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
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Inertia is the retaliation of an object to change in its velocity
Answer:
Silver Acetate would be the Limiting Reagent.
Explanation:
The balance chemical equation for the given double displacement reaction is as;
HCl + AgC₂H₃O₂ → AgCl + HC₂H₃O₂
Step 1: <u>Calculate Moles of Starting Materials:</u>
Moles of HCl:
Moles = Mass / M.Mass
Moles = 72.9 g / 36.46
Moles = 1.99 moles
Moles of AgC₂H₃O₂:
Moles = 150 g / 166.91 g/mol
Moles = 0.898 moles
Step 2: <u>Find out Limiting reagent as:</u>
According to balance chemical equation.
1 mole of HCl reacts with = 1 mole of AgC₂H₃O₂
So,
1.99 moles of HCl will react with = X moles of AgC₂H₃O₂
Solving for X,
X = 1.99 mol × 1 mol / 1 mol
X = 1.99 mol of AgC₂H₃O₂
Hence, to completely consume 1.99 moles of Hydrochloric acid we will require 1.99 moles of Silver Acetate, But, we are provided with only 0.898 moles of Silver Acetate. This means Silver Acetate will consume first in the reaction therefore, it is the LIMITING REAGENT.
Conjugate base pairs are acid and bases having common features. These features are the equal gain or loss of protons of the pairs. Conjugate pairs should always be one base and one acid. One would not exist without the other. Conjugate acids are the substances that gains protons while conjugates bases are those that loses protons. <span>The substances in the equilibrium reaction that is given is identified as follows:
HCO3^- + H2O <-----> CO3^2- + H3O^+
acid base conjugate base conjugate acid
HCO3^- ion is an intermediate molecule of CO2 and CO3^2-. When we add OH- to HCO3^-, we produce CO3^2-. And when we add H+ to HCO3, we produce CO2. </span>
Answer:
65.08 g.
Explanation:
- For the reaction, the balanced equation is:
<em>2AlCl₃ + 3Br₂ → 2AlBr₃ + 3Cl₂,</em>
2.0 mole of AlCl₃ reacts with 3.0 mole of Br₂ to produce 2.0 mole of AlBr₃ and 3.0 mole of Cl₂.
- Firstly, we need to calculate the no. of moles of 36.2 grams of AlCl₃:
<em>n = mass/molar mass</em> = (36.2 g)/(133.34 g/mol) = <em>0.2715 mol.</em>
<u><em>Using cross multiplication:</em></u>
2.0 mole of AlCl₃ reacts with → 3.0 mole of Br₂, from the stichiometry.
0.2715 mol of AlCl₃ reacts with → ??? mole of Br₂.
∴ The no. of moles of Br₂ reacts completely with 0.2715 mol (36.2 g) of AlCl₃ = (0.2715 mol)(3.0 mole)/(2.0 mole) = 0.4072 mol.
<em>∴ The mass of Br₂ reacts completely with 0.2715 mol (36.2 g) of AlCl₃ = no. of moles of Br₂ x molar mass</em> = (0.4072 mol)(159.808 g/mol
) = <em>65.08 g.</em>
