# How many grams of manganese (IV) oxide are needed to make 5.6 liters of a 2.1 M solution?

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To calculate the grams of manganese (IV) oxide needed, you can use the formula:

[ \text{{grams}} = \text{{molarity}} \times \text{{volume}} \times \text{{molar mass}} ]

Given: Molarity (( M )) = 2.1 M Volume (( V )) = 5.6 liters Molar mass of manganese (IV) oxide (( \text{{MnO}}_2 )) = 86.94 g/mol

Using the formula:

[ \text{{grams}} = 2.1 , \text{{mol/L}} \times 5.6 , \text{{L}} \times 86.94 , \text{{g/mol}} ]

[ \text{{grams}} = 1027.872 , \text{{grams}} ]

So, 1027.872 grams of manganese (IV) oxide are needed to make 5.6 liters of a 2.1 M solution.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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