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Aneli [31]
3 years ago
8

Which of the following best describes a single-replacement reaction?

Chemistry
2 answers:
balandron [24]3 years ago
8 0

Answer: B. One element takes the place of another in a compound.

Explanation:

Single replacement reaction is a chemical reaction in which more reactive element displaces the less reactive element from its salt solution.

Example: Mg+2HCl\rightarrow MgCl_2+H_2

Thus magnesium when dropped into an aqueous solution of HCl releases hydrogen gas by replacing hydrogen from its position.

A. A compound breaks apart into separate elements: Decomposition

Example: Li_2CO_3\rightarrow Li_2O+CO_2 Double displacement reaction is one in which exchange of ions take place.  

B. One element takes the place of another in a compound: Single replacement

C. Two elements switch places in two compounds: Double displacement reaction

Example: 2NaOH(aq)+(NH_4)_2SO_4(aq)\rightarrow 2NH_4OH(aq)+Na_2SO_4(aq)

D.  Two elements combine to form a compound: Synthesis reaction

Example: 2Mg+O_2\rightarrow 2MgO  

blondinia [14]3 years ago
6 0

Answer

B. One element takes the place of another in a compound

Explanation

A single-replacement reaction occurs when one element is substituted for another element in a compound.The  reacting materials are normally pure elements such as a metal or hydrogen gas with an aqueous compound.A new aqueous compound plus a different pure element are formed as products.

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When a neutral atom loses an electron, it forms an ion with a positive charge called a _________? Question 1 options:
Black_prince [1.1K]
It should be C) cation
8 0
3 years ago
Read 2 more answers
How many moles of helium are needed to fill a balloon to a volume of 5.3 L at 22 ℃ and 632 mmHg?
son4ous [18]

Answer:

0.18 moles

Explanation:

Applying,

PV = nRT................... Equation 1

Where P = pressure, V = volume, n = number of moles, R = molar gas constant, T = temperature.

make n the subject of the equation

n = PV/RT............... Equation 2

Given: V = 5.3 L, T = 22 °C = (22+272) K = 295 K, P = 632 mmHg = (0.00131579×632) = 0.8316 atm,  R = 0.083 L.atm/K.mol

Substitute these values into equation 2

n = (0.8316×5.3)/(0.083×295)

n = 0.18 moles

6 0
3 years ago
How many total atoms are in 3.3 moles of potassium sulfide, K2S?
marissa [1.9K]
The answer is 363.99 g
3 0
2 years ago
What is the composition, in atom percent, of an alloy that consists of a) 5.5 wt% Pb and b) 94.5 wt% of Sn? Assume that the atom
Anastaziya [24]

Answer : The percent composition of Pb and Sn in atom is, 3.21 % and 96.8 % respectively.

Explanation :

First we have to calculate the number of atoms in 5.5 wt% Pb and 94.5 wt% of Sn.

As, 207.2 g of lead contains 6.022\times 10^{23} atoms

So, 5.5 g of lead contains \frac{5.5}{207.2}\times 6.022\times 10^{23}=1.59\times 10^{22} atoms

and,

As, 118.71 g of lead contains 6.022\times 10^{23} atoms

So, 94.5 g of lead contains \frac{94.5}{118.71}\times 6.022\times 10^{23}=4.79\times 10^{23} atoms

Now we have to calculate the percent composition of Pb and Sn in atom.

\% \text{Composition of Pb}=\frac{\text{Atoms of Pb}}{\text{Atoms of Pb}+\text{Atoms of Sn}}\times 100

\% \text{Composition of Pb}=\frac{1.59\times 10^{22}}{(1.59\times 10^{22})+(4.79\times 10^{23})}\times 100=3.21\%

and,

\% \text{Composition of Sn}=\frac{\text{Atoms of Sn}}{\text{Atoms of Pb}+\text{Atoms of Sn}}\times 100

\% \text{Composition of Sn}=\frac{4.79\times 10^{23}}{(1.59\times 10^{22})+(4.79\times 10^{23})}\times 100=96.8\%

Thus, the percent composition of Pb and Sn in atom is, 3.21 % and 96.8 % respectively.

6 0
3 years ago
Draw the major organic substitution product(s) for (2R,3S)-2-bromo-3-methylpentane reacting with the given nucleophile. Indicate
Andrew [12]

Answer:

(2R,3S)-2-ethoxy-3-methylpentane

and

(2S,3S)-2-ethoxy-3-methylpentane

Explanation:

For this case, we will have  CH_3CH_2O^- as nucleophile. Also, this compound is also in excess. So, we will have as solvent CH_3CH_2OH a protic solvent. Therefore the Sn1 reaction would be favored.

The first step would be the carbocation formation followed by the attack of the nucleophile. In this case both isomers would be produced: R and S (see figure).

7 0
3 years ago
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