Suppose you combine two substances but no chemical reaction occurs. Which of the following best describes the result?

If you sketch a Lewis dot structure of I3-, you'll perceive that the center I atom formulates two bonds, and also has three non-bonding pairs of electrons. That's an over-all of five electron pairs - beyond an octet. Lots of compounds disrupt the octet rule that means, but the first long-row elements like F can't because the lower elements like iodine can use an obtainable d-orbital to house the additional pair of electrons (sp3d hybridization), but F just doesn't have a d-orbital that it can use.

4 0

3 years ago

please hurry! If 3.87g of powdered aluminum oxide is placed in a container containing 5.67g of water, what is the limiting react

VikaD [51]

Answer:

Explanation:

Given parameters:

Mass of aluminium oxide = 3.87g

Mass of water = 5.67g

Unknown:

Limiting reactant = ?

Solution:

The limiting reactant is the reactant in short supply in a chemical reaction. We need to first write the chemical equation and convert the masses given to the number of moles.

Using the number of moles, we can ascertain the limiting reactants;

Al₂O₃ + 3H₂O → 2Al(OH)₃

Number of moles;

Number of moles = $\frac{mass}{molar mass}$

molar mass of Al₂O₃ = (2x27) + 3(16) = 102g/mole

number of moles = $\frac{3.87}{102}$ = 0.04mole

molar mass of H₂O = 2(1) + 16 = 18g/mole

number of moles = $\frac{5.67}{18}$ = 0.32mole

From the reaction equation;

1 mole of Al₂O₃ reacted with 3 moles of H₂O

0.04 mole of Al₂O₃ will react with 3 x 0.04 mole = 0.12 mole of H₂O

But we were given 0.32 mole of H₂O and this is in excess of amount required.

This shows that Al₂O₃ is the limiting reactant

6 0

3 years ago

Urea, (NH2)2CO, is a product of metabolism of proteins. An aqueous solution is 37.2% urea by mass and has a density of 1.032 g/m

Feliz [49]

Answer:

The molarity of urea in this solution is 6.39 M.

Explanation:

Molarity (M) is <em>the number of moles of solute in 1 L of solution</em>; that is

$molarity = moles of solute ÷ liters of solution$

To calculate the molality, we need to know the number of moles of urea and the volume of solution in liters. We assume 100 grams of solution.

Our first step is to calculate the moles of urea in 100 grams of the solution,

using the molar mass a conversion factor. The total moles of 100g of a 37.2 percent by mass solution is

60.06 g/mol ÷ 37.2 g = 0.619 mol

Now we need to calculate the volume of 100 grams of solution, and we use density as a conversion factor.

1.032 g/mL ÷ 100 g = 96.9 mL

This solution contains 0.619 moles of urea in 96.9 mL of solution. To express it in molarity, we need to calculate the moles present in 1000 mL (1 L) of the solution.

0.619 mol/96.9 mL × 1000 mL= 6.39 M

Therefore, the molarity of the solution is 6.39 M.

4 0

3 years ago