Explain why mineral oil which is a mixture of hydrocarbons does not dissolve in water

1 answer:

3 0

Oil molecules like to stick to other oil molecules more than they like to stick to water molecules. This is because mineral oil has nonpolar molecules and water has polar ones.

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PH scale is a measure which helps us to compare the strength of various acids and alkalis.

nevsk [136]

Solutions with pH less than 7 are acidic in nature.

Therefore, a substance of pH 6 will be a weak acid.

6 0

3 years ago

Generally how does the first ionization energy vary as the atomic number increases going across a period

Vlad1618 [11]

Across a period I.E increases progressively from left to right

Explanation:

The trend of the first ionization energy is such that across a period I.E increases from left to right due to the decreasing atomic radii caused by the increasing nuclear charge. This not compensated for by successive electronic shells.

Ionization energy is a measure of the readiness of an atom to lose an electron.
The lower the value, the easier it is for an atom to lose an electron.
Elements in group I tend to lose their electrons more readily whereas the halogens hold most tightly to them.
The first ionization energy is the energy needed to remove the most loosely bonded electron of an atom in the gaseous phase.

Learn more:

Ionization energy brainly.com/question/6324347

#learnwithBrainly

3 0

4 years ago

A 19.0 g piece of metal is heated to 99.0 °C and then placed in 150 mL of water at 21°C. The temperature of the water rose to 23

Anna35 [415]

Answer:

$0.8696\ \text{J/g}^{\circ}\text{C}$

Explanation:

$m_m$ = Mass of metal = 19 g

$c_m$ = Specific heat of the metal

$\Delta T_m$ = Temperature difference of the metal = $99-23=76^{\circ}\text{C}$

V = Volume of water = 150 mL = $150\ \text{cm}^3$

$\rho$ = Density of water = $1\ \text{g/cm}^3$

$c_w$ = Specific heat of the water = 4.186 J/g°C

$\Delta T_w$ = Temperature difference of the water = $23-21=2^{\circ}\text{C}$

Mass of water

$m_w= \rho V\\\Rightarrow m_w=1\times 150\\\Rightarrow m_w=150\ \text{g}$

Heat lost will be equal to the heat gained so we get

$m_mc_m\Delta T_m=m_wc_w\Delta T_w\\\Rightarrow c_m=\dfrac{m_wc_w\Delta T_w}{m_m\Delta T_m}\\\Rightarrow c_m=\dfrac{150\times 4.186\times 2}{19\times 76}\\\Rightarrow c_m=0.8696\ \text{J/g}^{\circ}\text{C}$