# QuestionArrange the elements according to atomic radius.Largest radius to Smallest radiusAnswer BankKCaGaGeAsScBrKr

## Answers

Answer 1
Answer:

The elements according to the decreasing atomic radius are arranged as-

K, Ca, Sc, Ga, Ge, As, Br, Kr

An atomic radius is half the distance between adjacent atoms of the same element in a molecule. It is a measure of the size of the element’s atoms, which is typically the mean distance from the nucleus centre to the boundary of its surrounding shells of the electrons.

An atom gets larger as the number of electronic shells increase; therefore the radius of atoms increases as you go down a certain group in the periodic table of elements. The atomic radius decreases on moving from left to right across a period.

Thus the elements according to the decreasing atomic radius are arranged as -

K, Ca, Sc, Ga, Ge, As, Br, Kr

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Answer 2
Answer: Largest

K
Ca
Ga
Ge
As
Br
Kr
Smallest

## Related Questions

A cool, yellow-orange flame is used to heat the crucible. Would this affect the mass of the crucible? If so, how?

### Answers

Answer:

yes

Explanation:

Usually, it would not affect the crucible, but depending on the temperature of the flame the enamel of the crucible may begin to melt and stick to the metal object being used to handle the crucible. This tiny amount that is melted off can cause very small changes in the original mass of the crucible, which although it is almost unnoticeable it is still there. Therefore, the answer to this question would be yes.

### Final answer:

Using a cool, yellow-orange flame to heat the crucible does not directly affect its mass, but can lead to the burning off or decomposition of any impurities or residues present.

### Explanation:

When a cool, yellow-orange flame is used to heat the crucible, it does not directly affect the mass of the crucible. The color of the flame is an indication of the temperature and the type of fuel being burned.

However, if there are impurities or residues in the crucible, the heat from the flame can cause them to burn off or decompose, which may slightly affect the mass of the crucible.

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Find the pH during the titration of 20.00 mL of 0.1000 M butanoic acid, CH3CH2CH2COOH (Ka = 1.54 x 10^-5), with 0.1000 M NaOH solution after the following additions of titrant. (a) 10.00 mL: pH =
(b) 20.10 mL: pH =
(c) 25.00 mL: pH =

### Answers

Answer:

pH after the addition of 10 ml NaOH = 4.81

pH after the addition of 20.1 ml NaOH = 8.76

pH after the addition of 25 ml NaOH = 8.78

Explanation:

(1)

Moles of butanoic acid initially present = 0.1 x 20 = 2 m moles  = 2 x 10⁻³ moles,

Moles of NaOH added = 10 x 0.1 = 1 x 10⁻³ moles

CH₃CH₂CH₂COOH + NaOH ⇄ CH₃CH₂CH₂COONa + H₂O

Initial conc.            2 x 10⁻³                 1 x 10⁻³           0

Equilibrium             1 x 10⁻³                   0                  1 x 10⁻³

Final volume = 20 + 10 = 30 ml = 0.03 lit

So final concentration of Acid =

Final concentration of conjugate base [CH₃CH₂CH₂COONa]

Since a buffer solution is formed which contains the weak butanoic acid and conjugate base of that acid .

Using Henderson Hasselbalch equation to find the pH

### Final answer:

During titration of butanoic acid with NaOH, we can calculate the pH at various points using the Henderson-Hasselbalch equation for buffer scenarios. After 10.00mL of NaOH, the pH will be 4.74. After 20.10 mL, the pH will be 8.27, and after 25.00 mL, the pH will be 12.30.

### Explanation:

This involves calculating the pH at various stages during a titration procedure. Here the titration involves a weak acid, butanoic acid, with a strong base, NaOH. We can simplify the reaction as follows: CH₃CH₂CH₂COOH + OH- --> CH₃CH₂CH₂COO- + H₂O.

(a) After 10.00 mL of NaOH is added, the system isn't at equivalence. Here, the reaction hasn't fully completed and a buffer solution is present. Using the Henderson-Hasselbalch equation, we can find the pH: pH = pKa + log([base]/[acid]). After calculating, we can find pH = 4.74.

