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Naddik [55]
3 years ago
11

A) indicate the volume of the liquid and its shape after pouring it into the beaker

Chemistry
1 answer:
Naily [24]3 years ago
6 0

Answer:

a). Volume is 30 cm³ and it has a indefinite shape.

Volume of the liquid is measured using a conical flask on the right.

c). Drinking water is a homogeneous mixture though it is a compound. And it has a indefinite shape though it takes up shape of a container.

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Chemistry Help (Naming Oxyacids Chart)
guapka [62]

Answer:

HClO₃ /chloric acid /suffix -ic/ ClO₃⁻ (chlorate)

HClO₂/ chlorous acid/ suffix -ous/ ClO₂⁻ (chlorite)

HNO₃ /nitric acid /suffix -ic/ NO₃⁻ (nitrate)

HNO₂/ nitrous acid/ suffix -ous/ NO₂⁻ (nitrite)

Explanation:

Chlorine has 4 positive oxidation numbers to form oxyacids: +1, +3, +5 and +7.

  • When it uses the oxidation number +5, it forms HClO₃, which is named chloric acid, with the suffix -ic. When it loses an H⁺, it forms the oxyanion ClO₃⁻ (chlorate).
  • When it uses the oxidation number +3, it forms HClO₂, which is named chlorous acid, with the suffix -ous. When it loses an H⁺, it forms the oxyanion ClO₂⁻ (chlorite).

Nitrogen has 2 positive oxidation numbers to form oxyacids: +3 and +5.

  • When it uses the oxidation number +5, it forms HNO₃, which is named nitric acid, with the suffix -ic. When it loses an H⁺, it forms the oxyanion NO₃⁻ (nitrate).
  • When it uses the oxidation number +3, it forms HNO₂, which is named nitrous acid, with the suffix -ous. When it loses an H⁺, it forms the oxyanion NO₂⁻ (nitrite).
7 0
3 years ago
What is the entropy change for the freezing process of 1 mole of liquid methanol at its freezing temperature (–97.6˚C) and 1 atm
Rudiy27

Answer : The value of change in entropy for freezing process is, -18.07 J/mol.K

Explanation :

Formula used :

\Delta S=\frac{\Delta H_{freezing}}{T_f}

where,

\Delta S = change in entropy

\Delta H_{fus} = change in enthalpy of fusion = 3.17 kJ/mol

As we know that:

\Delta H_{fus}=-\Delta H_{freezing}=-3.17kJ/mol=-3170J/mol

T_f = freezing point temperature = -97.6^oC=273+(-97.6)=175.4K

Now put all the given values in the above formula, we get:

\Delta S=\frac{\Delta H_{freezing}}{T_m}

\Delta S=\frac{-3170J/mol}{175.4K}

\Delta S=-18.07J/mol.K

Therefore, the value of change in entropy for freezing process is, -18.07 J/mol.K

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