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dexar [7]
3 years ago
11

Which of the following are not single-displacement reactions?

Chemistry
1 answer:
Goshia [24]3 years ago
6 0

Answer:

\boxed{\text{B and C }}

Explanation:

In a single-displacement reaction, one element exchanges partners with another element in a compound.

\textbf{A. } \rm Fe + 2HCl \longrightarrow FeCl_2 + H_2

This is a single-displacement reaction, because the element Fe exchanges partners with H in HCl.

\textbf{B. } \rm KOH + HNO_3 \longrightarrow H_2O + KNO_3

This is not a single-displacement reaction, because it is a reaction between two compounds.

This is a double displacement reaction in which the K⁺ and H⁺ cations change partners with the anions.

\textbf{C. } \rm Na_2S + 2HCl \longrightarrow 2NaCl + H_2S

This is not a single-displacement reaction. It is another double displacement reaction, in which the Na⁺ and H⁺ cations change partners with the anions.

\textbf{D. } \rm Ca + 2HOH \longrightarrow Ca(OH)_2 + H_2

This is a single-displacement reaction, because the element Ca exchanges partners with H in H₂O.

\boxed{\textbf{B and C }} are not single-displacement reactions.

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The green pigment required for photosynthesis is:
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C. chlorophyll



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3 years ago
A student mixes 33.0 mL of 2.70 M Pb ( NO 3 ) 2 ( aq ) with 20.0 mL of 0.00157 M NaI ( aq ) . How many moles of PbI 2 ( s ) prec
GalinKa [24]

<u>Answer:</u> The moles of precipitate (lead (II) iodide) produced is 1.57\times 10^{-5} moles

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}     .....(1)

  • <u>For lead (II) nitrate:</u>

Molarity of lead (II) nitrate solution = 2.70 M

Volume of solution = 33.0 mL = 0.033 L   (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

2.70M=\frac{\text{Moles of lead (II) nitrate}}{0.033L}\\\\\text{Moles of lead (II) nitrate}=(2.70mol/L\times 0.0330L)=0.0891mol

  • <u>For NaI:</u>

Molarity of NaI solution = 0.00157 M

Volume of solution = 20.0 mL = 0.020 L

Putting values in equation 1, we get:

0.00157M=\frac{\text{Moles of NaI}}{0.020L}\\\\\text{Moles of NaI}=(0.00157mol/L\times 0.0200L)=3.14\times 10^{-5}mol

For the given chemical reaction:

Pb(NO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaNO_3(aq.)

By Stoichiometry of the reaction:

2 moles of NaI reacts with 1 mole of lead (II) nitrate

So, 3.14\times 10^{-5} moles of NaI will react with = \frac{1}{2}\times 3.14\times 10^{-5}=1.57\times 10^{-5}mol of lead (II) nitrate

As, given amount of lead (II) nitrate is more than the required amount. So, it is considered as an excess reagent.

Thus, NaI is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of NaI produces 1 mole of lead (II) iodide

So, 3.14\times 10^{-5} moles of NaI will produce = \frac{1}{2}\times 3.14\times 10^{-5}=1.57\times 10^{-5}moles of lead (II) iodide

Hence, the moles of precipitate (lead (II) iodide) produced is 1.57\times 10^{-5} moles

4 0
3 years ago
This investigation is a controlled experiment, so you can change only one variable at a time. Which reactant do you think Koby s
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Answer:

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What is the scientific notation of 36000x10 ex.10?
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Answer:

3.6 times 10^4

Explanation:

Scientific notation is between 1-9. So, we move 36000 to 4 decimal places. SO it would be 3.6 times 10^4. Scientific Notation always has the base of 10 . Enjoy :)

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4 years ago
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