(b) After 20.10 mL NaOH is added, the system reaches past equivalence. The pH can be determined by finding pOH using the remaining OH- concentration and then subtracting from 14. After the calculation, the pH = 8.27.

(c) For 25.00 mL of NaOH, the system is beyond equivalence and extra OH- ions increase pH. The pH calculation is like previous step and the result will be pH = 12.30.

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What is the molarity of a solution prepared from 25.0 grams of methanol (CH3OH, density = 0.792 g/mL) with 100.0 milliliters of ethanol (CH3CH2OH)? Assume the volumes are additive.

### Final answer:

The molarity of a solution prepared from 25.0 grams of methanol and 100.0 milliliters of ethanol is approximately 7.80 M.

### Explanation:

This is a question about calculating molarity, which is a measure of concentration using moles per liter. To calculate the molarity of a methanol in ethanol, we first have to convert the mass of methanol into moles. The molar mass of methanol (CH3OH) is about 32.04 g/mol. Therefore, 25.0 g of methanol equals about 0.780 moles (25.0 g ÷ 32.04 g/mol).

Next, the volume of ethanol needs to be converted from milliliters to liters. Thus, 100.0 mL becomes 0.100 L. Finally, the molarity is calculated by dividing the moles of methanol by the volume of the ethanol in liters, resulting in a molarity of approximately 7.80 M (0.780 moles ÷ 0.100 L).

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How many liters of the antifreeze ethylene glycol [CH2(OH)CH2(OH)] would you add to a car radiator containing 6.50 L of water if the coldest winter temperature in your area is -10.°C? (The density of ethylene glycol is 1.11 g/mL. Assume the density of water at -10.°C is 1.00 g/mL.)

### Answers

Answer:

Around 2.0 L of ethylene glycol needs to be added to the car radiator

Explanation:

The depression in freezing point ΔTf of a solution is directly proportional to its molality (m), i.e.

From the given information:

= freezing pt of solution = -10.0 C

= freezing pt of pure solvent = 0 C

Kf = freezing pt depression constant = 1.86 C/m

i = 1 for ethylene glycol antifreeze

Volume of water = 6.50 L = 6500 ml

Density of water = 1.00 g/ml

Therefore mass of water =

Molar mass of ethylene glycol = 62 g/mol

Mass of ethylene glycol needed =

Density of ethylene = 1.11 g/ml

Therefore, volume needed =

16 = m x 1.86
m = 8.60 = moles solute / 6.50 Kg

moles solute = 55.9

mass solute = 55.9 x 62.068 g/mol=3470 g

V = 3479/ 1.11 =3126 mL= 3.13 L

delta T = 8.60 x 0.512 =4.40
boling point = 104.4 °C

Hope this helps.

Many calculators use photocells to provide their energy. Find the maximum wavelength needed to remove an electron from silver (Φ = 7.59 x 10⁻¹⁹ J). Is silver a good choice for a photocell that uses visible light?

### Answers

Answer:

= 261 nm

Silver is not a good choice.

Explanation:

Where,

h is Plank's constant having value

c is the speed of light having value

is the wavelength of the light

Given that:- Energy =

= 261 nm

Visible range has a spectrum of 380 to 740 nm

So, Silver is not a good choice.

### Final answer:

The maximum wavelength needed to remove an electron from silver is approximately 262 nm. Silver is not a good choice for a photocell that uses visible light.

### Explanation:

To find the maximum wavelength needed to remove an electron from silver, we can use the work function of silver, which is Φ = 4.73 eV. The threshold wavelength for observing the photoelectric effect in silver can be calculated using Equation 6.16, which is λ = hc/Φ. Substituting the given values, we have λ = (1240 eV⋅nm) / (4.73 eV), which gives us a threshold wavelength of approximately 262 nm. Since visible light ranges from 400 to 700 nm, silver is not a good choice for a photocell that uses visible light.

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4. define a tsunami

### Answers

Answer:

A high wave caused by either an earthquake, submarine landslide, or other disturbance that occurs.

Explanation:

Answer:

a Big Big wave

Explanation:

A tsunami is a Big Big wave