Answer:
Explanation:
When the amount of H2O2 is doubled while KI is kept constant, the rate of reaction doubles.
When the amount of KI is doubled and the amount of H2O2 is halved, the rate stays nearly constant.
2H2O2 (aq) → O2(g) + 2H2O (l) ------------- first order kinetics reaction.
Catalysts are KI, FeCl3 only, KCl is not a catalyst. Order: KI < MnO2 < Pb < FeCl3.
H2O2 + I– -> IO– + H2O (Step 1)
H2O2 + IO– -> I– + H2O + O2 (Step 2)
It can be seen that the iodine ion (provided by the KI solution) is a product as well as a reactant.
02(g)2Fe? (aq) + 2 H(a) 2 H 2 Fe3 (aq) H2O2(aq) + 2 Fe,Taq) H02(aq) 2 Fe (aq) 2 H (aq)
Answer:
= 61.25 g
= 88.75 g
Explanation:
= 
= 50 g
⇒ 
= 
= 1.25 (moles)
2NaOH + H2SO4 ⇒ Na2SO4 + 2H2O
2 : 1 : 1 : 2
1.25 (moles)
⇒ 
= 1.25 × 1 ÷ 2 = 0.625 (moles) ⇒ 
= 0.625 × 98 = 61.25 g
= 1.25 × 1 ÷ 2 = 0.625 (moles) ⇒ = 0.625 × 142 = 88.75 g
1.062 mol/kg.
<em>Step 1</em>. Write the balanced equation for the neutralization.
MM = 204.22 40.00
KHC8H4O4 + NaOH → KNaC8H4O4 + H2O
<em>Step 2</em>. Calculate the moles of potassium hydrogen phthalate (KHP)
Moles of KHP = 824 mg KHP × (1 mmol KHP/204.22 mg KHP)
= 4.035 mmol KHP
<em>Step 3</em>. Calculate the moles of NaOH
Moles of NaOH = 4.035 mmol KHP × (1 mmol NaOH/(1 mmol KHP)
= 4.035 mmol NaOH
<em>Step 4</em>. Calculate the mass of the NaOH
Mass of NaOH = 4.035 mmol NaOH × (40.00 mg NaOH/1 mmol NaOH)
= 161 mg NaOH
<em>Step 5</em>. Calculate the mass of the water
Mass of water = mass of solution – mass of NaOH = 38.134 g - 0.161 g
= 37.973 g
<em>Step 6</em>. Calculate the molal concentration of the NaOH
<em>b</em> = moles of NaOH/kg of water = 0.040 35 mol/0.037 973 kg = 1.062 mol/kg
The wavelength and frequency of light are closely related. The higher the frequency, the shorter the wavelength. ... The equation that relates wavelength and frequency for electromagnetic waves is: λν=c where λ is the wavelength, ν is the frequency and c is the speed of light.
