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Step2247 [10]
3 years ago
5

Of the elements calcium, Ca, beryllium, Be, barium, Ba, and strontium, Sr, which has the largest atomic radius? Explain your ans

wer in terms of trends in the periodic table.
Chemistry
2 answers:
Nezavi [6.7K]3 years ago
8 0
The largest atomic radius has Ba,because they are all metals of 2nd group,but as you know when you go down upright in the periodic table the metallic features getting stronger but also orbital layers increasing,and you can say more layers larger atomic radius.Hope I helped,sorry for English
Travka [436]3 years ago
7 0

Explanation:

When we move down a group then there occurs an increase in the size elements due to the addition of more number of electrons into new shells.

Hence, it means that atomic radius of the elements also increases.  

For the given group 2A elements, beryllium is at the top so it has smaller atomic radius whereas barium is at the bottom.

Therefore, we can conclude that barium has the largest atomic radius.

You might be interested in
Thirty (30.0) milliliters of ethane (C2H6) occupy a container at STP. If the container is heated to 300.0 K and the pressure red
wariber [46]

Answer:

44.6millilitres

Explanation:

Using the general gas law equation as follows:

P1V1/T1 = P2V2/T2

Where;

P1 = initial pressure (atm)

V1 = initial volume (L)

T1 = initial temperature (K)

P2 = final pressure (atm)

V2 = final volume (L)

T2 = final temperature (K)

According to this question;

V1 = 30mL

T1 = 273K (STP)

P1 = 1 atm (STP)

V2 = ?

T2 = 300K

P2 = 75.0 kPa = 75 × 0.00987 = 0.74atm

Using P1V1/T1 = P2V2/T2

1 × 30/273 = 0.74×V2/300

30/273 = 0.74V2/300

Cross multiply

300 × 30 = 273 × 0.74V2

9000 = 202.02V2

V2 = 9000/202.02

V2 = 44.55

V2 = 44.6millilitres.

6 0
3 years ago
What is the freezing point (in degrees Celcius) of 3.75 kg of water if it contains 189.9 g of C a B r 2?
blondinia [14]

Answer:

The freezing point of the solution is -1.4°C

Explanation:

Freezing point decreases by the addition of a solute to the original solvent, <em>freezing point depression formula is:</em>

ΔT = kf×m×i

<em>Where Kf is freezing point depression constant of the solvent (1.86°C/m), m is molality of the solution (Moles CaBr₂ -solute- / kg water -solvent) and i is Van't Hoff factor.</em>

Molality of the solution is:

-moles CaBr₂ (Molar mass:

189.9g ₓ (1mol / 199.89g) = 0.95 moles

Molality is:

0.95 moles CaBr₂ / 3.75kg water = <em>0.253m</em>

Van't hoff factor represents how many moles of solute are produced after the dissolution of 1 mole of solid solute, for CaBr₂:

CaBr₂(s) → Ca²⁺ + 2Br⁻

3 moles of ions are formed from 1 mole of solid solute, Van't Hoff factor is 3.

Replacing:

ΔT = kf×m×i

ΔT = 1.86°C/m×0.253m×3

ΔT = 1.4°C

The freezing point of water decreases in 1.4°C. As freezing point of water is 0°C,

<h3>The freezing point of the solution is -1.4°C</h3>

<em />

4 0
3 years ago
Read 2 more answers
Notice that "SO4" appears in two different places in this chemical equation. SO 2−4 is a polyatomic ion called "sulfate." What n
kramer
3CaCl₂ + Al₂(SO₄)₃ = 3CaSO₄↓ + 2AlCl₃
8 0
3 years ago
In a sample consisting of 1.00 mol nabr and 0.300 mol ki, what is the mass percent of iodine? (a) 24.9% (b) 32.6% (c) 47.2% (d)
AlexFokin [52]
The quantities you have to know are the molar masses and the molecular stoichiometric ratios. The solution for this problem is as follows:

Mass Percent of Iodine = (Mass of Iodine/Mass of NaBr and KI)*100

Mass of NaBr = 1 mol NaBr * 102.89 g/mol = 102.89 g
Mass of KI = 0.3 mol KI * 166 g/mol = 49.8 g
Mass of KI = 0.3 mol KI * 1 mol I/1mol KI * 1 mol I₂/2 mol I * 253.81 g/mol = 38.07 g

Mass %I₂ = 38.07/(102.89+49.8) * 100 = <em>24.9%</em>
6 0
4 years ago
Hydrogen, when combined with nitrogen in a chemical reaction, forms ammonia. Which type of chemical reaction is this an example
larisa86 [58]

Answer: Combination reaction

Explanation:

Hydrogen (H2) combines with nitrogen (N2) to form a single compound, ammonia (NH3) in a combination reaction.

The chemical equation is as follows:

N2(g) + 3H2(g) --> 2NH3(g)

7 0
3 years ago
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