Since water molecules are polar. Unequal sharing of electrons between the elements that make it. The hydrogen atoms would experience a slight partial positive charge as the electron density is more close to the oxygen atom making it partially negative.
Answer:
The freezing point of the solution is - 4.39 °C.
Explanation:
We can solve this problem using the relation:
<em>ΔTf = (Kf)(m),</em>
where, ΔTf is the depression in the freezing point.
Kf is the molal freezing point depression constant of water = -1.86 °C/m,
density of water = 1 g/mL.
<em>So, the mass of 575 mL is 575 g = 0.575 kg.</em>
m is the molality of the solution (m = moles of solute / kg of solvent = (465 g / 342.3 g/mol)/(0.575 kg) = 2.36 m.
<em>∴ ΔTf = (Kf)(m</em>) = (-1.86 °C/m)(2.36 m) = <em>- 4.39 °C.</em>
<em>∵ The freezing point if water is 0.0 °C and it is depressed by - 4.39 °C.</em>
<em>∴ The freezing point of the solution is - 4.39 °C.</em>
Answer:
The force exerted on it.
Explanation:
As the law of motion states, an object at rest will remain at rest an object in motion will remain in motion unless acted upon by another force.
Answer:
Chlorine is more likely to steal a valence electron from sodium.
Explanation:
Sodium is number 11 on the periodic table with one valence electron. Belonging to the first group, it's one of the alkali metal, which are known to be highly reactive. Chlorine is number 17 with seven valence electrons, and it's in the second-to-last group of halogens--also very reactive.
Considering that elements with one valence electron are just about 100% likely to give up electrons to reach a stable state, sodium would be the element that is more likely to lose its valence electron to chlorine. In other words, chlorine would be the electron thief.
Answer:
64799.4 J
Explanation:
The following data were obtained from the question:
Mass (M) = 1.05 kg = 1.05 x 1000 = 1050g
Specific heat capacity (C) = 0.9211 J/g°C
Initial temperature (T1) = 23°C
Final temperature (T2) = 90°C
Change in temperature (ΔT) = T2 – T1 =
90°C – 23°C = 67°C
Heat required (Q) =....?
The heat required to increase the temperature of the kettle can b obtain as follow:
Q = MCΔT
Q = 1050 x 0.9211 x 67
Q = 64799.4 J
Therefore, 64799.4 J of heat is required to increase th temperature of the kettle from 23°C to 90°C.
