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ASHA 777 [7]
1 year ago
10

Give the name, atomic symbol, and group number of the element with each Z value, and classify it as a metal, metalloid, or nonme

tal:
(e) Z = 42
Chemistry
1 answer:
Viktor [21]1 year ago
5 0

Atomic number = 42

Name = Molybdenum

Atomic symbol  = Mo

Group number = VI(B)6

Mo = Metal

How to find all valves of Z=42?

Here, we are going to find out the name, symbol and group numbers element with the following Z value and their classification as a metal, metalloid, or nonmetal.

The atomic number of the element is 42

Therefore, the name of the element is Molybdenum

The atomic symbol of the element is Mo

The group number of  is VI(B)6

Mo is a metal

Hence, the element is  Molybdenum

Learn more about atomic symbol here :

brainly.com/question/930789

#SPJ4

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Complete and balance the following neutralization reaction Ca(OH)2 + H2SO4
Debora [2.8K]

Answer: Ca(OH)2 (aq) + H2SO4 (aq) ---------->  CaSO4(aq) + 2H2O(l)

Explanation:

Since this is a neutralization reaction, the end product would be salt and water. In this equation Calcium will displace hydrogen from the acid because it is more reactive, resulting in the formation of CaSO4 (salt), while the displaced H2 molecule combines with OH molecules to form water.

The equation of the reaction is thus;

Ca(OH)2 (aq) + H2SO4 (aq) ---------->  CaSO4(aq) + H2O(l), in other to balance it, we add ''2'' to the water molecule in the right hand side of the equation.

Balance equation is  

Ca(OH)2 (aq) + H2SO4 (aq) ---------->  CaSO4(aq) + 2H2O(l)

6 0
3 years ago
What orbital is represented by the transition metals in period four
andreev551 [17]

The name transition metal refers to the position in the periodic table of elements. The transition elements represent the successive addition of electrons to the d atomic orbitals of the atoms. In this way, the transition metals represent the transition between group 2 (2A) elements and group 13 (3A) elements.

8 0
3 years ago
If 316 mL nitrogen is combined with 178 mL oxygen, what volume of N2O is produced at constant temperature and pressure if the re
lord [1]

Answer;

=259 ml

Explanation;

-According to Gay Lussac's Law of Combining Volumes when gases react, they do so in volumes which have a simple ratio to one another, and to the volume of the product formed if gaseous, provided the temperature and pressure remain constant.

-Thus; from the volume of nitrogen and oxygen gases; we have; 316 / 178 = 1.775 moles of nitrogen gas per mole of oxygen gas.

-Therefore, nitrogen gas is the limiting reactant, and for each mole of nitrogen gas used, we will get 1 mole of N2O. This means the resulting volume of N2O with 100% yield will be the same as the volume of nitrogen gas used, thus, 100% yield will produce 316 mL.

However, with 82% yield the volume would be; 316 × 82/100 =259 ml

Therefore; the volume of N2O at 82% yield will be 259 ml

7 0
3 years ago
Read 2 more answers
A sample of pure lithium chloride contains 16% lithium by mass. What is the % lithium by mass in a sample of pure lithium carbon
Anna35 [415]

Answer:

Percentage lithium by mass in Lithium carbonate sample = 19.0%

Explanation:

Atomic mass of lithium = 7.0 g; atomic mass of Chlorine = 35.5 g; atomic mass of carbon = 12.0 g; atomic mass of oxygen = 16.0 g

Molar mass of lithium chloride, LiCl = 7 + 35.5 = 42.5 g

Percentage by mass of lithium in LiCl = (7/42.5) * 100% = 16.4 % aproximately 16%

Molar mass of lithium carbonate, Li₂CO₃ = 7 * 2 + 12 + 16 * 3 =74.0 g

Percentage by mass of lithium in Li₂CO₃ = (14/74) * 100% = 18.9 % approximately 19%

Mass of Lithium carbonate sample = 2 * 42.5 = 85.0 g

mass of lithium in 85.0 g Li₂CO₃ = 19% * 85.0 g = 16.15 g

Percentage by mass of lithium in 85.0 g Li₂CO₃ = (16.15/85.0) * 100 % = 19.0%

Percentage lithium by mass in Lithium carbonate sample = 19.0%

3 0
3 years ago
What is the PH of a 1.3×(10)^-9 M HBr solution?
olga nikolaevna [1]

pH solution = 8.89

<h3>Further explanation</h3>

Given

The concentration of HBr solution = 1.3 x 10⁻⁹ M

Required

the pH

Solution

HBr = strong acid

General formula for strong acid :

[H⁺]= a . M

a = amount of H⁺

M = molarity of solution

HBr⇒H⁺ + Br⁻⇒ amount of H⁺ = 1 so a=1

Input the value :

[H⁺] = 1 x  1.3 x 10⁻⁹

[H⁺] = 1.3 x 10⁻⁹

pH = - log [H⁺]

pH = 9 - log 1.3

pH = 8.89

7 0
2 years ago
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