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Marina CMI [18]
3 years ago
12

Iron III sulfide is classified as what type

Chemistry
2 answers:
zhuklara [117]3 years ago
8 0
I would say it is a
3 plus ion
Oxana [17]3 years ago
5 0

Answer:

ionic compound

Explanation:

that is the answer if you meant what type of compound.

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Okay So Blink your eyes. Was work done? Explain your answer.
Yuri [45]

Answer:

no work was needed. Blinking does not require using energy sometimes we blink without thinking.

Explanation:

7 0
3 years ago
Read 2 more answers
Which class of organic compounds has molecules that contain nitrogen atoms?
ziro4ka [17]
2) amine is the correct answer 
7 0
3 years ago
B. if 6.73 g of na2co3 is dissolved in enough water to make 250. ml of solution, what is the molar concentration of sodium carbo
BARSIC [14]

Answer:- [Na_2CO_3]=0.254M , [Na^+]=0.508 M , [CO_3^2^-]=0.254M

Solution:- We are asked to calculate the molarity of sodium carbonate solution as well as the sodium and carbonate ions.

Molarity is moles of solute per liter of solution. We have been given with 6.73 grams of sodium carbonate and the volume of solution is 250.mL.  Grams are converted to moles and mL are converted to L and finally the moles are divided by liters to get the molarity of sodium carbonate.

Molar mass of sodium carbonate is 105.99 gram per mol. The calculations for the molarity of sodium carbonate are shown below:

\frac{6.73gNa_2CO_3}{250.mL}(\frac{1mol}{105.99g})(\frac{1000mL}{1L})

= 0.254MNa_2CO_3

So, molarity of sodium carbonate solution is 0.254 M.

sodium carbonate dissociate to give the ions as:

Na_2CO_3(aq)\rightarrow 2Na^+(aq)+CO_3^2^-

There is 1:2 mol ratio between sodium carbonate and sodium ion. So, the molarity of sodium ion will be two times of sodium carbonate molarity.

[Na^+]=2(0.254M) = 0.508 M

There is 1:1 mol ratio between sodium carbonate and carbonate ion. So, the molarity of carbonate ion will be equal to the molarity of sodium carbonate.

[CO_3^2^-]=0.254M


5 0
3 years ago
There are over 100 elements, but only
elena-14-01-66 [18.8K]

Answer:

Symbol_Element

He_Helium

Li_Lithium

Be_Beryllium

B_Boron

C_Carbon

N_Nitrogen

O_Oxygen

F_Fluorine

Ne_Neon

Na_Sodium

Mg_Magnesium

Al_Aluminum

Si_Silicon

P_Phosphorous

S_Sulfur

Cl_Chlorine

Ar_Argon

K_Potassium

Ca_Calcium

Sc_Scandium

Ti_Titanium

V_Vanadium

Cr_Chromium

Mn_Manganese

Fe_Iron

Co_Cobalt

Ni_Nickel

Cu_Copper

Zn_Zinc

Ga_Gallium

Ge_Germanium

As_Arsenic

Se_Selenium

Br_Bromine

6 0
3 years ago
Consider the compound hydroxyapatite, Ca10(PO4)6(OH)2. Name each ion present, give the charge of each ion and show that the form
emmainna [20.7K]
Ca₁₀(PO₄)₆(OH)₂  or  Ca(OH)₂·3Ca₃(PO₄)₂

PO₄³⁻   phosphate ion
OH⁻     oxyhydroxide ion
Ca²⁺     calcium ion

10*(+2)  +  6*(-3)  +  2*(-1) = 0
 10Ca²⁺      6PO₄³⁻      2OH⁻
7 0
3 years ago
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