Answer:
59.8%
Explanation:
First find the Mr of manganese (III) nitrate.
Mr of Mn(NO₃)₃ = 54.9 + (14 × 3) + (16 × 3 × 3) = <u>240.9</u>
Since we have to find the percentage composition of oxygen, we need to find the Mr of oxygen in the compound, which is:
Mr of (O₃)₃ = (16 × 3) × 3 = <u>144</u>
Now we can find percentage composition / percentage by mass of oxygen.
% composition = 
× 100
% composition = 
× 100 = <u>59.776%</u>
∴ % compostion of oxygen in maganese(III)nitrate is 59.8% (to 3 significant figures).
Water is <u>not wet</u> because the word wet is a form of liquid/water saturated/soaking the object.
Justification for your answer
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Chlorine is less reactive than fluorine because the outer electrons in a chlorine atom are further from the nucleus than the outer electrons in a fluorine atom. It is harder for a chlorine atom to gain an electron than it is for a fluorine atom.
There are three things to consider every single time relative reactivity is unknown; atomic radius, shielding, and number of electrons. The reactivity is the halogens ability to gain an electron, so number of electrons already in the atom plays a vital role. Chlorine has more electrons so repels a reacting electron with greater force than fluorine, making it less likely to react.
Fluorine also has fewer electron shells than chlorine, so there are fewer electrons between the positive nucleus and the reacting electron to essentiallly block, or weaken, the electromagnetic attraction. This is shielding. Lastly, fluorine is much smaller molecule than chlorine, and the shorter distance, or radius, between the nucleus and the electron again makes it more likely to attract the electron and react to gain a noble gas configuration.
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Answer:
This solution has a volume of 98.4 mL
Explanation:
Step 1: Data given
Molarity of AgClO4 solution = 1.27 mol/L
Number of moles AgClO4 = 125 mmol = 0.125 mol
Molar mass of AgClO4= 207.32 g/mol
Step 2: Calculate volume of the 1.27 M solution
Molarity = moles / volume
Volume = moles / molarity
Volume = 0.125 moles / 1.27 mol /L
Volume = 0.0984 L = 98.4 mL
This solution has a volume of 98.4 mL
