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3 years ago

This is about rocks and/or minerals

Chemistry

2 answers:

Vitek1552 [10]3 years ago

8 0

Graphite is most likely to break apart with cleavage due to the fact that the carbon atoms in graphite,from sheets, could split easily in layers. Graphite had cleavage as the weak bonds in the layers can easily break. It’s unlikely to fracture as fractures usually break in jagged pieces. Atoms in graphite are arranged in layers.

vampirchik [111]3 years ago

5 0

I need the same answer

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What does stoichiometry use to relate moles of one molecule to moles of<br> another?

klio [65]

Answer:

The answer is B

Explanation:

6 0

3 years ago

What is the element of 1s^22s^2p^63s^23p^64s^23d^104p^65s^24d^105p^1

77julia77 [94]

The answer is Indium

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3 years ago

How many grams of o2 are necessary to react with 212 G of mg 2mg+o2➡2mgo?

gulaghasi [49]

We are given an equation 2Mg+O2-> 2MgO and a starting chemical Mg of about 212 g. In order to solve for the amount of O2 needed, we need the molecular weight of Mg and O2.
Molecular weight:
Mg=24.305 g/mol
O2=16(2)=32 g/mol

Note that for every 1 mol of O2, the amount of Mg must be 2 mol.
So,
g O2 = 212 g Mg x1mo Mgl/24.305 g Mg x1mol O2 /2 mol Mg x 32 g O2/mol O2

gO2=139.56 g

Therefore, 139.56 g of O2 is needed for every 212 g Mg.

7 0

4 years ago

is the general formula of a certain hydrate. When 256.3 g of the compound is heated to drive off the water, 214.2 g of anhydrous

Marizza181 [45]

Answer:

The general formula of the hydrate is Caa Seb Oc. nH2O. Based on the given information, the weight of the hydrated compound is 256.3 grams, the weight of the anhydrous compound is 214.2 grams.

Therefore, the weight of water evaporated is 256.3 g - 214.2 g = 42.1 grams

The molecular weight of water is 18 gram per mole. So, the number of moles of water will be,

Moles of water = weight of water/molecular weight

= 42.1 grams / 18 = 2.3

The given composition of calcium is 21.90 %. So, the concentration of calcium in anhydrous compound is,

= 214.2 * 0.2190 = 46.91 grams

The given composition of Se is 43.14 %. So, the concentration of selenium in anhydrous compound is,

= 214.2 * 0.4314 = 92.40 grams

The given composition of oxygen is 34.97%, So, the concentration of oxygen in anhydrous compound is,

= 214.2 * 0.3497 = 74.91 grams

The molecular weight of Ca is 40.078, the obtained concentration is 46.91 grams, stoichiometry will be, 46.91/40.078 = 1.17

The molecular weight of Se is 78.96, the obtained concentration is 92.40, stoichiometry will be,

92.40/78.96 = 1.17

The molecular weight of Oxygen is 15.999, the concentration obtained is 74.91, the stoichiometry will be,

74.91/15.999 = 4.68.

Thus, the formula becomes, Ca1.17. Se1.1e O4.68. 2.3H2O, the closest actual component is CaSeO4.2H2O

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4 years ago

1. Which of these is the basic unit of mass?

jeka57 [31]

1.)B 2.)C 3.)A 4.)B 5.)C

7 0

3 years ago

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