Answer: 0.0725ppm
Explanation:
133.4g of MgBr2 dissolves in 1.84L of water.
Therefore Xg of MgBr2 will dissolve in 1L of water. i.e
Xg of MgBr2 = 133.4/1.84 = 72.5g
The concentration of MgBr2 is 72.5g/L = 0.0725mg/L
Recall,
1mg/L = 1ppm
Therefore, 0.0725mg/L = 0.0725ppm
The number of moles of nitrate dissolved in water if 6.15 mole of strontium nitrate is : 12.3 moles
<u>Given data :</u>
Moles of strontium nitrate = 6.15 moles
<h3>Ionization of Strontium Nitrate </h3>
When strontium nitrate (
) is placed in water it will dissociates into its ions because strontium nitrate is an ionic compound.
Next step : Represent the ionization of strontium nitrate

From the Stoichiometry of the chemical reaction
I mole of strontium nitrate = 2 moles of Nitrate ions
6.15 mole of strontium nitrate = 2 * 6.15
= 12.3 moles of Nitrate
Hence we can conclude that The number of moles of nitrate dissolved in water if 6.15 mole of strontium nitrate is : 12.3 moles.
Learn more about Stoichiometry of chemical reaction : brainly.com/question/27058367
The empirical formula for the citric acid is C₆H₈O₇
<h3>Data obtained from the question </h3>
Divide by their molar mass
C = 37.51 / 12 = 3.126
H = 4.2 / 1 = 4.2
O = 58.29 / 16 = 3.643
Divide by the smallest
C = 3.126 / 3.126 = 1
H = 4.2 / 3.126 = 1.34
O = 3.643 / 3.126 = 1.17
Multiply through by 6 to express in whole number
C = 1 × 6 = 6
H = 1.34 × 6 = 8
O = 1.17 × 6 = 7
Thus, the empirical formula for the citric acid is C₆H₈O₇
Learn more about empirical formula:
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