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3 years ago

If you had 6 moles of CaCl2 and 5 moles of Na3PO4, which of these would be the limiting and excess reactant

Chemistry

1 answer:

faltersainse [42]3 years ago

8 0

Answer:

Na3PO4 is excess reactant, CaCl2 is limiting reactant.

Explanation:

3CaCl2 + 2Na3PO4 ---> Ca3(PO4)2 + 6NaCl

from reaction : 3 mol 2 mol

given: 6 mol 5 mol (X)

X = (6*2)/3 = 4 mol Na3PO4

For 6 mol CaCl2 we need 4 mol Na3PO4, but we have 5 mol Na3PO4,

Na3PO4 is excess reactant, so CaCl2 is limiting reactant.

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Two species of frog mate in the same pond. One breeds in early summer and one in late summer. This is an example of what kind of

MatroZZZ [7]

Answer:

Pre-zygotic, temporal separation

Explanation:

Reproductive isolation mechanism is of two types:

Prezygotic mechanism
Postzygotic mechanism

Prezygotic mechanism isolation occurs before fertilization and helpful in preventing formation of fertile offspring.

In frog external fertilization occurs. In the external fertilization, eggs and sperms are released in water and fertilization occur outside the water.

Prezygotic isolating mechanisms may include behavioral isolation, temporal isolation, mechanical isolation, gametic isolation and habitat isolation.

Temporal separation in reproduction is the sexual activity in the same geographical range but in different periods.

Therefore, the given reproductive isolation is pre-zygotic, temporal separation.

4 0

3 years ago

If 12.5 grams of strontium hydroxide is reacted with 150 mL of 3.5 M carbonic acid, identify the limiting reactant.

vesna_86 [32]

Answer:

Sr(OH)2

Explanation:

We'll begin by calculating the number of mole of carbonic acid in 150mL of 3.5 M carbonic acid solution. This is illustrated below:

Molarity = 3.5M

Volume = 150mL = 150/1000 = 0.15L

Mole of carbonic acid, H2CO3 =..?

Mole = Molarity x Volume

Mole of carbonic acid, H2CO3 = 3.5 x 0.15 = 0.525 mole.

Next, we shall convert 0.525 mole of carbonic acid, H2CO3 to grams.

Mole of H2CO3 = 0.525 mole

Molar mass of H2CO3 = (2x1) + 12 + (16x3) = 62g/mol.

Mass of H2CO3 =..?

Mass = mole x molar mass

Mass of H2CO3 = 0.525 x 62 = 32.55g

Next, we shall write the balanced equation for the reaction. This is given below:

Sr(OH)2 + H2CO3 → SrCO3 + 2H2O

Next, we shall determine the mass of Sr(OH)2 and H2CO3 that reacted from the balanced equation. This is illustrated below:

Molar mass of Sr(OH)2 = 88 + 2(16 + 1) = 88 + 2(17) = 122g/mol

Mass of Sr(OH)2 from the balanced equation = 1 x 122 = 122g

Molar mass of H2CO3 = (2x1) + 12 + (16x3) = 62g/mol.

Mass of H2CO3 from the balanced equation = 1 x 62 = 62g.

From the balanced equation above, 122g of Sr(OH)2 reacted with 62g of H2CO3.

Finally, we shall determine the limiting reactant as follow:

From the balanced equation above, 122g of Sr(OH)2 reacted with 62g of H2CO3.

Therefore, 12.5g of Sr(OH)2 will react with = (12.5 x 62)/122 = 6.35g.

We can see evidently from the calculations made above that it will take 6.35g out 32.55g of H2CO3 to react with 12.5g of Sr(OH)2. Therefore, Sr(OH)2 is the limiting reactant and H2CO3 is the excess reactant

5 0

3 years ago

Read 2 more answers

What are the units of k in the following rate law?

Ludmilka [50]

Answer:

B. $\frac{1}{M^{2} s }$

Explanation:

The unit for rate is M/s while the unit for each molecule should be M. You can find the unit for k by putting the units for rate and the molecules into the equation

rate= k{X][Y]

M/s= k * $M^{2}$ * $M^{1}$