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WARRIOR [948]
3 years ago
14

Given any two elements within a group, is the element with the larger atomic number likely to have a larger or smaller atomic ra

dius than the other element
Chemistry
1 answer:
zhenek [66]3 years ago
3 0

Answer:

Atomic radius of sodium = 227 pm

Atomic radius of potassium = 280 pm

Explanation:

Atomic radii trend along group:

As we move down the group atomic radii increased with increase of atomic number. The addition of electron in next level cause the atomic radii to increased. The hold of nucleus on valance shell become weaker because of shielding of electrons thus size of atom increased.

Consider the example of sodium and potassium.

Sodium is present above the potassium with in same group i.e, group one.

The atomic number of sodium is 11 and potassium 19.

So potassium will have larger atomic radius as compared to sodium.

Atomic radius of sodium = 227 pm

Atomic radius of potassium = 280 pm

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How much pv work is done in kilojoules for the reaction of 0.68 mol of h2 with 0.34 mol of o2 at atmospheric pressure if the vol
dsp73
The delta H of -484 kJ is the heat given off when 2 moles of H2 react with 1 mole of O2 to make 2 moles of H2O. You don't have anywhere near that much reactants, only 1/4 as much

<span>actual delta H = 0.34  moles H2 x (-484 kJ / 2 moles H2) = 823 kJ </span>

<span>delta E = delta H - PdeltaV = 823 kJ - 0.41 kJ = 822 kJ</span>
3 0
4 years ago
How do i know if a reaction is exo/endothermic?? help please.
goblinko [34]

Answer:

Explanation: in exothermic reaction heat is released out

In endothermic reaction heat is absorbed

3 0
3 years ago
What fraction of a Sr-90 sample remains unchanged after 87.3 years
jolli1 [7]
The answer is 1/8.

Half-life is the time required for the amount of a sample to half its value.
To calculate this, we will use the following formulas:
1. (1/2)^{n} = x,
where:
<span>n - a number of half-lives
</span>x - a remained fraction of a sample

2. t_{1/2} = \frac{t}{n}
where:
<span>t_{1/2} - half-life
</span>t - <span>total time elapsed
</span><span>n - a number of half-lives
</span>
The half-life of Sr-90 is 28.8 years.
So, we know:
t = 87.3 years
<span>t_{1/2} = 28.8 years

We need:
n = ?
x = ?
</span>
We could first use the second equation, to calculate n:
<span>If:
t_{1/2} = \frac{t}{n},
</span>Then: 
n = \frac{t}{ t_{1/2} }
⇒ n = \frac{87.3 years}{28.8 years}
⇒ n=3.03
<span>⇒ n ≈ 3
</span>
Now we can use the first equation to calculate the remained amount of the sample.
<span>(1/2)^{n} = x
</span>⇒ x=(1/2)^3
⇒x= \frac{1}{8}<span>
</span>
8 0
4 years ago
Observación cualitativa
arlik [135]
The answer:
La planta tiene las hojas secas.
3 0
3 years ago
Please help me :( would really appreciate whoever that helps me thank you so much!
sergij07 [2.7K]

Answer:

Option e and f are possible

Explanation:

Since we know that 0.264 gallon = 1L ⇒ we change this in all equations.

8.08 L * A  =

a) 8.08 L * 0.264 gal / 8.08 L   = 0.264 gal = 1L

b) 8.08 L * 8.08 gallon / 0.264L = (8,08/ 0.264) L/0.264 L = 30.61 / 0.264 = 115.93 L *8.08 L

c) 8.08 L * 8.08 gallon / 1L = (8,08/ 0.264) L/1 L = 30.61 L * 8.08 L

d) 8.08 L * 0.264 L / 1 gallon =8.08 L * 0.264 L / 1 gal

e) 8.08 L * 0.264 gallon / 1L = 8.08 L *1L / 1L = 8.08 L

f) 8.08 L * 1L / 0.264 gallon = 8.08L * 1L / 1L = 8.08 L

The last 2 options are possible ( e and f )

7 0
3 years ago
Read 2 more answers
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