The majority of elements are A.metals B.metalloids C.noble gases D.nonmetals

As you move down a group, you will recall that the radius increases. Why do you think an increase in atomic radius would result

Assoli18 [71]

Answer:

Because as we move down the group size/ radius of atom increases due to increase in shells and screening effect. The hold of nucleus on valence electrons decreases and effective nuclear charge decreases so it is easy to remove electron from greater sized atoms as compared to smaller sized aroms.

Energy required to remove electron is ionization energy so greater sized atoms have low ionization energy.

Explanation:

8 0

3 years ago

What is the activity series of metals?

steposvetlana [31]

Answer:

C.A list of metals arranged in order of reactivity

7 0

3 years ago

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Chromatography separates gas or liquid solutions using a stationary phase. True False

Strike441 [17]

The correct answer for the given statement above would be TRUE. It is true that chromatography separates gas or liquid solutions using a stationary phase. Chromatography is the separation of mixture that distributes components to separate between two phases which includes stationary phase and the other is mobile phase. Hope this answer helps.

8 0

3 years ago

0.0500 mol of gas occupies a cylinder which is sealed on top by a moveable piston. The piston is circular, with a mass of 30.0 k

Viefleur [7K]

Answer:

The workdone by both N₂ and neon gas is 49.3 J

The change in internal energy of N₂ and neon gas is 125.6 J and 73.54 J respectively

The heat for N₂ and neon gas is 171.9 J and 122.84 J respectively.

Explanation:

Given that:

number of moles = 0.05 mole

mass of the piston = 30 kg

diameter = 5.00 cm = 0.05 m

Area (A) = πr²

$Area (A) = \pi*(\frac{0.05}{2})^2 \\ \\ Area (A) = 0.0019635 \\ \\ Area (A) = 19.635*10^{-4} \ m^2$

The piston is said to move from 30 cm - 40 cm

So, the change in volume ΔV is calculated as:

$=(40-30)*10^{-2} *19.635*10^{-4}$

$= 1.9635*10^{-4} \ m^3$

Outside the cylinder; the pressure $P_{air}= 1 \ atm$ = 101325 Pa

Thus, workdone $w_1$ = PΔV

= $101325*1.9635*10^{-4}$

= 19.90 J

The gravitational work $w_2 = mgh$

Given that the height (h) = 10 cm = 0.1 m

Then; $w_2 = 30*9.8*0.1$

$w_2 = 29.4 \ J$

The total workdone $w_{total}$ for both cases is:

$w_{total } =w_1 + w_2$

$w = (19.90 + 29.4) \ J$

$w =49.3 \ J$

The pressure of gas inside the cylinder is determined as:

$P_{in}.A = P_{out}.A +mg$

$(P_{in}-P_{out}) = \frac{mg}{A} \\ \\ P_{in} -10^5 = \frac{30*9.8}{19.635*10^{-4}} \\ \\ P_{in} = 149732.6203+10^5 \\ \\ P_{in} = 2.497*10^5 \ Pa$

a). assuming that the gas is N₂.

$C_v =\frac{5}{2}R$

Thus, the change in internal energy ΔU is given as:

$\delta U = nC_v \delta T \\ \\ \delta U = n* \frac{5}{2}R \delta T \\ \\ \delta U = \frac{5}{2}nR \delta T$

Since $P_{in} \delta V = nR \delta T ; \ Then$ ;

$\delta \ U = \frac{5}{2} P_{in} \delta V \\ \\ \delta \ U = \frac{5}{2}*2.497*10^5 *1.9635*10^{-4} \\ \\ \delta \ U = 122.57 \ J$

ΔU ≅ 125.6 J

The heat Q = ΔU + W

Q = (122.6 + 49.3) J

Q = 171.9 J

b) In Neon gas:

$C_v = \frac{3}{2}R$

∴

change in internal energy is;

$\delta U = nC_v \delta T \\ \\ \delta U = n* \frac{3}{2}R \delta T \\ \\ \delta U = \frac{3}{2}P_{in}.V$

$\delta U = \frac{3}{2}*2.497*10^5*1.9635*10^{-4}$

ΔU = 73.54 J

The heat Q = ΔU + W

Q = (73.54 + 49.3) J

Q = 122.84 J

5 0

3 years ago

Which of these is an example of a monotreme, a mammal that lays eggs?

Bas_tet [7]

Answer: it is the 4th option in there

Explanation:

4 0

3 years ago