Answer:
k = 1.3 x 10⁻³ s⁻¹
Explanation:
For a first order reaction the integrated rate law is
Ln [A]t/[A]₀ = - kt
where [A] are the concentrations of acetaldehyde in this case, t is the time and k is the rate constant.
We are given the half life for the concentration of acetaldehyde to fall to one half its original value, thus
Ln [A]t/[A]₀ = Ln 1/2[A]₀/[A]₀= Ln 1/2 = - kt
- 0.693 = - k(530s) ⇒ k = 1.3 x 10⁻³ s⁻¹
<span>People with protanopia are unable to sense any ‘red’ light, people with deuteranopia do not sense ‘green’ light and people with tritanopia cannot sense ‘blue’ light. If a person perceives the color green, then the yellow sensitive nerves must work somewhat effectively since green is a combination of yellow and blue. Red-sensitive nerves are most likely not responding properly for this person. The answer is C.</span>
Same as balancing a regular chemical reaction! Please see the related question to the bottom of this answer for how to balance a normal chemical reaction. This is for oxidation-reduction, or redox reactions ONLY! These instructions are for how to balance a reduction-oxidation, or redox reaction in aqueous solution, for both acidic and basic solution. Just follow these steps! I will illustrate each step with an example. The example will be the dissolution of copper(II) sulfide in aqueous nitric acid, shown in the following unbalanced reaction: CuS (s) + NO 3 - (aq) ---> Cu 2+ (aq) + SO 4 2- (aq) + NO (g) Step 1: Write two unbalanced half-reactions, one for the species that is being oxidized and its product, and one for the species that is reduced and its product. Here is the unbalanced half-reaction involving CuS: CuS (s) ---> Cu 2+ (aq) + SO 4 2- (aq) And the unbalanced half-reaction for NO 3 - is: NO 3 - (aq) --> NO (g) Step 2: Insert coefficients to make the numbers of atoms of all elements except oxygen and hydrogen equal on the two sides of each half-reaction. In this case, copper, sulfur, and nitrogen are already balanced in the two half-reaction, so this step is already done here. Step 3: Balance oxygen by adding H 2 O to one side of each half-reaction. CuS + 4 H 2 O ---> Cu 2+ + SO 4 2- NO 3 - --> NO + 2 H 2 O Step 4: Balance hydrogen atoms. This is done differently for acidic versus basic solutions. . For acidic solutions: Add H 3 O + to each side of each half-reaction that is "deficient" in hydrogen (the side that has fewer H's) and add an equal amount of H 2 O to the other side. For basic solutions: add H 2 O to the side of the half-reaction that is "deficient" in hydrogen and add an equal amount of OH - to the other side. Note that this step does not disrupt the oxygen balance from Step 3. In the example here, it is in acidic solution, and so we have: CuS + 12 H 2 O ---> Cu 2+ + SO 4 2- + 8 H 3 O + . NO 3 - + 4 H 3 O + --> NO + 6 H 2 O Step 5: Balance charge by inserting e - (electrons) as a reactant or product in each half-reaction. Oxidation: CuS + 12 H 2 O ---> Cu 2+ + SO 4 2- + 8 H 3 O + + 8 e - . Reduction: NO 3 - + 4 H 3 O + + 3 e - --> NO + 6 H 2 O . Step 6: Multiply the two half-reactions by numbers chosen to make the number of electrons given off by the oxidation step equal to the number taken up by the reduction step. Then add the two half-reactions. If done correctly, the electrons should cancel out (equal numbers on the reactant and product sides of the overall reaction). If H 3 O + , H 2 O, or OH - appears on both sides of the final equation, cancel out the duplication also. Here the oxidation half-reaction must be multiplied by 3 (so that 24 electrons are produced) and the reduction half-reaction must by multiplied by 8 (so that the same 24 electrons are consumed). 3 CuS + 36 H 2 O ---> 3 Cu 2+ + 3 SO 4 2- + 24 H 3 O + + 24 e - 8 NO 3 - + 32 H 3 O + + 24 e - ---> 8 NO + 48 H 2 O Adding these two together gives the following equation: 3 CuS + 36 H 2 O + 8 NO 3 - + 8 H 3 O + ---> 3 Cu 2+ + 3 SO 4 2- + 8 NO + 48 H 2 O Step 7: Finally balancing both sides for excess of H 2 O (On each side -36) This gives you the following overall balanced equation at last: 3 CuS (s) + 8 NO 3 - (aq) + 8 H 3 O + (aq) ---> 3 Cu 2+ (aq) + 3 SO 4 2- (aq) + 8 NO (g) + 12 H 2 O (l)
2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O
c₁=2.00 mol/L
v₁=0.25 L
v₂=2.00 L
c₂-?
n(NaOH)=c₂v₂
n(H₂SO₄)=c₁v₁
n(NaOH)=2n(H₂SO₄)
c₂v₂=2c₁v₁
c₂=2c₁v₁/v₂
c₂=2*2.00*0.25/2.00=0.5 mol/L
0.5 M NaOH
Explanation:
Dipole moment is defined as the measurement of the separation of two opposite electrical charges.
is a bent shaped molecule with a dipole moment of 1.87.
is also a bent shaped molecule with a dipole moment of 1.10.
is a also a bent shaped molecule and has a negligible dipole moment.
has a dipole moment of 0.29.
Therefore, given molecules are arranged according to their increasing dipole moment as follows.
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