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3 years ago

How many atoms are in a 591 g sample of gold?

Chemistry

1 answer:

skelet666 [1.2K]3 years ago

3 0

Answer:

You use the fact that 1 mole of any substance contains exactly 6.022⋅1023 atoms or molecules of that substance - this is known as Avogadro's number. In your case, 1 mole of gold will have exactly 6.022⋅1023 atoms of gold.

So it would be answer 3690 atoms or 3.60x1025

Explanation:

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

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3 years ago

What are the respective hybridizations of the imine nitrogen and carbonyl carbon of the CPV B-state as it is depicted in the pas

horsena [70]

Answer:

C) sp2 and sp2

Explanation:

The hybridization depens on the ammount and type of bonds the atom analized has in the molecule.

For example:

A C atom bonded to 4 H atoms has a sp3 hybridization.
A C atom bonded to 2 H atoms and to 1 C with a double bond (like in ethene) has a sp2 hybridization
A C bonded to 1 H and 1 C with a triple bond (like in ethyne) has a sp hybridization.

Analyzing the type and amount of unions of the nitrogen and the carbonyl you will be able to determine the hybridization.

In the imine, the N atom has a double bond to a C and a simple bond two other C, plus the lone pair of electrons (counts as a bond) so it will have a sp2 hybridization.

In the carbonyl, the C has two simple bonds to other C and a double bond to an oxygen atom. It will also have a sp2 hybridization

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Answer:

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Explanation:

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slavikrds [6]

Answer:

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Explanation:

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VMariaS [17]

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