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SCORPION-xisa [38]
2 years ago
12

You have a new box of breakfast cereal that contains cereal bits and fluffy marshmallows bits. your brother shakes the box and t

hen pours himself a bowl of marshmallows bits, leaving you with a bowl of only cereal bits. what form of separation technique did he undertake? what property does this separation technique depend on?
Chemistry
2 answers:
Tom [10]2 years ago
5 0
This is the process of sieving, in which the heavier particles settle at the bottom and the lighter ones are retained at the top.
LenKa [72]2 years ago
5 0
The separation technique is called sieving. The separation technique depends on the different particle size of the cereal bits and marshmallow bits. <span />
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Which of the following species is not formed through a termination reaction in the chlorination of methane? Which of the followi
krek1111 [17]

Explanation:

When we add chlorine to a substance or compound then this process is known as chlorination.

For example, a process of chlorination is as follows.

Initiation : Cl_{2} \overset{light}{\rightarrow} 2Cl*          

where, Cl* is a free radical.

Propagation:

    Cl* + CH_{4} \rightarrow HCl + *CH_{3}

    CH_{3} + Cl_{2} \rightarrow CH_{3}Cl + Cl*

Termination:

    2Cl* \rightarrow Cl_{2}

    Cl* + *CH_{3} \rightarrow CH_{3}Cl

    2*CH_{3} \rightarrow CH_{3}-CH_{3}

Thus, we can conclude that out of the given options H_{2} is not formed through a termination reaction in the chlorination of methane.

5 0
2 years ago
150.0 grams of AsF3 were reacted with 180.0 g of CCl4 to produce AsCl3 and CCl2F2. The theoretical yield of CCl2F2 produced, in
Kamila [148]
The chemical reaction would be written as 

2 AsF3<span> + 3 CCl4 = 2 AsCl3 + 3 CCl2F2
</span>
We use the given amounts of the reactants to first find the limiting reactant. Then use the amount of the limiting reactant to proceed to further calculations.

150 g AsF3 ( 1 mol / 131.92 g) = 1.14 mol AsF3
180 g CCl4 (1 mol / 153.82 g) = 1.17 mol CCl4

 Therefore, the limiting reactant would be CCl4 since it would be consumed completely. The theoretical yield would be:

1.17 mol CCl4 ( 3 mol CCl2F2 / 3 mol CCl4 ) = 1.17 mol CCl2F2

7 0
1 year ago
If 15.6 g of hydrate are heated and only 11.7 g of anhydrous salt remain, calculate the % of water lost.
Elodia [21]

Answer:

Percent loss of water = 25%

Explanation:

Given data:

Mass of hydrated salt = 15.6 g

Mass of anhydrous salt = 11.7 g

Percentage of water lost = ?

Solution:

First of all we will calculate the mass of water in hydrated salt.

Mass of water =  Mass of hydrated salt - Mass of anhydrous salt

Mass of water = 15.6 g - 11.7 g

Mass of water = 3.9 g

Now we will calculate the percentage.

Percent loss of water = mass of water / total mass × 100

Percent loss of water = 3.9 g/ 15.6 g × 100

Percent loss of water = 25%

8 0
1 year ago
Roundup, an herbicide manufactured by Monsanto, has the formula C3H8NO5P. How many moles of molecules are there in a 669.1-g sam
Vedmedyk [2.9K]

Answer:

2.4 ×10^24 molecules of the herbicide.

Explanation:

We must first obtain the molar mass of the compound as follows;

C3H8NO5P= [3(12) + 8(1) + 14 +5(16) +31] = [36 + 8 + 14 + 80 + 31]= 169 gmol-1

We know that one mole of a compound contains the Avogadro's number of molecules.

Hence;

169 g of the herbicide contains 6.02×10^23 molecules

Therefore 669.1 g of the herbicide contains 669.1 × 6.02×10^23/ 169 = 2.4 ×10^24 molecules of the herbicide.

7 0
1 year ago
Calculate the maximum concentration (in m) of silver ions (ag+) in a solution that contains 0.025 m of co32-. the ksp of ag2co3
Helen [10]
Equilibrium equation is

<span>Ag2CO3(s) <---> 2 Ag+(aq) + CO32-(aq) </span>

<span>From the reaction equation above, the formula for Ksp: </span>

<span>Ksp = [Ag+]^2 [CO32-] = 8.1 x 10^-12 </span>
<span>You know  [CO32-], so you can solve for [Ag+] as: </span>
<span>(8.1 x 10^-12) = [Ag+]^2 (0.025) </span>
<span>[Ag+]^2 = 3.24 x 10^-10 </span>
<span>[Ag+] = 1.8 x 10^-5 M </span>
5 0
2 years ago
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