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Monica [59]
3 years ago
6

In which material are the particles arranged in a regular geometric pattern?

Chemistry
1 answer:
Illusion [34]3 years ago
6 0
 I think only solids arrange in a regular geometric pattern, so i2
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Arrange these molecules in order of decreasing mass: C​6​H​14​ , NO​2​ , Fe​2​O​3 ​, and CaCO​
babunello [35]

The following is the arrangement of the given molecules in decreasing order of mass: \text{ Fe2O3} > \text{C6H14} > \text{CaCO} > \text{NO2}

<u>Solution:</u>

First inorder to arrange the elements in descending order of their mass we have to calculate the molecular mass of each element. The calculation is as follows:

<u>Mass of C6H14:</u>

C\rightarrow6\times12.01 = 72.06

H\rightarrow14\times1.008 = 14.112

<em>Mass of C6H14 is 86.172</em>

<u>Mass of NO2:</u>

N\rightarrow1\times14.0067 = 14.0067

O\rightarrow2\times15.9994 = 31.9988

<em>Mass of NO2 is 46.0055</em>

<u>Mass of Fe​2​O3:</u>

Fe\rightarrow2\times55.845 = 111.69

O\rightarrow3\times15.9994 = 47.9982

<em>Mass of Fe2O3 is 159.6882</em>

<u>Mass of CaCO:</u>

Ca\rightarrow1\times40.078 = 40.078

C\rightarrow1\times12.01 = 12.01

O\rightarrow1\times15.9994 = 15.9994

<em>Mass of CaCO will be 68.0874</em>

So, the order will be 159.6882>86.172>68.0874>46.0055 which is \text{ Fe2O3} > \text{C6H14} > \text{CaCO} > \text{NO2}

6 0
3 years ago
How many atoms in total are there in 7.35 mol of magnesium oxide (MgO)<br> molecules?
andreev551 [17]

Answer:

4.42 x 10^24

Explanation:

3 0
3 years ago
Lead (Pb) is used to make a number of different alloys. What is the mass of lead present in an alloy containing 0.15 mol of lead
Vesna [10]
Lead is 207.2 g/mol.
0.15 mol (\frac{207.2g}{mol}) = 31.08 g Pb
3 0
3 years ago
Why might it be useful to determine the elements that a planet or moon is made up of?
GuDViN [60]

to see if its copatible with our environment or to see if there is any unkown element, otherwise i dont know

if wrong very sorry

5 0
3 years ago
How many atoms are in 0.075 mol of titanium?
Annette [7]
The mass of titanium is = 47,867 g/1mol

Applying the rule of avrogado

1mol _______ 6,023 × 10^(23) at

0,075mol ___ x


X . 1mol = 0,075mol . 6,023 . 10^(23)at

X = 0,075 . 6,023 . 10^(23) at

X = 4,51 . 10^(22) atoms

Hope this helps
7 0
3 years ago
Read 2 more answers
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