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2 years ago

The hybridization for the B in Bcl3 is ?

Chemistry

1 answer:

Phoenix [80]2 years ago

6 0

Answer:

sp²

Explanation:

You need to look at how many electron orbitals around the atom. Looking at the structure below, you can see that there are three electron orbitals. This gives you an sp² hybridization.

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What element has 4 valence electrons and 5 electron shells?

marta [7]

Answer: Tin (Sn)

Explanation: The electron configuration for tin (Sn) is shown in the picture. It's last electrons are:

5s^2 4d^10 5p^2

The valence electrons are in the 5th electron shell and include 2 each in the 5s and 5p orbitals.

8 0

2 years ago

Drug testing if a person do not use drugs how cam he test positive on hair drug test

Hitman42 [59]

For example, certain medications may influence the results of the test.

Drug test can show the presence of illicit or prescription drugs on hair samples, although the person has not used these drugs.

Some of false positive drug use is because of this medications: antihistamines, decongestants, antibiotics, antidepressants, analgesics and antipsychotics.

7 0

3 years ago

A 50/50 blend of engine coolant and water (by volume) is usually used in an automobile's engine cooling system. If a car's cooli

Diano4ka-milaya [45]

Answer:

$\large \boxed{109.17 \, ^{\circ}\text{C}}$

Explanation:

Data:

50/50 ethylene glycol (EG):water

V = 4.70 gal

ρ(EG) = 1.11 g/mL

ρ(water) = 0.988 g/mL

Calculations:

The formula for the boiling point elevation ΔTb is

$\Delta T_{b} = iK_{b}b$

i is the van’t Hoff factor — the number of moles of particles you get from 1 mol of solute. For EG, i = 1.

1. Moles of EG

$\rm n = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1 gal}} \times \dfrac{\text{1000 mL}}{\text{1 L}} \times \dfrac{\text{1.11 g}}{\text{1 mL}} \times \dfrac{\text{1 mol}}{\text{62.07 g}} = \text{159 mol}$

2. Kilograms of water

$m = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1 gal}} \times \dfrac{\text{998 g}}{\text{1 L}} \times \dfrac{\text{1 kg}}{\text{1000 g}} = \text{8.88 kg}$

3. Molal concentration of EG

$b = \dfrac{\text{159 mol}}{\text{8.88 kg}} = \text{17.9 mol/kg}$

4. Increase in boiling point

$\rm \Delta T_{b} = iK_{b}b = 1 \times 0.512 \, \, ^{\circ}\text{C} \cdot kg \cdot mol^{-1} \, \times 17.9 \cdot mol \cdot kg^{-1} = 9.17 \, ^{\circ}\text{C}$

5. Boiling point

$\rm T_{b} = T_{b}^{\circ} + \Delta T_{b} = 100.00 \, ^{\circ}\text{C} + 9.17 \, ^{\circ}\text{C} = \mathbf{109.17 \, ^{\circ}C}\\\rm \text{The boiling point of the solution is $\large \boxed{\mathbf{109.17 \, ^{\circ}C}}$}$

7 0

3 years ago

ASAPP Subtract to find the temperature changes

mamaluj [8]

Hello, I would like to help you, but I really don't understand the question

4 0

2 years ago

.<br> How many moles are there in 8.5 x 1024 molecules of sodium sulfate, Na2SO4?

NeTakaya

Answer: To solve this question, we need to use the Avogadro's Number, which is a constant first discovered by Amadeo Avogadro, an Italian scientist. He discovered that in a mole of a substance, there are 6,02*10²³ molecules. Using this relationship, we apply the following conversion factor:

So, 8,50 * 10²⁴ molecules of Na₂SO₃ represent 14,12 moles of Na₂SO₃

Explanation:

5 0

2 years ago