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Nesterboy [21]
3 years ago
14

What is the oxidation state of each element in FeBr2?

Chemistry
1 answer:
maxonik [38]3 years ago
3 0
Fe: 2+ 
Br: 1-
Checked on the periodic table
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What would be the concentration of a solution resulting when you dissolve 4.725g of nacl (molar mass = 58.45) in water to make a
aliya0001 [1]
<span>There are a number of ways to express concentration of a solution. This includes molarity. It is expressed as the number of moles of solute per volume of the solution.  To convert the mass of the solute to moles, we use the molar mass of the substance. We calculate as follows:

MOlarity = 4.725 g ( 1 mol / 58.45 g ) / .5 L = 0.162 mol / L</span>
3 0
3 years ago
If 20.0g of water at 56.2℃ absorbs 195 J of heat, what is the final temperature? *
Gnom [1K]

Answer

Explanation:

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3 0
2 years ago
What is the mass of H2O produced when 14.0 grams of H2 reacts completely with 2.0 grams of O2?
Vladimir79 [104]
Balanced chemical equation:

2 H2 + 1 O2 = 2 H2O

4 g H2  -------> 32 g O2 -----------> 36 g H2O
   ↓                       ↓                             ↓
14.0 g ---------> 2.0 g O2 ----------> mass H2O ?

32 * mass H2O = 2.0 * 36

32 * mass H2O = 72

mass of H2O = 72 / 32

mass of H2O = 2.25 g

hope this helps!.


4 0
3 years ago
Calculate the volume of a 0.323-mol sample of a gas at 265 K and<br> 0.900 atm.
Katarina [22]

Answer:

V = 7.8 L

Explanation:

Message

5 0
3 years ago
The solubility product of PbI2 is 7.9x10^-9. What is the molar solubility of PbI2 in distilled water?
Agata [3.3K]

<u>Answer:</u> The molar solubility of PbI_2 is 1.25\times 10^{-3}mol/L

<u>Explanation:</u>

Solubility is defined as the maximum amount of solute that can be dissolved in a solvent at equilibrium.

Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio.

The balanced equilibrium reaction for the ionization of calcium fluoride follows:

PbI_2\rightleftharpoons Pb^{2+}+2I^-

                s       2s

The expression for solubility constant for this reaction will be:

K_{sp}=[Pb^{2+}][I^-]^2

We are given:

K_{sp}=7.9\times 10^{-9}

Putting values in above equation, we get:

7.9\times 10^{-9}=(s)\times (2s)^2\\\\7.9\times 10^{-9}=4s^3\\\\s=1.25\times 10^{-3}mol/L

Hence, the molar solubility of PbI_2 is 1.25\times 10^{-3}mol/L

3 0
3 years ago
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