Answer: Once the fingerprints are captured and described, judicial physicists use a categorization structure to label the ruling class. The three elementary mark on finger patterns is Whorl, Arch, and Loop. There are more intricate categorization methods that further decay the pattern to plain arches or make camp arches. Loops grant permission be branching or ulnar.
Reaction of current interest is:
<span>CS2(l) + Cl2(g) </span>→ <span>CCl4(l) + S2Cl2(l)
While balancing the chemical reaction, care must be taken that number of atoms of reactant side is equal to number of reactant on product side.
In present case, There is 1 'C' atom of both reactant and product side
There are 2 'S' atoms on both reactant and product side.
However, there are 2 Cl atoms of reactant side, but 6 Cl atoms on product side.
Hence multiplying Cl2 by 3, would equal the number of Cl atoms on both the sides.
Thus, the balanced reaction is
</span>CS2(l) + 3 Cl2(g) → CCl4(l) + S2Cl2(l)
Answer:
1. S+ O2-->So2..... synthesis reaction or direct combination.
2.. precipitation reaction
3. Displacement reaction
4. Decomposition reaction
Answer:-
0.633 moles
Explanation:-
The first step to solving this problem is to write the balanced chemical equation.
The balanced chemical equation for this reaction is
3 CaCO3 + 2 FePO4 = Ca3(PO4)2 + Fe2(CO3)3
From the balanced chemical equation, we can see
3 moles of CaCO3 react with 2 moles of FePO4.
In the question we are told there are 1.9 moles of CaCO3.
So 1.9 moles of CaCO3 react with (2 moles x 1.9 moles )/ 3 moles= 1.267 moles of FePO4.
The question tells us there are 2.7 moles of FePO4.
So there is excess FePO4.
Thus our limiting reagent is CaCO3. It will determine how much Ca3(PO4)2 is formed.
From the balanced chemical equation, we can see
3 moles of CaCO3 gives 1 mole of Ca3(PO4)2
1.9 moles of CaCO3 gives (1 moles x 1.9 moles ) / 3 moles= 0.633 moles of Ca3(PO4)2.
0.633 moles of Ca3 (PO4)2 can be formed when 1.9 moles of CaCO3 react with 2.7 moles of FePO4
Answer: 10 moles/kg.
Explanation:
Given, Mass of solute = 24.2 g
Molar mass of solute = 24.2 g/mol
Mass of solvent = 100.0g = 0.1 kg [1 g=0.001 kg]
Hence, the molality of the resulting solution is 10 moles/kg.
