Somebody answer please a picture is attached!??

Determine whether or not each ion contributes to water hardness.

Virty [35]

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.

4 0

2 years ago

Carbon monoxide (CO) is a poisonous gas because it binds very strongly to the oxygen carrier hemoglobin in blood. A concentratio

velikii [3]

Answer: The amount of CO that is occupied in the room is $1.98\times 10^3L$

Explanation:

We are given:

Concentration of CO = $8.00\times 10^2ppm=800pm$ by volume

This means that $800\mu L\text{ or }800\times 10^{-6}$ of CO is present in 1 L of blood

To calculate the volume of cuboid, we use the equation:

$V=lbh$

where,

V = volume of cuboid

l = length of cuboid = 10.99 m

b = breadth of cuboid = 18.97 m

h = height of cuboid = 11.89 m

$V=10.99\times 18.97\times 11.89=2478.83m^3$

Converting this into liters, by using conversion factor:

$1m^3=1000L$

So, $2478.83m^3=2.479\times 10^6L$

Applying unitary method:

In 1 L of blood, the amount of CO present is $800\times 10^{-6}$

So, in $2.479\times 10^6L$ of blood, the amount of CO present will be = $\frac{800\times 10^{-6}}{1}\times 2.479\times 10^{6}=1983.2L=1.98\times 10^3L$

Hence, the amount of CO that is occupied in the room is $1.98\times 10^3L$

8 0

2 years ago

it was observed that the atoms of a substance were arranged in a closely packed structure. What is most likely the state of matt

Kryger [21]

Answer:

Solid

Explanation:

Atoms are closely packed in a "Solid" substance, So, the substance is a solid.

5 0

3 years ago

At a particular temperature, Kp 0.25 for the reaction a. A ask containing only N2O4 at an initial pressure of 4.5 atm is allowed

Alex

Answer:

a. pNO₂ = 1 atm pN₂O₄ = 4 atm

b. pNO₂ = 1 atm pN₂O₄ = 4 atm

c. It does not matter.

Explanation:

From the information given in this question we know the equilibrium involved is

N₂O₄ (g) ⇄ 2 NO₂ (g)

with Kp given by

Kp = p NO₂²/ p N₂O₄ = 0.25

We know that if we place 4.5 atm of N₂O₄ is placed in a flask, a quantity x is going to be consumed producing 2x atm of NO₂ and we can setup the following equation:

0.25 = p NO₂²/ p N₂O₄ = (2x)² / (4.5 - x)

0.25 x (4.5 - x) = 4x²

4x² + 0.25 x - 1.125 = 0

after solving this quadratic equation, we get two roots

x₁ = 0.5

x₂ = -0.56

the second root is physically impossible, and the partial pressures for x₁ = 0.5 will be

pNO₂ = 2 x 0.5 atm = 1.0 atm

pN₂O₄ = (4.5 - 0.5) atm = 4.0 atm

Similarly for part b, we get the equilibrium equation

0.25 = (9- 2x)² / x

0.25x = 81 - 36x + 4x²

the roots of this equation are:

x₁ = 5.0625

x₂ = 4

the first root is physically impossible since it will give us a negative partial pressure of N₂O₄ :

p N₂O₄ = 9 - 2(5.0625) = -1.13

the second root give us the following partial pressures:

p N₂O₄ = (9 - 2x4) atm = 1 atm

p NO₂ = 4 atm

The partial pressures are the same, it does not matter from which direction an equilibrium position is reached since what is essential is that the partial pressures of the gasses N₂O₄ and NO₂ obey the equilibrium equation.

8 0

2 years ago

How many significant digits are in the volume measurement 0.010 mL? A) 1 B) 2 C) 3 D) 4 E) none of the above

fgiga [73]

Answer:

The correct answer is option B.

Explanation:

Significant figures :These are figures in a number which expresses the value of the magnitude of a quantity to a specific degree of accuracy is known as significant digits.

Significant digits in the volume measurement 0.010 mL are:

$0.0\bar{10}$ = 2 significant figures

All non-zero numbers are always significant.
All zero’s preceding the first integers are never significant.
All zero’s after the decimal point are always significant.

6 0

3 years ago