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Molodets [167]
3 years ago
5

Minerals are grouped into mineral classes primarily on a basis of _____. the number of cleavage directions present chemistry, sp

ecifically the cations within the chemical formula chemistry, specifically the anions within the chemical formula hardness; hard, soft, and medium are the three primary classes
Chemistry
1 answer:
KonstantinChe [14]3 years ago
3 0
I think minerals are grouped into mineral classes primarily on a basis of specifically the anions within the chemical formula. Anions are the negatively charged ions in a mineral. These anions are larger than cations and are mostly attracted to positive ions.
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Complete the table for ion charge based upon their losing or gaining electrons in the outer shell. (Use the periodic table as ne
PolarNik [594]

Answer:

Explanation:

Group one:

The elements of group one shows +1 charge because these all are metals and lose their one valance electron.

Hydrogen lithium sodium potassium rubidium cesium francium

Group 2:

The elements of group two shows +2 charge because these all alkali metals and lose their two valance electrons.

beryllium magnesium calcium strontium barium radium

Group 3:

The elements of group three-B shoes +3 charge by losing three valance electrons.

Scandium yttrium lanthanum actinium

Group 4:

The elements of group 4th A and 4th B lose four electrons or gain four electrons to complete the octet and shows +4 or -4 charge.

Group 5:

Group 5th elements gain three electrons and shows -3 charge to complete the 8 electrons. (octet).

It involve the elements of group 5th A.

Group 6:

The elements of group 6A gain two electrons to complete the octet and shows -2 charge.

Group 7:

The elements of group 7A gain one electron to complete the octet and shows -1 charge.

Group 8:

The elements of group 8A are noble gases and have complete octet. That's why shows 0 charge.

4 0
3 years ago
Below are 5 sets of potential solutes for you to compare. Both members of each pair are very soluble in water. If you had equal
Gelneren [198K]

Answer:

Option D. AlCl₃, MgC₂

Explanation:

We need to dissociate all the salts, to determine the i. (Van't Hoff factor).

The salt who has the highest value, will be the better conductor of electricity

CsCl  → Cs⁺  +  Cl⁻   i = 2

CaCl → Ca²⁺ +  Cl⁻   i = 2

CaS → Ca²⁺ +  S⁻²    i = 2

Li₂S → 2Li⁺  +  S⁻²    i = 3

KBr →  K⁺  +  Br⁻       i = 2

AlCl₃ →  Al³⁺ + 3Cl⁻   i = 4

MgC₂  →  Mg²⁺  +  2C⁻   i = 3

KI  →  K⁺  +  I⁻     i = 2

K₂S  →  2K⁺  + S⁻²   i = 3

The biggest i, is in pair D.

6 0
4 years ago
Temperature is a measure of how much heat you have in, or around you. is it true or false
allochka39001 [22]

Your answer is

true

hope it helps you

7 0
3 years ago
Elimination of the pharmaceutical IV antibiotic gentamicin follows first-order kinetics. If the half-life of gentamicin is 1.5 h
Genrish500 [490]

Explanation:

The given data is:

The half-life of gentamicin is 1.5 hrs.

The reaction follows first-order kinetics.

The initial concentration of the reactants is 8.4 x 10-5 M.

The concentration of reactant after 8 hrs can be calculated as shown below:

The formula of the half-life of the first-order reaction is:

k=\frac{0.693}{t_1_/_2}

Where k = rate constant

t1/2=half-life

So, the rate constant k value is:

k=\frac{0.693}{1.5 hrs}

The expression for the rate constant is :

k=\frac{2.303}{t} log \frac{initial concentration}{concentration after time "t"}

Substitute the given values and the k value in this formula to get the concentration of the reactant after time 8 hrs is shown below:

\frac{0.693}{1.5 hrs} =\frac{2.303}{8 hrs} x log \frac{8.4x10^-^5}{y} \\ log \frac{8.4x10^-^5}{y} =1.604\\\frac{8.4x10^-^5}{y}=10^1^.^6^0^4\\\frac{8.4x10^-^5}{y}=40.18\\y=\frac{8.4x10^-^5}{40.18} \\=>y=2.09x10^-^6

Answer: The concentration of reactant remains after 8 hours is 2.09x10^-6M.

5 0
3 years ago
4. How many grams of ammonium carbonate are needed to decompose in order to produce
Thepotemich [5.8K]

Answer:

14.23g of (NH4)2CO3

Explanation:

We'll begin by writing the balanced equation for the reaction.

(NH4)2CO3 –> (NH4)2O + CO2

Next,, we shall determine the mass of (NH4)2CO3 that decomposed and the mass of CO2 produced from the balanced equation. This is illustrated below:

Molar mass of (NH4)2CO3 = 2[14+(4x1)] + 12 + (16x3)

= 2[14 +4] + 12 + 48

= 2[18] + 60 = 96g/mol

Mass of (NH4)2CO3 from the balanced equation = 1 x 96 = 96g

Molar mass of CO2 = 12 + (2x16) = 44g/mol

Mass of CO2 from the balanced equation = 1 x 44 = 44g.

Summary:

From the balanced equation above,

96g of (NH4)2CO3 decomposed to produce 44g of CO2.

Finally, we can determine the mass of (NH4)2CO3 that decomposed to produce 6.52g of CO2 as follow:

From the balanced equation above,

96g of (NH4)2CO3 decomposed to produce 44g of CO2.

Therefore, Xg of (NH4)2CO3 will decompose to produce 6.52g of CO2 i.e

Xg of (NH4)2CO3 = (96 x 6.52)/44

Xg of (NH4)2CO3 = 14.23g

Therefore, 14.23g of (NH4)2CO3 is needed to produce 6.52g of CO2.

4 0
3 years ago
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