Rutherfordium has atomic number of 84 how many protons and electrons does it have

Can a physical change actually change what a substance is?

zlopas [31]

No in that case it would be a chemical change

4 0

3 years ago

How to find molar mass

Shkiper50 [21]

Answer:

Multiply the element's atomic mass by the number of atoms of that element in the compound. This will give you the relative amount that each element contributes to the compound. For hydrogen chloride, HCl, the molar mass of each element is 1.007 grams per mole for hydrogen and 35.453 grams per mole for chlorine.

3 0

3 years ago

Is there an N−Cl bond in solid ammonium chloride?

Tems11 [23]

When ammonia is reacted with HCl it abstracts proton from acid and forms Ammonium Ion and Chloride Ion.

NH₃ + HCl → ⁺NH₄ + Cl⁻ (simply Written NH₄Cl)
Structure,
The structure of Ammonium Chloride is among those structures which contains all three types of bonding's, i.e.

Ionic Bond

Covalent Bond

Coordinate Covalent Bond

Three Hydrogen atoms previously bonded with Nitrogen are covalent in nature. The new incoming proton from HCl forms co-ordinate covalent bond with Nitrogen and Chloride Ion containing negative charge make Ionic Bond with the positive Ammonium Ion. In question, if the line between Nitrogen and Chlorine atom is assumed covalent then it is incorrect. Structure is shown below,

5 0

2 years ago

At the start of a measurement, a radioisotope has 10,000 unstable nuclei. over 2 days 5,000 of these unstable nuclei undergo rad

Black_prince [1.1K]

The half life of a radioactive element is the time needed to the element to decay and reach the half amount of the initial amount. Here we have a radioisotope element which decays its half from 10,000 to 5,000 in two days. Therefore, its half life is 2 days.

6 0

3 years ago

Will give lots of points if answered correctly. Determine the kb for chloroform when 0.793 moles of solute in 0.758 kg changes t

Liono4ka [1.6K]

Answer: The value of $K_{b}$ for chloroform is $3.62^{o}C/m$ when 0.793 moles of solute in 0.758 kg changes the boiling point by 3.80 °C.

Explanation:

Given: Moles of solute = 0.793 mol

Mass of solvent = 0.758

$\Delta T_{b} = 3.80^{o}C$

As molality is the number of moles of solute present in kg of solvent. Hence, molality of given solution is calculated as follows.

$Molality = \frac{no. of moles}{mass of solvent (in kg)}\\= \frac{0.793 mol}{0.758 kg}\\= 1.05 m$

Now, the values of $K_b$ is calculated as follows.

$\Delta T_{b} = i\times K_{b} \times m$

where,

i = Van't Hoff factor = 1 (for chloroform)

m = molality

$K_{b}$ = molal boiling point elevation constant

Substitute the values into above formula as follows.

$\Delta T_{b} = i\times K_{b} \times m\\3.80^{o}C = 1 \times K_{b} \times 1.05 m\\K_{b} = 3.62^{o}C/m$

Thus, we can conclude that the value of $K_{b}$ for chloroform is $3.62^{o}C/m$ when 0.793 moles of solute in 0.758 kg changes the boiling point by 3.80 °C.

7 0

3 years ago