Answer:
5200 ppm
Explanation:
As per the definition, parts per million of a contaminant is a measure of the amount of mass of contaminant present per million amount of the solution. It is denoted by ppm.
Given in the question,
Water = 250 ml = 250 g
Lead = 1.30 g
So,
ppm of Lead = 
= 
= 5200 ppm
So, as calculated above, there is 5200 ppm of lead present in 250 ml of water.
Answer:
- metal sulfate
- metal sulfate
- copper sulfate
- copper nitrate
- copper chloride
- copper phosphate
- hydrochloric acid, water
- Potassium, sulfuric acid, water
(Correct me if I am wrong)
Answer:
Limiting reactant: O2
grams NO2 produced = 230.276 g NO2
grams of NO unused = 26.67 gNO
Explanation:
2NO + O2 --> 2NO2
Step 1: Determine the molar ratio NO:O2
molar ratio NO:O2 = 5.895: 2.503 = 2.35
stoichiometric molar ratio NO:O2 = 2:1
So, O2 is the limiting reactant.
Step2: Determine the grams of NO2:
?g NO2 = moles O2 x (2moles NO2/1 mol O2) x (MM NO2/ 1 mol NO2) = 2.503 x 2 x 46 = 230.276 g NO2
Step 3: Determine the amount of excess reagent unreacted
moles excess NO reacted = moles O2 x (2 moles NO/1 mol O2) = 2.503 x 2 = 5.006 moles NO reacted
moles NO unreacted = total moles NO - moles NO reacted = 5.895-5.006 =0.889 moles NO unreacted
mass NO unreacted = moles NO unreacted x MM NO = 0.889 x 30 =26.67 g NO unreacted
Answer:
Fe(NO₃)₃ + 3KSCN → Fe(SCN)₃ + 3KNO₃
Explanation:
Chemical equation:
Fe(NO₃)₃ + KSCN → Fe(SCN)₃ + KNO₃
Balanced Chemical equation:
Fe(NO₃)₃ + 3KSCN → Fe(SCN)₃ + 3KNO₃
Type of reaction:
It is double displacement reaction.
In this reaction the anion or cation of both reactants exchange with each other. In given reaction the cation Fe⁺³ exchange with cation K⁺.
The given reaction equation is balanced so there are equal number of atoms of each elements are present on both side of equation and completely hold the law of conservation of mass.
Double replacement:
It is the reaction in which two compound exchange their ions and form new compounds.
AB + CD → AC +BD
Anthracene is a polycyclic aromatic hydrocarbon with chemical formula C₁₄H₁₀. The number of fused rings in Anthracene are three in number. This compound is colorful and is used in the formation of different dyes due to its property of deloclization of pi electrons. All the carbon atoms in Anthracene are sp² hybridized with a trigonal planar structure hence, the Anthracene is planar in nature.
Number of Sigma Bonds:
There are 26 sigma bonds (colored in Blue) in Anthracene among which 10 sigma bonds are between carbon and hydrogen atoms while the remaining are between the carbon atoms.
Number of Pi-Bonds:
There are 7 pi bonds in Anthracene (colored in red). All pi bonds are present between carbon and carbon atoms.
Number of Electrons in Sigma Bonds:
As one sigma bond is formed by 2 electrons hence, 26 sigma bonds will be formed by 52 electrons.
Number of Electrons in Pi Bonds:
As one pi bond is formed by the side wise overlap of two p orbitals hence one pi bond is formed by two electrons so, 7 pi bonds will be formed by 14 electrons.
