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inn [45]
3 years ago
9

Find surface phenomena from the following order: spherical form of drops, crystal growth, absorption, osmosis, colligative prope

rties, diffusion, adsorption, hemosorption.
Chemistry
1 answer:
Volgvan3 years ago
7 0

Answer:

spherical form of drops, crystal growth, absorption, osmosis, colligative properties, diffusion, adsorption, hemosorption.

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2. Name each of the following ionic compounds A. K2O B. Cacl2. C. Mg3N2 D. NaCIO E. KNO3​
Ludmilka [50]
A. Potassium oxide
B. Calcium chloride
C. Magnesium nitride
D. Sodium hypochlorite
E. Potassium nitrate
3 0
3 years ago
Ammonia and sulfuric acid react to form ammonium sulfate. Determine the starting mass of each reactant if 20.3 g of ammonium sul
strojnjashka [21]
Hope you are able to see

8 0
3 years ago
What happens when light passes through a solution? perform an activity with a homogenous to expalin what happen when light pass
Brilliant_brown [7]

Answer:

light bends and makes effects in the water

Explanation:

6 0
2 years ago
How many atoms are in 8.28 moles of aluminum?
Lilit [14]

Answer:

49.86 × 10²³  atoms of Al

Explanation:

Given data:

Number of moles of Al = 8.28 mol

Number of atoms = ?

Solution:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

For example,

18 g of water = 1 mole = 6.022 × 10²³ molecules of water

1.008 g of hydrogen = 1 mole = 6.022 × 10²³ atoms of hydrogen

For 8.28 moles of Al:

1 mole = 6.022 × 10²³  atoms of Al

8.28 mol×6.022 × 10²³  atoms / 1mol

49.86 × 10²³  atoms of Al

4 0
3 years ago
Calculate the work (kJ) done during a reaction in which the internal volume expands from 20 L to 43 L against an outside pressur
Alenkinab [10]

Answer:

\large \boxed{\text{-10.0 kJ}}

Explanation:

1. Calculate the work

w = - pΔV = -4.3 atm × (43 L - 20 L) = -4.3 × 23 L·atm = -98.9 L·atm

2. Convert litre-atmospheres to joules

w = \text{-98.9 L\cdot$atm } \times \dfrac{\text{101.3 J}}{\text{1 L$\cdot$atm }} = \text{-10000 J} = \textbf{-10.0 kJ}\\\\\text{The work done is $\large \boxed{\textbf{-10.0 kJ}}$}

The negative sign indicates that the work was done against the surroundings.

4 0
3 years ago
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