All categories

3 years ago

Enter the atomic symbol, including mass number and atomic number, for iodine-130.

Chemistry

1 answer:

WINSTONCH [101]3 years ago

6 0

<u>Answer:</u> The atomic symbol of the given element is $_{53}^{130}\textrm{I}$

<u>Explanation:</u>

The general isotopic representation of an element is given as: $_Z^A\textrm{X}$

where,

Z represents the atomic number of the element

A represents the mass number of the element

X represents the symbol of an element

For the given isotope: 130-iodine

Mass number = 130

Atomic number = 53

Hence, the atomic symbol of the given element is $_{53}^{130}\textrm{I}$

You might be interested in

If there are no changes in the oxidation state of the reactants or products of a particular reaction, that reaction is not a red

dmitriy555 [2]

Yes, that is true. in order for it to be a redox reaction, both oxidation and reduction must be occurring.

3 0

3 years ago

Predict the products of below reaction, and whether the solution at equilibrium will be acidic, basic, or neutral.

kati45 [8]

Answer: The product of the given reaction is $HNO_{3}$ and the solution at equilibrium will be acidic.

Explanation:

When two or more chemical substances react together then it forms new substances and these new substances are called products.

For example, $3N_{2}O_{5} + 3H_{2}O \rightarrow 6HNO_{3}$

This shows that nitric acid $(HNO_{3})$ is the product formed and it is an acidic substance.

Hence, the solution at equilibrium will be acidic in nature.

Thus, we can conclude that the product of the given reaction is $HNO_{3}$ and the solution at equilibrium will be acidic.

8 0

3 years ago

Number of Mg(OH)2 formula units in 7.40 moles of Mg(OH)2.

-Dominant- [34]

The answer is 4.45 × 10²⁴ units.

To calculate this, we will use Avogadro's number which is the number of units (atoms, molecules) in 1 mole of substance:
6.02 × 10²³ units per 1 mole

So, we need a proportion:
If 6.02 × 10²³ units are in 1 mole, how many units will be in 7.40 moles:
6.02 × 10²³ units : 1 mole = x : 7.40 moles

After crossing the products:
1 mole * x = 7.40 moles * 6.02 × 10²³ units
x = 7.40 * 6.02 × 10²³ units
x = 44.5 × 10²³ units = 4.45× 10²⁴ unit

5 0

2 years ago

Before tackling this problem, be sure you know how to find the antilog of a number using a scientific calculator.

dybincka [34]

<h2>Question:- </h2>

A solution has a pH of 5.4, the determination of [H+].

<h2>Given :- </h2>

pH:- 5.4
pH = - log[H+]

<h2>To find :- concentration of H+</h2>

<h2>Answer:- Antilog(-5.4) or 4× 10-⁶</h2>

<h2>Explanation:- </h2><h3>Formula:- pH = -log H+ </h3>

Take negative to other side

-pH = log H+

multiple Antilog on both side

(Antilog and log cancel each other )

Antilog (-pH) = [ H+ ]

New Formula :- Antilog (-pH) = [+H]

Now put the values of pH in new formula

Antilog (-5.4) = [+H]

we can write -5.4 as (-6+0.6) just to solve Antilog

Antilog ( -6+0.6 ) = [+H]

Antilog (-6) × Antilog (0.6) = [+H]

$Antilog (-6) = {10}^{ - 6} , \\ Antilog (0.6) = 4$

put the value in equation

${10}^{ - 6} \times 4 = [H+] \\ 4 \times {10}^{ - 6} = [H+]$

7 0

2 years ago

Read 2 more answers

At room temperature (20°C} and pressure, the density of air is 1.189 g/L. An object will float in air if its density is less tha

Alekssandra [29.7K]

Explanation:

$Density =\frac{Mass}[Volume}$

Density of the air ,d= 1.189 g/L

(a) Density of the evacuated ball

Mass of the ball ,m = 0.12 g

Volume of the ball = $V=560 cm^3=560 ml=0.560 L$

$D =\frac{0.12 g}{0.560 L}=0.214 g/L$

D<d, teh evacuated ball will flaot in air.

(b) Density of the evacuated ball D = 0.214 g/L

Density of carbon dioxide gas = $d_1=1.830 g/L$

Mass of the carbon dioxide gas :

$1.830 g/L\times 0.560 L=1.0248 g$

Total density of filled ball with carbon dioxide gas:

$\frac{0.12 g+1.0248 g}{0.560 L}==2.044 g/L$

The ball filled with carbon dioxide will not float in the air because total density of filled ball is greater than the density of an air.

(c) Density of the evacuated ball D = 0.214 g/L

Density of hydrogen gas = $d_2=0.0899 g/L$

Mass of the hydrogen gas :

$1.830 g/L\times 0.560 L=0.050344 g$

Total density of filled ball with hydrogen gas:

$\frac{0.12 g+0.050344 g}{0.560 L}==0.3041 g/L$

The ball filled with hydrogen will float in the air because total density of filled ball is lessor than the density of an air.

(d) Density of the evacuated ball D = 0.214 g/L

Density of oxygen gas = $d_3=1.330 g/L$

Mass of the oxygen gas :

$1.330 g/L\times 0.560 L=1.7448 g$

Total density of filled ball with oxygen gas:

$\frac{0.12 g+1.7448 g}{0.560 L}=1.5442 g/L$

The ball filled with oxygen will not float in the air because total density of filled ball is greater than the density of an air.

(e) Density of the evacuated ball D = 0.214 g/L

Density of nitrogen gas = $d_4=1.165 g/L$

Mass of the nitrogen gas :

$1.165 g/L\times 0.560 L=0.6524 g$

Total density of filled ball with nitrogen gas:

$\frac{0.12 g+0.6524 g}{0.560 L}==1.3792 g/L$

The ball filled with nitrogen will not float in the air because total density of filled ball is greater than the density of an air.

f) Mass must be added to sink the ball = m

Density of ball > Density of the air ; to sink the ball.

$\frac{0.12g +m}{0.560L}>1.189 g/L$

m > 0.54584 g

For any case weight added to ball to make it sink in an air should be grater than the value of 0.54584 grams.

5 0

2 years ago