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3 years ago

Why is Neon in the same group of the periodic table as Helium?

1 answer:

3 0

Because they have a full outer shell and they are both noble gases...Because the outer shell has attained stable electronics configuration of octate rule this it is with noble gas as helium. Hope I helped!

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A gas is contained in a thick-walled balloon. When the pressure change from 100 kPa to 90.0 kPa, the volume changes from 2.50 L

Orlov [11]

The temperature change is calculated using the combined gas law
that is P1V1/T1 =P2V2/T2
P1= 100KPa
P2=90kpa
v1= 2.50 L
v2= 3.75 L
T1= 303 K
T2=?

T2 is therefore = P2V2T1/P1V1
=( 90 x 3.75 x303)/ (100 x2.50) = 409.05 K

3 0

3 years ago

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A student mixed together 8 g of sugar, 5.20 g of salt

maw [93]

its C) 113.21g

u just add all the masses

7 0

4 years ago

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Name two compounds with more exothermic lattice energies than scandium oxide and justify your choice.

N76 [4]

Answer:

1 . $Al_2O_3$

2. $TiO_2$

Explanation:

The more stable the ionic compound, the more is it lattice energy.

The more the charge on the cation and the anion, the greater is the lattice energy.
The less the size of the cation and the anion, the greater is the lattice energy.

Scandium oxide ( $Sc_2O_3$ ) is an oxide in which $Sc^{3+}$ behaves as cation and $O^{2-}$ behaves as anion.

The compounds which has higher lattice energy than scandium oxide are:

1 . $Al_2O_3$

This is because the charge are same on the cation and the anion as in the case of the Scandium oxide but the size of the cation $Al^{3+}$ is smaller than $Sc^{3+}$ . Thus, this corresponds to higher lattice energy.

2. $TiO_2$

This is because the charge on the cation $Ti^{4+}$ is greater than that of $Sc^{3+}$ and also the size of the cation $Ti^{4+}$ is smaller than $Sc^{3+}$ . Thus, this corresponds to higher lattice energy.

3 0

3 years ago

The radioactivity due to carbon-14 measured in a piece of a wood from an ancient site was found to produce 20 counts per minute

ss7ja [257]

Answer:

The answer is "17200 years".

Explanation:

Given:

$A = 20 \ \frac{counts}{minute}\\\\A_{o} = 160\ \frac{counts}{minute}$

Let the half-life of carbon-14, is beta emitter, is $T = 5730\ years$

Constant decay $\ w = \frac{0.693}{ T}$

$= 1.209 \times 10^{-4} \ \frac{1}{year}\\$

The artifact age $t= ?$

$A = A_{o} e^{-wt} \\\\e^{-wt} = \frac{A}{A_{o}}\\\\-wt = \ln \frac{A}{A_{o}}\\\\= -2.079\\\\t = 1.7199 \times 10^{4} \ years\\\\\sim \ 17200\ years\\$

5 0

3 years ago

Calculate the density of a 20.015 g object that occupies 5.44 cm3 .

Scilla [17]

Answer:

$density = \frac{mass}{volume} \\ \\ density = \frac{20.015}{5.44} \\ \\ { \boxed{ \boxed{density = 3.679 \: g {cm}^{ - 3} }}}$

5 0

3 years ago

Read 2 more answers