68 POINTS WILL GIVE BRAINliest!!!

HBr can be added to an alkene in the presence of peroxides (ROOR). What function does the peroxide serve in this reaction

RSB [31]

Answer:

Radical chain initiator

Explanation:

The peroxide here serves as a radical chain initiator. In the field of chemistry the radical initiatives are those substances that are used in industrial processes like polymer synthesis. These initiatives have weak bonds generally and they're mostly used to create free radicals. These radicals are atoms that have odd numbers of electrons. Peroxide is an example of such.

6 0

2 years ago

Fill in the blank for the following statement: ________ is the movement of rocks, sand, or materials from a place?

Verizon [17]

Answer:

Erosion.

Explanation:

It can be as small as a grain of sand or as large as a boulder. Sediment moves from one place to another through the process of erosion. Erosion is the removal and transportation of rock or soil. Erosion can move sediment through water, ice, or wind.

3 0

2 years ago

Read 2 more answers

What do Potassium and Krypton have in common and what are their differences based on their location on the periodic table?

Kamila [148]

Potassium (K) and Krypton (Kr) are in the same period, so they have the same number of electron shells. Potassium is more reactive than Krypton because it only has one valence electron.

K and Kr are both in Period 4, so they each have the n = 1, 2, 3, and 4 shells. Kr is less reactive than K because it already has a complete octet.

“Potassium (K) and Krypton (Kr) are in the same period, so they have the same number of electron shells. Krypton is more reactive than Potassium because it has a full outer shell” is incorrect because having a full outer shell means that the element is unreactive.

“Potassium (K) and Krypton (Kr) are in the same group, so they have the same number of valence electrons. They have different numbers of electron shells” is incorrect for three reasons:

1. K is in Group 1 and Kr is in Group 18.

2. K has one valence electron, and Kr has eight valence electrons.

3. They have the same number of electron shells.

“Potassium (K) and Krypton (Kr) are in the same group, so they have the same number of valence electrons. They have different numbers of electron shells” is incorrect for two reasons:

1. K and Kr have different numbers of valence electrons.

2. They have the same number of electron shells.

8 0

2 years ago

Which of the following reactions could be used to extract the lead from the lead nitrate?

alexgriva [62]

Answer:

Option b. Decomposition

Followed by a reduction process using charcoal

Explanation:

Lead can be obtained from lead nitrate by thermal decomposition of lead nitrate as shown below:

2Pb(NO3)2 —> 2PbO + 4NO2 + O2

The PbO obtained is reduced by charcoal(C) to obtain the metallic Pb as shown below:

2PbO + C —> Pb + CO2

4 0

3 years ago

Aluminum reacts with sulfur gas to produce aluminum sulfide. a) What is the limiting reactant? What is the excess reagent? b) Ho

Sophie [7]

Answer:

a) Limiting: sulfur. Excess: aluminium.

b) 1.56g Al₂S₃.

c) 0.72g Al

Explanation:

Hello,

In this case, the initial mass of both aluminium and sulfur are missing, therefore, one could assume they are 1.00 g for each one. Thus, by considering the undergoing chemical reaction turns out:

$2Al(s)+3S_2(g)\rightarrow 2Al_2S_3(s)\\$

a) Thus, considering the assumed mass (which could be changed based on the one you are given), the limiting reagent is identified as shown below:

$n_S^{available}=1.00gS_2*\frac{1molS_2}{64gS_2} =0.0156molS_2\\n_S^{consumed\ by \ Al}=1.00gAl*\frac{1molAl}{27gAl}*\frac{3molS_2}{2molAl}=0.0556molS_2$

Thereby, since there 1.00g of aluminium will consume 0.0554 mol of sulfur but there are just 0.0156 mol available, the limiting reagent is sulfur and the excess reagent is aluminium.

b) By stoichiometry, the produced grams of aluminium sulfide are:

$m_{Al_2S_3}=0.0156molS_2*\frac{2molAl_2S_3}{3molS_2} *\frac{150gAl_2S_3}{1molAl_2S_3} =1.56gAl_2S_3$

c) The leftover is computed as follows:

$m_{Al}^{excess}=(0.0556-0.0156)molS_2*\frac{2molAl}{3molS_2}*\frac{27gAl}{1molAl} =0.72 gAl\\$

NOTE: Remember I assumed the quantities, they could change based on those you are given, so the results might be different, but the procedure is quite the same.

Best regards.

7 0

2 years ago