><em>hope </em><em>this</em><em> helps</em>
Answer:
stronger/ brighter, thats why red is first on the rainbow
Explanation:
Answer:
552 g of LiNO₃
Explanation:
From the question given above, the following data were obtained:
Volume of solution = 2 L
Molarity of LiNO₃ = 4 M
Mass of LiNO₃ =?
Next, we shall determine the number of mole of LiNO₃ in the solution. This can be obtained as follow:
Volume of solution = 2 L
Molarity of LiNO₃ = 4 M
Mole of LiNO₃ =?
Molarity = mole /Volume
4 = mole of LiNO₃ / 2
Cross multiply
Mole of LiNO₃ = 4 × 2
Mole of LiNO₃ = 8 moles
Finally, we shall determine the mass of of LiNO₃ needed to prepare the solution. This is can be obtained as follow:
Mole of LiNO₃ = 8 moles
Molar mass of LiNO₃ = 7 + 14 + (16×3)
= 7 + 14 + 48
= 69 g/mol
Mass of LiNO₃ =?
Mole = mass /Molar mass
8 = Molar mass of LiNO₃ /69
Cross multiply
Molar mass of LiNO₃ = 8 × 69
Molar mass of LiNO₃ = 552 g
Thus, 552 g of LiNO₃ is needed to prepare the solution.
Answer:
It donates a hydrogen ion
Explanation:
Under the Bronsted-Lowry definition of an acid, acids are protons donors which donate the H+ ion, or the hydrogen ion.