For snow to fall to the ground, the temperature must be cold both up in the clouds where snowflakes form, and down at ground level. If the air near ground level is too warm, the snow will melt on its way down, changing to rain or freezing rain. Moisture is needed to form clouds and precipitation.
"if it is tested in a controlled setting with repeated results" is the statement among the choices given in the question that best describes that can possibly make this scientific claim valid. The correct option among all the options that are given in the question is the first option or option "A". I hope the answer has helped you.<span>
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Benzene, C6H6 is miscible with hexane (C6H14) but iron (IIl) nitrate, and Sulfuric acid, H2SO4Fe(NO3)3 are not miscible with hexane (C6H14)
Since organic or non polar dissolves in nom polar solvent.here both benzene and hexane (C6H14) are non polar solvents.But iron (IIl) nitrate, Fe(NO3)3 and Sulfuric acid, H2SO4 are inorganic and polar, therefore cannot dissolve with non polar solvent like hexane.
Miscible is a flowery heard that oil and water aren't very miscible substances, while seltzer and orange juice are miscible and scrumptious.Miscible beverages are also described as liquids that can blend to form a homogeneous solution. Miscible drinks generally blend with out limit, meaning they may be soluble at all quantities.
Miscible approach the materials blend completely. If substances are miscible, they're also absolutely soluble in each other no matter the order of creation. for example, tetrahydrofuran and water are miscible.
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The molarity of the acid given the data from the question is 0.30 M
<h3>Balanced equation </h3>
2HNO₃ + Ba(OH)₂ —> Ba(NO₃)₂ + 2H₂O
From the balanced equation above,
- The mole ratio of the acid, HNO₃ (nA) = 2
- The mole ratio of the base, Ba(NO₃)₂ (nB) = 1
<h3>How to determine the molarity of the acid</h3>
From the question given above, the following data were obtained:
- Volume of acid, HNO₃ (Va) = 39.7 mL
- Volume of base, Ba(NO₃)₂ (Vb) = 24 mL
- Molarity of base, Ba(NO₃)₂ (Cb) = 0.250 M
- Molarity of acid, HNO₃ (Ma) =?
MaVa / MbVb = nA / nB
(Ma × 39.7) / (0.25 × 24) = 2
(Ma × 39.7) / 6 = 2
Cross multiply
Ma × 39.7 = 6 × 2
Ma × 39.7 = 12
Divide both side by 39.7
Ma = 12 / 39.7
Ma = 0.30 M
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Answer:
1.30464 grams of glucose was present in 100.0 mL of final solution.
Explanation:
Moles of glucose = 
Volume of the solution = 100 mL = 0.1 L (1 mL = 0.001 L)
Molarity of the solution = 
A 30.0 mL sample of above glucose solution was diluted to 0.500 L:
Molarity of the solution before dilution = 
Volume of the solution taken = 
Molarity of the solution after dilution = 
Volume of the solution after dilution= 
Mass glucose are in 100.0 mL of the 0.07248 mol/L glucose solution:
Volume of solution = 100.0 mL = 0.1 L
Moles of glucose = 
Mass of 0.007248 moles of glucose :
0.007248 mol × 180 g/mol = 1.30464 grams
1.30464 grams of glucose was present in 100.0 mL of final solution.
