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kenny6666 [7]
3 years ago
15

Why are hydrogen bonds important for the properties of water?

Chemistry
1 answer:
zavuch27 [327]3 years ago
5 0
Water has the special type of attraction called Hydrogen bonding.  The bonds between the Hydrogen and the Oxygen in each water molecule make a super dipole because the Oxygen atom is way more electronegative than the hydrogen atom.  These OH bonds can then be attracted to other H2O molecules.  If you have ever poured water up to the brim and there is  little bit of water that is poking above the top, hydrogen bonding keeps those water molecules from spilling
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Pls help need it now as I don't know the answer
ioda

Answer:

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3 0
3 years ago
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1. Addition of which of the following will increase the solubility of CaCO3 in water? Consider the equilibrium process:
kolbaska11 [484]

Answer: HCl

Explanation:

calcium carbonate dissolves in HCl acid producing CO 2 gas. It will not dissolve in pure water. The Ksp for calcium carbonate in water is 3.4 x 10-9 moldm-3 which is very low. What takes place here is actually a chemical reaction:

CaCO 3 (s) + 2HCl(aq) → CaCl 2 (aq) + H 2CO 3(aq)

This reaction accounts for the solubility of the Calcium carbonate in HCl and not in pure water.

6 0
3 years ago
Why are ions charged particles?
Deffense [45]

Answer:

An ion is a charged atom or molecule. It is charged because the number of electrons do not equal the number of protons in the atom or molecule. An atom can acquire a positive charge or a negative charge depending on whether the number of electrons in an atom is greater or less then the number of protons in the atom.

Explanation:

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8 0
3 years ago
In an electrically heated boiler, water is boiled at 140°C by a 90 cm long, 8 mm diameter horizontal heating element immersed in
RideAnS [48]

Explanation:

The given data is as follows.

Volume of water = 0.25 m^{3}

Density of water = 1000 kg/m^{3}

Therefore,  mass of water = Density × Volume

                       = 1000 kg/m^{3} \times 0.25 m^{3}

                       = 250 kg  

Initial Temperature of water (T_{1}) = 20^{o}C

Final temperature of water = 140^{o}C

Heat of vaporization of water (dH_{v}) at 140^{o}C  is 2133 kJ/kg

Specific heat capacity of water = 4.184 kJ/kg/K

As 25% of water got evaporated at its boiling point (140^{o}C) in 60 min.

Therefore, amount of water evaporated = 0.25 × 250 (kg) = 62.5 kg

Heat required to evaporate = Amount of water evapotaed × Heat of vaporization

                           = 62.5 (kg) × 2133 (kJ/kg)

                           = 133.3 \times 10^{3} kJ

All this heat was supplied in 60 min = 60(min)  × 60(sec/min) = 3600 sec

Therefore, heat supplied per unit time = Heat required/time = \frac{133.3 \times 10^{3}kJ}{3600 s} = 37 kJ/s or kW

The power rating of electric heating element is 37 kW.

Hence, heat required to raise the temperature from 20^{o}C to 140^{o}C of 250 kg of water = Mass of water × specific heat capacity × (140 - 20)

                      = 250 (kg) × 40184 (kJ/kg/K) × (140 - 20) (K)

                     = 125520 kJ  

Time required = Heat required / Power rating

                       = \frac{125520}{37}

                       = 3392 sec

Time required to raise the temperature from 20^{o}C to 140^{o}C of 0.25 m^{3} water is calculated as follows.

                    \frac{3392 sec}{60 sec/min}

                     = 56 min

Thus, we can conclude that the time required to raise the temperature is 56 min.

4 0
3 years ago
Technetium-104 has a half life of 18.0 minutes. how much of a 165.0 g sample remains after 90.0 minutes?
11Alexandr11 [23.1K]

The mass of 165.0 g sample that remains after 90.0 minutes is 5.16 grams

calculation

lambda㏑2/18= 0.0385

m(t)= 165 x e( 0.0385 x90)  =5.16g

8 0
3 years ago
Read 2 more answers
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