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4 years ago

An atom of which element has the largest atomic radius?

Chemistry

2 answers:

sergejj [24]4 years ago

5 0

Answer:

Cesium

Explanation:

As you go down a Group in the Periodic Table from top to bottom, the number of energy levels or electron shells increases so the atomic radius of the elements increases.

In general, the atomic radius of elements decreases as you go across a Period from left to right.

This means, we would expect that the last element in group 1 to have the largest atomic radius. This is Francium. But the atom with the largest atomic radius is referred to as Cesium.

Why aren't francium atoms the biggest? The usual periodic trend for atomic size places larger atoms at the left of a row and towards the bottom of a column on the periodic table. It's no surprise that cesium is large. But shouldn't francium, in the next period with an even larger valence shell, be even larger?

The answer is "possibly, but we just don't know yet." Francium isn't easy to study. It's the least stable of the first 103 elements; the most stable Fr isotope has a half-life of just 22 minutes The distance between atoms in metallic francium has not yet been measured.

The element which has the largest known atomic radius is Cesium.

enot [183]4 years ago

4 0

You didn't post the whole question. However the answer would be : Mg, according to former questions in this website.

Hope this helps !

Photon

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A balloon that had a volume of 3.50 L at 25.0°C is placed in a hot room at 40.0°C. If the pressure remains constant at 1.00 atm,

belka [17]

To solve this we assume that the gas inside the balloon is an ideal gas. Then, we can use the ideal gas equation which is expressed as PV = nRT. At a constant pressure and number of moles of the gas the ratio T/V is equal to some constant. At another set of condition of temperature, the constant is still the same. Calculations are as follows:

T1 / V1 = T2 / V2

V2 = T2 x V1 / T1

V2 = 313.15 x 3.50 / 298.15

V2 = 3.68 L

6 0

3 years ago

Read 2 more answers

1.) 3.4 moles Magnesium are mixed with 5.6 moles of Hydrochloric Acid. How many moles of Hydrogen gas are produced?

love history [14]

Answer:

1. 2.8 moles of H₂

2. 7.38 moles of CO₂

3. 5.3 moles of O₂

4. 7.4 moles of KNO₃

Explanation:

Here are the steps to doing this:

1. Write the chemical equation of each reaction.

2. Balance the equation.

3. Find out the ratio between reactant and product

4. Determine the actual yield of your reactants.

5. The amount of product produced is determined by how much product the limiting reactant produces.

Let's do this!

1. Given: 3.4 moles of Magnesium(Mg) and 5.6 moles of Hydrochloric acid (HCl)

Equation:

Mg + 2HCl → MgCl₂ + H₂

Reactant to Product ratio

1 mole of Mg produces 1 mole of H₂ $\dfrac{1moleof Mg}{1mole of H_{2}}$

2 moles of HCl produces 1 mole of H₂ $\dfrac{2molesofHCl}{1mole of H_{2}}$

Determine actual yield of reactants

$3.4moles of Mg\times\dfrac{1moleofH_2}{1moleofMg}=3.4molesofH_{2}\\\\5.6moles ofHCl\times\dfrac{1moleofH_2}{2moleofHCl}=2.8molesofH_{2}$

Since 5.6 moles of HCl can only produce 2.8 moles of H₂, before it is used up, then this means that that is all the product this reaction can produce.

2. Given: 3.4 moles of C₃H₈ and 12.3 moles of oxygen gas (O₂)

Equation:

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

Reactant to Product ratio

1 mole of C₃H₈ produces 3 moles of CO₂ $\dfrac{1moleofC_{3}H_{8}}{3molesofCO_{2}}$

5 moles of O₂ produces 3 moles of CO₂ $\dfrac{5molesofO_{2}}{3moleofCO_{2}}$

Determine actual yield of reactants

$3.4molesofC_{3}H_{8}\times\dfrac{3molesofCO_{2}}{1moleofC_{3}H_{8}}=10.2molesofH_{2}$

$12.3molesofO_{2}\times\dfrac{3molesofCO_{2}}{5molesofO_{2}}=7.38molesofCO_{2}$

The answer is then 7.38 moles of CO₂

**3. 5.3 moles of H₂O

This one is a little bit different. It is asking how much of a reactant is needed to produce the amount of product given. For this, just write a balanced equation for the reaction and get the ratio of reactant to product and solve for the actual yield. Since it is only asking for oxygen gas, you just need to do that one.

Equation:

CH₄ + 2O₂ → CO₂ + 2H₂O

Reactant to Product ratio

$\dfrac{2molesofO_{2}}{2molesofH_{2}O}=\[tex]7.88molesofKI\times\dfrac{1moleofKNO_{3}}{1moleofKI}=7.88molesofKNO_{3}$