How would you balance: AlBr3 + K2SO4 --> KBr + Al2(SO4)3?
The potassium sulfate group can be multiplied by three to get three on either side. This is the first thing I notice: there are three sulfate groups on the product side and only one on the reactant side.
We multiply the potassium bromide by six instead, since this leaves us with six potassium atoms on the reactant side and only one on the product side.
We multiply the aluminum bromide on the right by two to finally leave two aluminum atoms on the reactant side and the product side, completing the balancing process. Currently, there are six bromine atoms on the right and only three on the right.
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2AlBr₃ + 3K₂SO₄ → 6KBr + Al₂(SO₄)₃
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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.
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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