All categories

2 years ago

Determine whether or not each ion contributes to water hardness.

Chemistry

1 answer:

Virty [35]2 years ago

4 0

Answer: The ion that contribute to water hardness are:

--> a. Ca2+

--> b. (HCO)3^- and

--> c. Mg2+

While K+ DOES NOT contribute to water hardness.

Explanation:

WATER in chemistry is known as a universal solvent. This is so because it is polar in nature and dissolves most inorganic solutes and some polar organic solutes to form aqueous solutions. It is composed of elements such as hydrogen and oxygen in the combined ratio of 2:1.

Water is said to be HARD if it does not lather readily with soap. There are two types of water hardness:

--> Permanent hardness: This is mainly due to the presence of CALCIUM and MAGNESIUM ions in the form of soluble tetraoxosulphate(VI) and chlorides. These ions are removed by adding washing soda or caustic soda.

--> Temporary hardness: This is due to the presence of calcium HYDROGENTRIOXOCARBONATES. It can be removed by boiling and using slaked lime.

Therefore from the above given ions, Ca2+,(HCO)3^- and Mg2+ contributes to water hardness.

You might be interested in

What is the SI unit of mass?<br> Select one:<br> a.Ounce<br> b.Pound<br> c.Gram<br> d.Newton

Varvara68 [4.7K]

the SI unit of mass is kilogram

8 0

3 years ago

Read 2 more answers

What are two systems that use radio waves

Goshia [24]

Mobile radio communication
Broadcasting
Navigation systems

7 0

3 years ago

A 44.0 g sample of an unknown metal at 99.0 oC was placed in a constant-pressure calorimeter of negligible heat capacity contain

tatiyna

Answer:

$C_m=0.474\frac{J}{g\°C}$

Explanation:

Hello.

In this case, since this is a system in which the water is heated up and the metal is cooled down in a calorimeter which is not affected by the heat lose-gain process, we can infer that the heat lost by the metal is gained be water, it means that we can write:

$Q_m=-Q_w$

Thus, in terms of masses, specific heats and temperatures we can write:

$m_mC_m(T_{eq}-T_m)=-m_wC_w(T_{eq}-T_w)$

Whereas the equilibrium temperature is the given final temperature of 28.4 °C and we can compute the specific heat of the metal as shown below:

$C_m=\frac{-m_wC_w(T_{eq}-T_w)}{m_m(T_{eq}-T_m)}$

Plugging the values in and since the density of water is 1.00 g/mL so the mass is 80.0g, we obtain:

$C_m=\frac{-80.0g*4.184\frac{J}{g\°C} (28.4\°C-24.0\°C)}{44.0g(28.4\°C-99.0\°C)}\\\\C_m=0.474\frac{J}{g\°C}$

Best regards!

6 0

2 years ago

6. How many moles of water would require 92.048 kJ of heat to raise its temperature from 34.0 °C to 100.0 °C? (3 marks)

scoray [572]

Taking into account the definition of calorimetry, 0.0185 moles of water are required.

<h3>Calorimetry</h3>

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

Sensible heat is defined as the amount of heat that a body absorbs or releases without any changes in its physical state (phase change).

So, the equation that allows to calculate heat exchanges is:

Q = c× m× ΔT

where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation.

<h3>Mass of water required</h3>

In this case, you know:

Heat= 92.048 kJ
Mass of water = ?
Initial temperature of water= 34 ºC
Final temperature of water= 100 ºC
Specific heat of water = 4.186 $\frac{J}{gC}$

Replacing in the expression to calculate heat exchanges:

92.048 kJ = 4.186 $\frac{J}{gC}$ × m× (100 °C -34 °C)

92.048 kJ = 4.186 $\frac{J}{gC}$ × m× 66 °C

m= 92.048 kJ ÷ (4.186 $\frac{J}{gC}$ × 66 °C)

<u><em>m= 0.333 grams</em></u>

<h3>Moles of water required</h3>

Being the molar mass of water 18 $\frac{g}{mole}$ , that is, the amount of mass that a substance contains in one mole, the moles of water required can be calculated as: