Answer:
it is iodine it seems very right
Reduction half reaction: Cu²⁺(aq) + 2e⁻ → Cu⁰(s).
Oxidation half reaction: NO₂⁻(aq) + H₂O(l) → NO₃⁻(aq) + 2H⁺(aq) + 2e⁻.
Balanced chemical reaction:
Cu²⁺(aq) + NO₂⁻(aq) + H₂O(l) → Cu(s) + NO₃⁻(aq) + 2H⁺(aq).
Copper is reduced from oxidation number +2 (Cu²⁺) to oxidation number 0 (Cu) and nitrogen is oxidized from oxidation number +3 (in NO₂⁻) to oxidation number +5 (in NO₃⁻).
Answer:
1. True
2. False
3. False
4. True
Explanation:
A lab is an abbreviation for laboratory and it can be defined as a building that is equipped with scientific equipments and apparatus, which are used for performing scientific experiments, manufacturing of chemicals and drugs, research, testing and analysis of hypothesis in order to establish a theory.
Generally, laboratory scientists, technicians, students and anyone going into a laboratory is prohibited from taking in edible items such as food and water.
Basically, all laboratories are required to place a visible warning sign around the lab prohibiting the bringing in and ingestion of food in any form.
1. True: Tobacco products of all kinds are prohibited in lab.
2. False: Beverages are allowed in lab as long as they are in containers with lids. It's completely prohibited from the lab so as to prevent food poisoning.
3. False: Chewing gum is okay in the lab as long as you don't blow bubbles.
4. True: Food items should not be brought into the lab.
Answer:
The answer to your question is below
Explanation:
Data
mass of CaCO₃ = 155 g
mass of HCl = 250 g
mass of CaCl₂ = 142 g
reactants = CaCO₃ + HCl
products = CaCl₂ + CO₂ + H₂O
1.- Balanced chemical reaction
CaCO₃ + 2HCl ⇒ CaCl₂ + CO₂ + H₂O
2.- Limiting reactant
molar mass of CaCO₃ = 40 + 12 + 48 = 100 g
molar mass of HCl = 2[1 + 35.5 ] = 73 g
theoretical proportion CaCO₃ /HCl = 100 / 73 = 1.37
experimental proportion CaCO₃ /HCl = 155 / 250 = 0.62
As the experimental proportion was lower than the theoretical proportion the limiting reactant is CaCO₃
3.-
Calculate the molar mass of CaCl₂
CaCl₂ = 40 + 71 = 111 g
100 g of CaCO₃ ------------------ 111 g of CaCl₂
155 g of CaCO₃ ----------------- x
x = (155 x 111) / 100
x = 17205 / 100
x = 172.05 g of CaCl₂
4.- percent yield
Percent yield = 142 / 172.05 x 100 = 82.5 %
5.- Excess reactant
100 g of CaCO₃ -------------------- 73 g of HCl
155 g of caCO₃ ------------------- x
x = (155 x 73)/100
x = 133.15 g
Mass of HCl = 250 - 133.15
= 136.9 g
