A dissolving liquid composed of polar molecules is a polar solvent.
The distinction of polar and non-polar liquids is important because the like dissolves like rule. This rule states that the solubility is greater when the polarity of the liquid is similar to the polarity of the solute.
So, to dissolve polar compounds (e.g. ionic compounds) you should use polar solvents (e.g. water).
Answer: polar solvent
Answer:
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Reproduction is defined as the production of individuals of the same species, that is the next generation of the species.
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There are basically two types of reproduction - asexual and sexual.
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Asexual reproduction involves only one parent and the offspring is
genetically similar to the parent ...
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The different types of asexual reproduction are fission, budding, fragmentation, spore formation and vegetative propagation.
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Sexual reproduction is a type of reproduction that involves a complex life cycle in which a gamete (such as a sperm or egg cell) with a single set of chromosomes combines with another to produce a organism composed of cells with two sets of chromosomes .
Asexual reproduction is a type of reproduction which does not involve the fusion of gametes or change in the number of chromosomes. The offspring that arise by asexual reproduction from a single cell or from a multicellular organism inherit the genes of that parent.
Explanation:
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The balanced reaction is as follows;
BiCl₂ + Na₂SO₄ --> 2NaCl + BiSO₄
this is a double displacement reaction
the oxidation number of Bi is +2 in both BiCl₂ and BiSO₄
oxidation number of Cl is -1 in both BiCl₂ and NaCl
oxidation number of Na is +1 in both Na₂SO₄ and NaCl
oxidation numbers of elements in SO₄²⁻ remains the same in both compounds.Therefore the oxidation state in any of the elements in the reaction doesn't change. Neither of the elements show an increase or decrease in the oxidation numbers .
Answer for this question is no element decreases its oxidation number.
FeBr₃ ⇒ limiting reactant
mol NaBr = 1.428
<h3>Further explanation</h3>
Reaction
2FeBr₃ + 3Na₂S → Fe₂S₃ + 6NaBr
Limiting reactant⇒ smaller ratio (mol divide by coefficient reaction)
211 g of Iron (III) bromide(MW=295,56 g/mol), so mol FeBr₃ :
186 g of Sodium sulfide(MW=78,0452 g/mol), so mol Na₂S :
Coefficient ratio from the equation FeBr₃ : Na₂S = 2 : 3, so mol ratio :
So FeBr₃ as a limiting reactant(smaller ratio)
mol NaBr based on limiting reactant (FeBr₃) :
Find the number of moles
C = n / V
C(Concentration) = 0.30 moles / L
V ( Volume) = 2 L
n = ??
n = C * V
n = 0.30 mol / L * 2 L
n = 0.60 mol
Find the molar mass
2Na = 23 * 2 = 46 grams
1S = 32 * 1 = 32 grams
O4 = 16 * 4 = 64 grams
Total = 142 grams / mol
Find the mass
n = given mass / molar mass
n = 0.06 mol
molar Mass = 142 grams / mol
given mass = ???
given mass = molar mass * mols
given mass = 142 * 0.6
given mass = 85.2 grams.
85.2 are in a 2 L solution that has a concentration of 0.6 mol/L
