A certain element has a melting point over 700 ∘C and a density less than 2.00 g/cm3. What is one possible identity for this element?

Answers

Answer 1
Answer: The only answer to an element that melts over 700 degrees Celcius and has a density of less than 2 g/cm3 is Beryllium. This element is rarely found anywhere in the entire universe. It's atomic number is 4 and it is created through stellar nucleosynthesis.
Answer 2
Answer:   The element is Beryllium

Related Questions

A student needs .575 moles of sodium chloride for an experiment. how many grams she she mass out
What is the pH of a Koh solution that has [H+]=1.87×10^-13M
Determine the mass of 5.20 moles of C6H12 (gram-formula mass = 84.2 grams/mole).
When dilute sulphuric acid is added to barium chloride solution then what happens in a balanced equation?
Compare the arrangements of individual particles in solids, liquids, and gases

Which polyatomic ion is found in the compound represented by the formula NaHCO3?(1) acetate (3) hydrogen sulfate
(2) hydrogen carbonate (4) oxalate

Answers

Answer: The correct option is 2.

Explanation: Polyatomic ions are defined as the ions in which two or more elements are covalently bonded to act as a single unit. They always carry some charge.

Examples of polyatomic ions are: OH^- known as hydroxide ion, SO_4^(2-) known as sulfate ion

We are given a compound NaHCO_3 having 2 ions which are Na^+ and HCO_3^-. these ions are named as sodium ions and hydrogen carbonate ions.

Therefore, the correct option is 2.

The correct answer is option 2. Hydrogen carbonate is the polyatomic ion in the compound given. A polyatomic ion is a charged species composed of covalently bonded atoms. Hydrogen carbonate, also known as bicarbonate, is  a polyatomic ion with the charge of -1.

Use the drop-down menus to identify the strongest intermolecular force that is likely to affect each of thesamples shown below.
Η Ο Η
TIL 1
H-C-C-C-H
1-CI
lodine monochloride, ICI
H
H
Acetone, C3H60
H-F
A mixture of water (H2O) and hydrogen fluoride
(HF)

Answers

Answer:

London dispersion, dipole-dipole interactions, hydrogen bonding

Explanation:

Please mark brainliest

Answer:

lodine monochloride- dipole-dipole interactions

Acetone-London dispersion

A mixture of water (H2O) and hydrogen fluoride-hydrogen bonding

Explanation:

Edg 2020 (Brainliest Lol)

Can someone help me find the theoretical value of methane? It is chemistry

Answers


Methane is lighter than air, having a specific gravity of 0.554. It is only slightly soluble in water. It burns readily in air, forming carbon dioxideand water vapour; the flame is pale, slightly luminous, and very hot.The boiling point of methane is −162 °C (−259.6 °F) and the meltingpoint is −182.5 °C (−296.5 °F). Methane in general is very stable, butmixtures of methane and air, with the methane content between 5 and 14percent by volume, are explosive. Explosions of such mixtures have beenfrequent in coal mines and collieries and have been the cause of manymine disasters.

Steel is made mostly of iron and carbon, though sometimes it can containa. tin.
b. nickel.
c. brass.
d. cadmium.

Answers

The correct answer is B. Steel is made mostly of iron and carbon, though sometimes it can contain nickel. Alloying elements are being added to steel to modify and enhance the characteristics of the steel. Other elements include manganese, chromium, boron, titanium, etc.

Answer:

Taskmasters is Correct. The answer is Choice B.

I just took the quiz and got it Correct.

Explain the reason for sodiums location in the periodic table , explaining both location in its group and in its period

Answers

Explanation:

It is known that the atomic number of sodium is 11 and its electronic distribution is 2, 8, 1.

As the number of rows (called periods) in the periodic table represent the number of shells present in an element.

Since, sodium element have only two shells. Therefore, sodium lies in row 2 of the periodic table.

Also, the elements which contain same number of valence electrons lie in the same group.

Therefore, sodium belongs to group 1 and elements of the group 1 are also known as alkali metals.

Thus, we can conclude that in the periodic table sodium lies in period 2 and group 1.

Sodium is part of the alkali metal group because it has properties that have periodic patterns relating to the rest of the group. It is in it's period because of it's energy level and the amount of electrons on that level. It is also apart of it's group because of how many valence electrons it has.

Reaction 2 occurs when an excess of 6 M HCI(aq) solution is added to 100. mL of NaOCl(aą) of unknown concentration. If the reaction goes to completion and 0.010 mol of Cl2(g) is produced, then what was the molarity of the NaOCl(aq) solution? (A) 0.0010 M
(B) 0.010 M
(C) 0.10 M
(D) 1.0 M

Answers

The molarity of the NaOCl solution is 0.10M

The correct answer to the question is Option C. 0.10 M

We'll begin by calculating the number of mole of NaOCl required to produce 0.010 mole of Cl₂.This can be obtained as follow:

NaOCl + 2HCl —> Cl₂ + NaCl + H₂O

From the balanced equation above,

1 mole of NaOCl reacted to produce 1 mole of Cl₂.

Therefore

0.010 mole of NaOCl will also react to produce 0.010 mole of Cl₂.

  • Finally,we shall determine the molarity of the NaOCl solution.

Mole of NaOCl = 0.01 mole

Volume = 100 mL = 100 / 1000 = 0.1 L

Molarity of NaOCl =?

Molarity = mole / Volume

Molarity of NaOCl = 0.01 / 0.1

Molarity of NaOCl = 0.1 M

Therefore, the molarity of the NaOCl solution is 0.1 M

Learn more: brainly.com/question/12755239

Answer:

Option C 0.10 M

Explanation:

you are not providing the reaction, but luckily we have data to write the equation. We know that we are taking HCl and NaOCl to produce Cl2, the reaction is as follow:

NaOCl + 2HCl ---------> Cl₂ + NaCl + H₂O

From this reaction, we know that we have excess of acid that is added to 100 mL (0.1 L) of NaOCl, and this produces 0.010 moles of Chlorine.

All we have to do here, is apply stechiometry. We already know that the acid is in excess, so the moles produced would be the moles of the limiting reactant, in this case, the NaOCl.

According to the reaction NaOCl and Cl2 has a mole ratio of 1:1, so the moles of Cl2 would be the same moles that NaOCl lost so:

moles Cl2 = moles NaOCl = 0.010 moles

Then the concentration:

M = 0.010 / 0.1

M = 0.10 M

This is the molarity of NaOCl