Draw the four structures of the compounds with molecular formula C5H10O that contain a carbon-carbon double bond, an unbranched

When two atoms with equal electronegativity bond together, they form nonpolar covalent bonds.

Your second statement mirrors the first; the second statement likely reads, "Bonds between two atoms with unequal electronegativity are termed polar covalent bonds."

The enthalpy change associated with the precipitation reaction is 84 kJ/mole

Why?

The chemical equation for the reaction can be written as

AgNO₃(aq) + NaCl (aq) → AgCl(s) + NaNO₃(aq)

To determine the enthalpy change, the following equation applies

To calculate the heat (Q):

Next, we need to calculate the number of moles involved in the reaction (n):

With these two values, we can substitute them into the first equation:

Have a great day!

Answer:6.94

Explanation:Atomic mass of Li = 6.94

To find the number of atoms: 12.6 g Li x 1 mole/6.94 g x 6.02x10^23 atom/mole = 1.09x10^24 atoms

Result:

94.7 %

Explanation:

The balanced reaction is:

2 S + 3 O₂ → 2 SO₃

The stoichiometric mole ratio is:

S: 2 moles

O₂: 3 moles

Moles are calculated as mass divided by molar mass:

n = w / m

where n = moles, w = mass, m = molar mass.

Given:

For sulfur: w = 6.0 g, molar mass = 32 g/mol, so n = 6 / 32 = 0.1871 mol

For oxygen: w = 5.0 g, molar mass = 32 g/mol, thus n = 5 / 32 = 0.15625 mol

Comparing to stoichiometric ratios, sulfur is in excess, so oxygen is the limiting reagent, controlling product formation.

Using proportions:

3 mol O₂ produce 2 mol SO₃, so 1 mol O₂ yields 2/3 mol SO₃.

Therefore, 0.15625 mol O₂ yields (2/3) × 0.15625 = 0.1042 mol SO₃.

Mass of SO₃ produced = n × molar mass = 0.1042 mol × 80 g/mol = 8.340 g

The percentage yield is actual yield divided by theoretical yield times 100:

Percent yield = (7.9 g / 8.340 g) × 100 = 94.7 %

The element represented by the nlx notation of 5d2 is hafnium, designated by the symbol hf. It is categorized as a chemical element with the atomic number 72. Hafnium has a chemical resemblance to zirconium and occurs in various zirconium minerals, whereas it is distinct from zirconium which shares some chemical similarities.