The molarity of the stock solution is 1.25 M.
<u>Explanation:</u>
We have to find the molarity of the stock solution using the law of volumetric analysis as,
V1M1 = V2M2
V1 = 150 ml
M1 = 0.5 M
V2 = 60 ml
M2 = ?
The above equation can be rearranged to get M2 as,
M2 = 
Plugin the values as,
M2 = 
= 1.25 M
So the molarity of the stock solution is 1.25 M.
The rate at which a radioactive<span> isotope decays is measured in </span>half-life. The termhalf-life<span> is defined as the time it takes for one-</span>half<span> of the atoms of a </span>radioactive material<span> to disintegrate. </span>Half-lives<span> for </span>various radioisotopes<span> can range from a few microseconds to billions of years.
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back at it again with that answer
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Answer:
68.3%
Explanation:
First, let us look at the equation of reaction involving silver and magnesium chloride:
2Ag + MgCl2 ----> 2AgCl + Mg
1 mole of MgCl2 is required to precipitate 2 moles of Ag completely from the solution. That is a ratio of 1 to 2.
Now, mole of MgCl2 used to precipitate all the Ag
= molarity x volume
= 2.19 M x 2.89/1000
= 0.0063291 mole
Since 1 mole of MgCl2 would always require 2 moles of Ag, 0.0063291 mole will therefore require:
0.0063291 x 2 = 0.0126 mole of Ag
This means that 0.0126 mole of Ag is present in stephanie.
Mass of silver in stephanie = mole x molar mass
= 0.0126 x 107.8682
= 1.365 g
Thus, 1.365 g of silver is present in 2.00 g sample of stephanie.
Mass percent of silver in stephanie = 1.365/2.00 x 100
= 68.25% = 68.3% to the correct number of significant figure.
The answer is A.
Pure substance is either an element or a compound, which elements in compund is chemically combined together. They cannot be separated by physical methods such as filtration or evaporation. Compounds can only be separated by chemical methods, which include using electricity (electrolysis) or applying heat.
One mole represents 6.022∙1023 separate entities, just like one dozen represents 12 objects. So, if there are 6.022∙1023 H2O molecules, that is the same as one mole of water.