Select all that apply.

Enter the atomic symbol, including mass number and atomic number, for iodine-125.

ICE Princess25 [194]

Answer:

Explanation:

Iodine - 125

The atomic symbol of iodine is ¹²⁵₅₃ I

The symbol for iodine is I

The atomic number of iodine is 53,

and the atomic mass of iodine is 125 .

The representation of the atomic symbol is as, the atomic mass is written in uppercase and the atomic number is written in lower case , followed by the symbol of the element .

Iodine is a radio active element , used for many biological process .

It is the second largest -lived radioisotope of iodine .

The first is iodine-129 .

3 0

2 years ago

The image shows the representation of an unknown element in the periodic table. A square is shown. Inside the square twelve is w

Ludmilka [50]

Answer: The correct answer is the mass number of the most common isotope of the element is 24.

Explanation:

We are given:

An element having atomic number 12 is magnesium and atomic mass of the element is 24.305

The image corresponding will be $_{24.305}\textrm{Mg}^{12}\\\text{Magnesium}$

The number '24.305' is the average atomic mass of magnesium element.

Average atomic mass is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$

Average atomic mass of magnesium = 24.305 amu

As, the average atomic mass of magnesium lies closer to the mass of Mg-24 isotope. This means that the relative abundance of this isotope is the highest of all the other isotopes.

The 'Mg-24' isotope is the most common isotope of the given element.

Hence, the correct answer is the mass number of the most common isotope of the element is 24.

4 0

2 years ago

We want to calculate the concentrations of all species in a 0.58 M Na 2 SO 3 (sodium sulfite) solution. The ionization constants

NeX [460]

Explanation:

Reaction equation is as follows.

$Na_{2}SO_{3}(s) \rightarrow 2Na^{+}(aq) + SO^{2-}_{3}(aq)$

Here, 1 mole of $Na_{2}SO_{3}$ produces 2 moles of cations.

$[Na^{+}] = 2[Na_{2}SO_{3}] = 2 \times 0.58$

= 1.16 M

$[SO^{2-}_{3}] = [Na_{2}SO_{3}]$ = 0.58 M

The sulphite anion will act as a base and react with $H_{2}O$ to form $HSO^{-}_{3}$ and $OH^{-}$ .

As, $K_{b} = \frac{K_{w}}{K_{a_{2}}}$

= $\frac{10^{-14}}{6.3 \times 10^{-8}}$

= $1.59 \times 10^{-7}$

According to the ICE table for the given reaction,

$SO^{2-}_{3} + H_{2}O \rightleftharpoons HSO^{-}_{3} + OH^{-}$

Initial: 0.58 0 0

Change: -x +x +x

Equilibrium: 0.58 - x x x

So,

$K_{b} = \frac{[HSO^{-}_{3}][OH^{-}]}{[SO^{2-}_{3}]}$

$1.59 \times 10^{-7} = \frac{x^{2}}{0.58 - x}$

$x^{2} = 1.59 \times 10^{-7} \times (0.58 - x)$

x = 0.0003 M

So, x = $[HSO^{-}_{3}] = [OH^{-}]$ = 0.0003 M

$[SO^{2-}_{3}]$ = 0.58 - 0.0003

= 0.579 M

Now, we will use $[HSO^{-}_{3}]$ = 0.0003 M

The reaction will be as follows.

$HSO^{2-}_{3} + H_{2}O \rightleftharpoons H_{2}SO_{3} + OH^{-}$

Initial: 0.0003

Equilibrium: 0.0003 - x x x

$K_{b} = \frac{x^{2}}{0.0003 - x}$

$K_{b} = \frac{K_{w}}{K_{a_{1}}}$

= $\frac{10^{-14}}{1.4 \times 10^{-2}}$

= $7.14 \times 10^{-13}$

Therefore, $7.14 \times 10^{-13} = \frac{x^{2}}{0.0003 - x}$

As, x <<<< 0.0003. So, we can neglect x.

Therefore, $x^{2} = 7.14 \times 10^{-13} \times 0.0003$

= $0.00214 \times 10^{-13}$

x = $0.0146 \times 10^{-6}$

x = $[OH^{-}] = [H_{2}SO_{3}]$ = $1.46 \times 10^{-8}$

$[H^{+}] = \frac{10^{-14}}{[OH^{-}]}$

= $\frac{10^{-14}}{0.0003}$

= $3.33 \times 10^{-11}$ M

Thus, we can conclude that the concentration of spectator ion is $3.33 \times 10^{-11}$ M.

5 0

3 years ago

Which of these is NOT a colloid?

SVETLANKA909090 [29]

Answer:C

Explanation:

Mountain Dew soda is not a colloid

8 0

2 years ago

- How many grams of nitrogen (N2) are needed to react with 435.2 g of oxygen (02) in the following equation?

Llana [10]

Explanation:

I think the answer is this for a better check mass- mass ratio in stoichiometry lesson

5 0

2 years ago