91.4 grams
91.4 grams of kcl in grams must be added to 500 ml of a 0. 15 m kcl solution to produce a 0. 40 m solution.
C = mol/volume
2.45M=mol/0.5L
2.45M⋅0.5L = mol
mol = 1.225
Convert no. of moles to grams using the atomic mass of K + Cl
1.225mol * 
mol=1.225
=1.225 mol . 
=1.225 . 74.6
=91.4g
therefore, 91.4 grams of kcl in grams must be added to 500 ml of a 0. 15 m kcl solution to produce a 0. 40 m solution.
What is 1 molar solution?
In order to create a 1 molar (M) solution, 1.0 Gram Molecular Weight of the chemical must be dissolved in 1 liter of water.
58.44 g make up a 1M solution of NaCl.
To learn more about molar solution visit:
brainly.com/question/10053901
#SPJ4
Answer:
(A)
Explanation:
Answer is (a) Both the assertion and the reason are correct and re!son is the correct explanation of the assertion
Answer:p waves hope that helps
Answer:
73.4% is the percent yield
Explanation:
2KClO₃ → 2KCl + 3O₂
This is a decomposition reaction, where 2 moles of potassium chlorate decompose to 2 moles of potassium chloride and 3 moles of oxygen.
We determine the moles of salt: 400 g . 1. mol /122.5g= 3.26 moles of KClO₃
In the theoretical yield of the reaction we say:
2 moles of potassium chlorate can produce 3 moles of oxygen
Therefore, 3.26 moles of salt, may produce (3.26 . 3) /2 = 4.89 moles of O₂
The mass of produced oxygen is: 4.89 mol . 32 g /1mol = 156.6g
But, we have produced 115 g. Let's determine the percent yield of reaction
Percent yield = (Produced yield/Theoretical yield) . 100
(115g / 156.6g) . 100 = 73.4 %
Answer:
Techniques and Tests
Qualitative analysis typically measures changes in color, melting point, odor, reactivity, radioactivity, boiling point, bubble production, and precipitation.