Answer:
f) is the answer please thanks
Answer:
A new Dana system of classification contains 78 different classes of minerals based on composition and then further classified by type and group. To be considered a mineral, a substance must be an inorganic, naturally formed solid, with a specific chemical formula and a fixed internal structure. To test whether something is a mineral, there are several identification tests to which the substance is subjected, including its resistance to scratching, its density in comparison to water, its color, the degree of light it reflects, the color of the powdered mineral, its breakage pattern and its crystalline form.
Explanation:
Because the the sand on the sea for is very compact and the water salt and chemicals make it compact making it stick together and fill up holes or cracks before going anyhere
The electronegativity of the element affects the ability of its compounds to dissolve in different solvents.
<h3>What is the meaning of electronegativity?</h3>
Electronegativity is a measure of an atom's ability to attract shared electrons to itself.
Polar bonds have a positive and negative side to them, and therefore can attract water dipoles and dissolve in water.
If the polarities of the solvent and solute match (both are polar or both are nonpolar), then the solute will probably dissolve.
If the polarities of the solvent and solute are different (one is polar, one is nonpolar), the solute probably won't dissolve.
Hence, option D is correct.
Learn more about electronegativity here:
brainly.com/question/14560699
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Answer:
ΔH⁰(11.4g NH₄NO₃) = -30.59Kj (4 sig. figs. ~mass of NH₄NO₃(s) given) (exothermic)
Explanation:
3NH₄NO₃(s) + C₁₀H₂₂(l) + 14O₂(g) => 3N₂(g) + 17H₂O(g) + 10CO₂(g)
ΔH⁰(f): 3(-365.6)Kj 1(-301)Kj 14(0)Kj 3(0)Kj 17(-241.8)Kj 10(-393.5)Kj
= -1096.8Kj = -301Kj = 0Kj = 0Kj = -4110.6Kj = -3930.5Kj
ΔHₙ°(rxn) = ∑
(ΔH˚(f)products) - ∑(ΔH˚(f)reactants)
= [3(0)Kj + 17(-241.8)Kj + (-393.5)Kj] - [(-(1096.8)Kj + (-301)Kj + (0)Kj]
= [-(8041.1) - (-1397.8)]Kj
= -6643.3Kj (for 3 moles NH₄NO₃ used in above equation)
∴ Standard Heat of Rxn = -6643.3Kj/3moles = -214.8Kj/mole NH₄NO₃(s)
ΔH°(rxn for 14.11g NH₄NO₃(s)) = (11.4g/80.04g·mol⁻¹)(-214.8Kj/mol) = 30.5937Kj ≅ 30.59Kj (4 sig. figs. ~mass of NH₄NO₃(s) given)
