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Leokris [45]
3 years ago
9

Trends in atomic radius of Period 3 elements

Chemistry
1 answer:
serious [3.7K]3 years ago
5 0

Answer:

Here's what I find  

Explanation:

A. Period or Group?

Na Mg Al Si

These elements are in Period 3.

B. Trend in atomic radius

\begin{array}{ccc}\textbf{Atom} &\textbf{At. No.} &\textbf{r/nm }\\\text{Na} & 11& 0.19 \\\text{Mg} & 12 & 0.14 \\\text{Al} & 13 & 0.12 \\\text{Si} & 14 & 0.11 \\\end{array}

The graph below shows that the atomic radius decreases as the atomic number increases, that is, as you go from left to right across the Period.

C. Explanation

As you go from left to right across a Period, you are adding electrons to the n = 3 shell. The electrons repel each other, so the atoms tend to increase in size.

However, you are also adding protons to the nucleus, so the attraction between the protons and the electrons increases. Each added electron can shield only about 0.3 of a proton's charge, so the nuclear attraction wins out and the atomic radius decreases.

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Answer:

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Explanation:

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3 years ago
Branches of chemistry with definition and examples in real life
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8 0
3 years ago
2.478 g of white phosphorus was used to make phosphine according to the equation: P₄(s) + 3OH⁻(aq) + 3H₂O(l) → PH₃(g) + 3H₂PO₂⁻(
prohojiy [21]

<u>Answer:</u> The amount of white phosphorus is 0.0199 moles

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

We are given:

Given mass of white phosphorus = 2.478 g

Molar mass of white phosphorus = 124 g/mol

Putting values in above equation, we get:

\text{Moles of white phosphorus}=\frac{2.478g}{124g/mol}=0.0199mol

Hence, the amount of white phosphorus is 0.0199 moles

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3 years ago
Why is sunlight considered a renewable resource?
allochka39001 [22]
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7 0
3 years ago
Read 2 more answers
The half-life of gold-198 is 2.7 days. After
Pie

Answer: 8.1 days

Explanation:

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant

t = age of sample

a = let initial amount of the reactant = x

a - x = amount left after decay process= \frac{x}{4} 

a) to find rate constant

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

t_{\frac{1}{2}}=\frac{0.693}{k}

k=\frac{0.693}{2.7days}=0.257days^{-1}

b) for completion of one fourth of reaction

t=\frac{2.303}{k}\log\frac{x}{\frac{x}{4}}

t=\frac{2.303}{0.257}\log{4}

t=8.1days

Thus after 8.1 days , one fourth of original amount will remain.

8 0
4 years ago
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