When you convert a measurement from SI to English, what changes? units , values , error , or resolution

In a reaction, chlorine accepts an electron to form Cl-. What is true about the atomic size of the chlorine ion?

Lynna [10]

Answer is: The atomic size of the chlorine ion is larger than the size of the chlorine atom.

Covalent radii of chlorine atom (Cl) is 0.099 nm and ionic radii of chlorine anion (Cl⁻) is 0.181 nm.

Difference between an chlorine atom and chlorine anion is the number of electrons that surround the nucleus.

Chlorine atom has 17 electrons and chlorine anion has 18 electrons.

6 0

4 years ago

Read 2 more answers

Consider the following ionization reaction.

77julia77 [94]

Answer:

See explanation

Explanation:

According to Bronsted-Lowry, an acid is a proton donor while a base is a proton acceptor.

Hence, if we consider the reaction above, we will notice that for each base there is a conjugate acid and for each acid there is a conjugate base.

For the acid HNO3, its conjugate base is NO3^- while for the acid H3O^+, its conjugate base is H2O.

4 0

3 years ago

What does the law of conservation of energy imply?

GenaCL600 [577]

Answer a is the correct one

4 0

3 years ago

Read 2 more answers

What is the temperature of liquid nitrogen?

motikmotik

Really darn cold lol

6 0

3 years ago

If 16.00 g of O₂ reacts with 80.00 g NO, how many the excess reactant are left over? (enter only the value, round to whole numbe

pishuonlain [190]

Answer:

50

Explanation:

We will need a balanced equation with masses, moles, and molar masses of the compounds involved.

1. Gather all the information in one place with molar masses above the formulas and masses below them.

Mᵣ: 30.01 32.00 46.01

2NO + O₂ ⟶ 2NO₂

Mass/g: 80.00 16.00

2. Calculate the moles of each reactant

$\text{moles of NO} = \text{80.00 g NO} \times \dfrac{\text{1 mol NO}}{\text{30.01 g NO}} = \text{2.666 mol NO}\\\\\text{moles of O}_{2} = \text{16.00 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.5000 mol O}_{2}$

3. Calculate the moles of NO₂ we can obtain from each reactant

From NO:

The molar ratio is 2 mol NO₂:2 mol NO

$\text{Moles of NO}_{2} = \text{2.333 mol NO} \times \dfrac{\text{2 mol NO}_{2}}{\text{2 mol NO}} = \text{2.333 mol NO}_{2}$

From O₂:

The molar ratio is 2 mol NO₂:1 mol O₂

$\text{Moles of NO}_{2} = \text{0.5000 mol O}_{2}\times \dfrac{\text{2 mol NO}_{2}}{\text{1 mol Cl}_{2}} = \text{1.000 mol NO}_{2}$

4. Identify the limiting and excess reactants

The limiting reactant is O₂ because it gives the smaller amount of NO₂.

The excess reactant is NO.

5. Mass of excess reactant

(a) Moles of NO reacted

The molar ratio is 2 mol NO:1 mol O₂

$\text{Moles reacted} = \text{0.500 mol O}_{2} \times \dfrac{\text{2 mol NO}}{\text{1 mol O}_{2}} = \text{1.000 mol NO}$

(b) Mass of NO reacted

$\text{Mass reacted} = \text{1.000 mol NO} \times \dfrac{\text{30.01 g NO}}{\text{1 mol NO}} = \text{30.01 g NO}$

(c) Mass of NO remaining

Mass remaining = original mass – mass reacted = (80.00 - 30.01) g = 50 g NO

5 0

3 years ago