A solution is made by dissolving 10.7 g of magnesium sulfate, MgSO4, in enough water to make exactly 100 mL of solution. Calculate the concentration (molarity) of MgSO4 in mmol/L (M).

[MgSO₄] = 890 mM/L

Explanation:

In order to determine molarity we need to determine the moles of solute that are in 1L of solution.

Solute: MgSO₄ (10.7 g)

Solvent: water

Solution: 100 mL as volume. (100 mL . 1L / 1000mL) = 0.1L

We convert the solute's mass  to moles → 10.7 g / 120.36 g/mol = 0.089 moles

Molarity (mol/L) → 0.089 mol/0.1L = 0.89 M

In order to calculate M to mM/L, we make this conversion:

0.89 mol . 1000 mmoles/ 1 mol = 890 mmoles

answer: you could do some invislibe ink.

explanation: i will tell you how to if u answer my question.

consider the reaction below.

 no rx

at 500 k, the reaction is at equilibrium with the following concentrations.

[pci5]= 0.0095 m

[pci3] = 0.020

[ci2] = 0.020 m

what is the equilibrium constant for the given reaction?

0.042

0.42

2.4

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