The % composition when 10g of magnesium combine with 4g of nitrogen is 71.43% magnesium and 28.57 % nitrogen
calculation
% composition = mass of an element / total mass x100
mass of magnesium = 10 g
mass of nitrogen = 4g
calculate the total mass used
= 10g of Magnesium + 4 g of nitrogen = 14 grams
% composition for magnesium is therefore = 10/14 x100 = 71.43 %
% composition for nitrogen is therefore = 4 /14 x100 = 28.57 %
The correct answer is <em>B. a Salt </em><em>because The reaction of an acid and a base is called a neutralization reaction because the properties of both the acid and base are diminished or neutralized when they react. A neutralization reaction is a reaction of an acid with a base in aqueous solution to produce water and a salt, as shown by the following equation:</em>
<em>acid + base → salt + water</em>
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<em>* Hopefully this helps:) Mark me the brainliest:) </em>
<em>∞ 234483279c20∞</em>
Answer:
If you see in the image above, there is an unbalance force applied while playing tug of war. Since it is 1 vs 2, there is a greater net force in the right side then the left side. If it was 2 vs 2 or 1 vs 1, then they are appling balance force. You can also see in the picture that the arrows are pointing outwards (--->) rather then inwards (<---) because you are pulling the rope not pushing the rope. If you add one person on the left side, then the newtons which is 20N will become to 35N and will be balanced, but since there in only 1 person, there is less force on the left side, the newtons gets subtracted having only 20N. Since you are pulling the rope, the friction is opposite (<---). Since you are pulling the rope, you are using Kinetic force and the rope stays in potential force since it stays constant.
Hope this helps, thank you :) and I am not sure about magnitude I think you can that since there is greater force on the right side, there is more magnitude there.
Answer:
2.5L [NaCl] concentrate needs to be 4.8 Molar solution before dilution to prep 10L of 1.2M KNO₃ solution.
Explanation:
Generally, moles of solute in solution before dilution must equal moles of solute after dilution.
By definition Molarity = moles solute/volume of solution in Liters
=> moles solute = Molarity x Volume (L)
Apply moles before dilution = moles after dilution ...
=> (Molarity X Volume)before dilution = (Molarity X Volume)after dilution
=> (M)(2.5L)before = (1.2M)(10.0L)after
=> Molarity of 2.5L concentrate = (1.2M)(10.0L)/(2.5L) = 4.8 Molar concentrate
