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Varvara68 [4.7K]
2 years ago
6

How many total atoms are in 3.3 moles of potassium sulfide, K2S?

Chemistry
1 answer:
marissa [1.9K]2 years ago
3 0
The answer is 363.99 g
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Determine whether each of the examples represents a colligative property or a non-colligative property. boiling point elevation
sleet_krkn [62]

Answer:

boiling point elevation - colligative property

color - non-colligative property

freezing point depression - colligative property

vapor pressure lowering - colligative property

density - non-colligative property

Explanation:

A colligative property is a property that depends on the number of particles present in the system.

Freezing point depression, boiling point elevation and vapour pressure lowering are all colligative properties of solutions.

Colour and density do not depend on the number of particles present hence they are not colligative properties.

7 0
3 years ago
What is the difference between a 1s orbital and a 2s orbital is that
iVinArrow [24]

Answer:

The main difference is their energy level, 2s orbital is higher than 1s orbital.

4 0
3 years ago
29.47 mL of a solution of the acid HBr is titrated, and 72.90 mL of 0.2500-M NaOH is required to reach the equivalence point. Ca
Klio2033 [76]

The original concentration of the acid solution is 6.175 \times 10^-4 mol / L.

<u>Explanation:</u>

Concentration is the ratio of solute in a solution to either solvent or total solution. It is expressed in terms of mass per unit volume

                        HBr + NaOH -----> NaBr + H2O

There is a 1:1 equivalence with acid and base.

Moles of NaOH = 72.90 \times 10^-3 \times 0.25

                          = 0.0182 mol.

[ HBr ] = moles of base / volume of a solution

          = 0.0182 / 29.47

          = 6.175 \times 10^-4 mol / L.

4 0
3 years ago
Which laws can be combined to form the ideal gas law?
slava [35]

Option 3- Avogadro's, Charles's and Boyle's


6 0
3 years ago
For the reaction o(g) + o2(g) → o3(g) δh o = −107.2 kj/mol given that the bond enthalpy in o2(g) is 498.7 kj/mol, calculate the
Aneli [31]
The reaction;
O(g) +O2(g)→O3(g), ΔH = sum of bond enthalpy of reactants-sum of food enthalpy of products.
ΔH = ( bond enthalpy of O(g)+bond enthalpy of O2 (g) - bond enthalpy of O3(g)
-107.2 kJ/mol = O+487.7kJ/mol =O+487.7 kJ/mol +487.7kJ/mol =594.9 kJ/mol
Bond enthalpy (BE) of O3(g) is equals to 2× bond enthalpy of O3(g) because, O3(g) has two types of bonds from its lewis structure (0-0=0).
∴2BE of O3(g) = 594.9kJ/mol
Average bond enthalpy = 594.9kJ/mol/2
=297.45kJ/mol
∴ Averange bond enthalpy of O3(g) is 297.45kJ/mol.
8 0
3 years ago
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