What happens is it makes water
Answer:
760 uM
Explanation:
<em>A biochemist carefully measures the molarity of magnesium ion in 47, mL of cell growth medium to be 97 uM. Unfortunately, a careless graduate student forgets to cover the container of growth medium and a substantial amount of the solvent evaporates. The volume of the cell growth medium falls to 6.0 mL. Calculate the new molarity of magnesium ion in the cell growth medium Be sure your answer has the correct number of significant digits.</em>
The problem here is that the amount of magnesium ion remains the same irrespective of the volume.
Amount of magnesium in the growth medium = <em>molarity x volume</em>
= 97 x 
x 47 x 
= 4.559 x
Then, the volume reduced to 6.0 mL, the new molarity becomes;
<em>molarity = mole/volume </em>
= 4.559 x /6 x = 7.598333 x 
M = 759.83333 uM
To the correct number of significant digits = 760 uM
Mass percent is the estimation of the mass of the element in a compound in percentage form. The mass % of manganese (Mn) in potassium permanganate is 34.76 %.
<h3>What is the mass percentage?</h3>
The mass percentage is defined by dividing the mass of the element by the molecular mass of the compound followed by multiplying it by 100%. It is given as,
Mass % = (Atomic Mass of element ÷ Molecular Mass) × 100
As it is known that,
Atomic Mass of manganese = 54.94 g/mol
Molecular Mass of potassium permanganate = 158.034 g/mol
Substituting values to calculate mass % as:
% Mn = (Atomic Mass of Mn ÷ Molecular Mass of KMnO₄) × 100
= (54.94 g/mol ÷ 158.034 g/mol) × 100
= 34.76 %
Therefore, 34.76% Mn is present in KMnO₄.
Learn more about mass percentage, here:
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Answer:
Mass of magnesium oxide formed = 35.1 g
Explanation:
Given data:
Mass of Mg = 20.9 g
Mass of O₂ = 15.2 g
Mass of magnesium oxide formed = ?
Solution:
Chemical equation:
2Mg + O₂ → 2MgO
Number of moles of Mg:
Number of moles = mass/molar mass
Number of moles = 20.9 g/ 24 g/mol
Number of moles = 0.87 mol
Number of moles of O₂:
Number of moles = mass/molar mass
Number of moles = 15.2 g/ 32 g/mol
Number of moles = 0.475 mol
Now we will compare the moles of MgO with magnesium and oxygen.
Mg : MgO
2 : 2
0.87 : 0.87
O₂ : MgO
1 : 2
0.475 : 2/1×0.475 = 0.95
Number of moles of MgO formed by Mg are less thus Mg will limiting reactant.
Mass of MgO:
Mass = number of moles × molar mass
Mass = 0.87 mol × 40.3 g/mol
Mass = 35.1 g
