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Llana [10]
3 years ago
12

Are the properties of Rubidium more similar to those of cesium or those of strontium?

Chemistry
1 answer:
fenix001 [56]3 years ago
5 0

More similar to Cesium

Explanation:

The properties of Rubidium are more similar to those of cesium compared to strontium.

Elements in the same group on the periodic table have similar chemical properties.

  • Rubidium and Cesium are located in the first group on the periodic table.
  • Other elements in this group are lithium, sodium, potassium and francium
  • Strontium belongs to the second group on the periodic table.
  • The first group have a ns¹ valence shells electronic configuration.
  • They are all referred to as alkali metals

Learn more:

Sodium brainly.com/question/6324347

Periodic table brainly.com/question/1704778

#learnwithBrainly

#learnwithBrainly

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timurjin [86]

Intermolecular forces of attraction hold the molecules together. These forces determine the physical properties of substances like melting and boiling points. There are five types of intermolecular forces: Hydrogen bonding, dipole-dipole interactions, ionic interactions, ion-dipole interactions and dispersion forces.

Hydrogen bonding is a stronger force of attraction between hydrogen atom and an electronegative atom (F, N, and O). So, water molecules exhibit hydrogen bonding.

In carbon dioxide molecules, although each C=O is polar the molecule as a whole will be non polar due to symmetry. Therefore, the only intermolecular forces in CO2 will be dispersion forces.

Hence, Hydrogen bonding exists between water molecules but not carbon dioxide molecules.

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3 years ago
How many liters of water can be made from 55 grams of oxygen gas and an excess of hydrogen at a pressure of 12.4 atm and a tempe
polet [3.4K]
First, we need the no.of moles of O2 = mass/molar mass of O2
                                                             = 55 g / 32 g/mol
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from the balanced equation of the reaction:
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we can see that the molar ratio between O2: H2O = 1: 2 
So we can get the no.of moles of H2O = 2 * moles of O2
                                                                  = 2 * 1.72 mol
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when P = 12.4 atm  & n H2O = 3.44 mol & R= 0.0821 & T = 85 + 273=358K

12.4 atm *V = 3.44 * 0.0821 * 358 = 8.15 L
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Answer:

Explanation:

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