I dont know what subject is this
An isotonic salt solution is 0.90% (w/w) nacl in water many grams of nacl are contained in 1.00 kg of such a solution Reconstruction debate that had federalism debate that had been an issue since the 1790s.
Reconstruction failed by most other measures: Radical Republican legislation ultimately failed to isotonic former slaves from white persecution and failed to engender fundamental changes to the social fabric of the South. When President Rutherford B. federalism debate that had been an issue since the 1790s almost mediately . Hayes removed federal troops from isotonic South in 1877, former Confederate officials and slave returned to With the support of a conservative Supreme Court, these newly empowered white southern politicians passed black codes, voter qualifications, and other anti-progressive legislation to reverse the rights that blacks had gained during Radical Reconstruction. The U.S. Supreme Court bolstered this federalism anti-progressive movement federalism with decisions in the Slaughterhouse Cases, the Civil Rights Cases, and United States v.
Learn more about Reconstruction on:
brainly.com/question/24761999
#SPJ4
Cu + 2H2SO4 ⟶ CuSO4 + SO2 + 2H20
In left hand side of the equation.
Cu = 1 atom
H = 4 atoms
S = 2 atoms
O = 8 atoms
In right hand side of the equation.
Cu = 1 atom
S = 2 atoms
0 = 8 atoms
H = 4 atoms
• All the atoms are balanced in the left and right side of the equation and it satisfies the law of conservation of mass.
• Equation is balanced and correct.
*See the attachment .
Answer:
O₂; KCl; 33.3
Explanation:
We are given the moles of two reactants, so this is a limiting reactant problem.
We know that we will need moles, so, lets assemble all the data in one place.
2KCl + 3O₂ ⟶ 2KClO₃
n/mol: 100.0 100.0
1. Identify the limiting reactant
(a) Calculate the moles of KClO₃ that can be formed from each reactant
(i)From KCl
(ii) From O₂
O₂ is the limiting reactant, because it forms fewer moles of the KClO₃.
KClO₃ is the excess reactant.
2. Moles of KCl left over
(a) Moles of KCl used
(b) Moles of KCl left over
n = 100.0 mol - 66.67 mol = 33.3 mol
