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3 years ago

I need an answer and explanation pls

Chemistry

1 answer:

OverLord2011 [107]3 years ago

8 0

Answer:

Cl2 + 2NaBr --> 2NaCl + Br2

Explanation:

This is a single displacement reaction where one side of the ionic compound switches with the other.

So, Cl2 + NaBr ---> NaCl + Br2

This isolates the Bromine and puts the Chlorine in it's place.

Then, balance out the equation like so and you should get

Cl2 + 2NaBr --> 2NaCl + Br2

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3 years ago

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The electron cloud model describes which of the following?

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3 years ago

The pKa of lactic acid is 3.9. A lactate buffer will be useful from pH values ________. The pKa of lactic acid is 3.9. A lactate

Vedmedyk [2.9K]

Answer:

Explanation:

The usefulness of a buffer is its ability to resist changes in pH when small quantities of base or acid are added to it. This ability is the consequence of having both the conjugate base and the weak acid present in solution which will consume the added base or acid.

This capacity is lost if the ratio of the concentration of conjugate base to the concentration of weak acid differ by an order of magnitude. Since buffers having ratios differing by more will have their pH driven by either the weak acid or its conjugate base .

From the Henderson-Hasselbach equation we have that

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3 years ago

Sodium phosphate and calcium chloride react to form sodium chloride and calcium phosphate. If you have 379.4 grams of calcium ch

barxatty [35]

Answer: 353.6 g.

Explanation:

1) Word equation (given)

Sodium phosphate + calcium choride → sodium chloride + calcium phosphate

2) Balanced chemical equation

2 Na₃PO₄ +3 CaCl₂ → 6 NaCl + Ca₃(PO4)₂

3) Mole ratio

2 moles Na₃PO₄ : 3 moles CaCl₂ : 6 moles NaCl : 1 mole Ca₃(PO4)₂

4) Formula to convert grams to moles

number of moles = mass in grams / molar mass

5) Number of moles of limiting reactant

a) mass in grams = 379.4 g CaCl₂

b) molar mass CaCl₂ = 110.98 g/mol

c) number of moles CaCl₂ = 379.4 g / 110.98 g/mol = 3.419 moles

6) Theoretical yield of Ca₃(PO4)₂

3 moles CaCl₂ / 1 mole Ca₃(PO4)₂ = 3.419 moles CaCl₂ / x ⇒

x = 3.419 × 1 / 3 = 1.140 moles Ca₃(PO4)₂

7) Convert 1.140 moles Ca₃(PO4)₂

a) molar mass Ca₃(PO4)₂ = 310.1767 g/mol

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3 years ago

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3 years ago