When a solute is dissolved in a solvent the solute raises the boiling point of the solvent. The boiling point of a liquid (a solvent) will be higher when another compound (solute) is added, such that a solution has a higher boiling point than a pure solvent. This takes place when a non-volatile solute, such as a salt, is added to a pure solvent such as water.
Answer:
The answer to your question is given after the questions so I just explain how to get it.
Explanation:
a)
Get the molecular weight of Phosphoric acid
H₃PO₄ = (3 x 1) + (31 x 1) + (16 x 4)
= 3 + 31 + 64
= 98 g
98 g ----------------- 1 mol
0.045 g --------------- x
x = (0.045 x 1) / 98
x = 0.045 / 98
x = 0.00046 moles or 4.6 x 10 ⁻⁴
b)
Molarity = 
Molarity = 
Molarity = 0.0013 or 1.31 x 10⁻³
c)
Formula C₁V₁ = C₂V₂
V₁ = C₂V₂ / C₁
Substitution
V₁ = (0.0013)(1) / 0.01
Simplification and result
V₁ = 0.0013 / 0.1
V₁ = 0.13 l = 130 ml
Answer:
See explanation
Explanation:
The reactivity of metals has a lot to do with their position in the electrochemical series. However, it is also known that metallic character decreases across the period. This implies that as we move from left to right along the periodic table. Sodium, magnesium, aluminum and silicon continues to decrease in metallic character. As a matter of fact, silicon is a metalloid and not a pure metal.
Sodium reacts with cold water to give a vigorous reaction,magnesium and aluminium reacts with steam at red heat.
Silicon does not react with water, even as steam, under normal conditions.
Answer:
Using Phosphoric acid will work perfectly for producing Hydrogen halides because its not an Oxidizing agent. ...
Using an ionic chloride and Phosphoric acid
H3PO4 + NaCl ==> HCl + NaH2PO4
H3PO4 + NaI ==> HI + NaH2PO4
H2SO4 + NaCl ==> HCl + NaHSO4
This method(Using H2So4) will work for all hydrogen hydrogen halide except Hydrogen Iodide and Hydrogen Bromide.
The Sulphuric acid won't be useful for producing Hydrogen Iodide because its an OXIDIZING AGENT. Whist producing the Hydrogen Iodide... Some of the Iodide ions are oxidized to Iodine.
2I-² === I2 + 2e-
Explanation:
