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goldenfox [79]
3 years ago
15

How do you think the atomic radii will change as electrons are added to a shell?

Chemistry
2 answers:
Korvikt [17]3 years ago
6 0
As electrons are added to the valence shell, an extra proton (i.e fundamental, positively charged nuclear particle) is added to the element's nucleus. As electrons add and Z the atomic number increases 1 by 1, nuclear charge WINS, and electronic radii contract.
kirill115 [55]3 years ago
3 0

Answer:

As electrons are added to the valence shell, an extra proton (i.e fundamental, positively charged nuclear particle) is added to the element's nucleus. As electrons add and Z the atomic number increases 1 by 1, nuclear charge WINS, and electronic radii contract.

Explanation:

Hope this helps you

Crown me as brainliest:)

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Which of the following are examples of physical change? Copper oxidizing Iron rusting Carbon dioxide sublimating Water evaporati
kari74 [83]
Carbon dioxide sublimation
water evaporate

physical change is just change between solid/liquid/gas, other two are chemical changes
6 0
3 years ago
Which is a factor in determining the average atomic mass of an element?
kvasek [131]
The factor in determining the average atomic mass of an element is: 
B or 2 relative abundance of each isotope because the by looking at how many protons , electrons and neutrons the most isotope is of the element has relative abundance. 
7 0
3 years ago
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An unknown piece of metal weighing 95.0 g is heated to 98.0°C. It is dropped into 250.0 g of water at 23.0°C. When equilibrium i
lutik1710 [3]

Answer:

C_{metal}=126.6\frac{J}{g\°C}

Explanation:

Hello!

In this case, when two substances at different temperature are put in contact and an equilibrium temperature is attained, we can evidence that the heat lost by the hot substance (metal) is gained by the cold substance (water) and we can write:

Q_{metal}=-Q_{water}

Which can be also written as:

m_{metal}C_{metal}(T_{EQ}-T_{metal})=-m_{water}C_{water}(T_{EQ}-T_{water})

Thus, since we need the specific heat of the metal, we solve for it as shown below:

C_{metal}=\frac{m_{water}C_{water}(T_{EQ}-T_{water})}{-m_{metal}(T_{EQ}-T_{metal})} \\\\C_{metal}=\frac{250.0g*4.184\frac{J}{g\°C}(29.0\°C-98.0\°C)}{95.0g(29.0\°C-23.0\°C)} \\\\C_{metal}=126.6\frac{J}{g\°C}

Best regards.

7 0
2 years ago
In this experiment when both crests combine they produce a __________ spot.
zvonat [6]

Answer:

meteor

Explanation:

6 0
3 years ago
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If a temperature increase from 10.0 ∘C to 22.0 ∘C doubles the rate constant for a reaction, what is the value of the activation
maria [59]

The activation energy barrier is 40.1 kJ·mol⁻¹

Use the Arrhenius equation

\ln( \frac{k_2 }{k_1 }) = (\frac{E_{a} }{R })(\frac{ 1}{T_1} - \frac{1 }{T_2 })\\

\ln( \frac{2k }{k}) = (\frac{E_{a} }{\text{8.314 J} \cdot \text{K}^{-1} \text{mol}^{-1} })(\frac{ 1}{\text{283.15 K}} - \frac{1 }{\text{295.15 K }})\\

\ln2 = (\frac{ E_{a} }{\text{8.314 J} \cdot \text{K}^{-1} \text{mol}^{-1}}) \times 1.436 \times10^{-4}\\

\ln2 = E_{a} \times 1.727 \times 10^{-5} \text{ mol} \cdot \text{J}^{-1}

E_{a} = \frac{\ln2 }{ 1.727 \times10^{-5}\text{ mol} \cdot \text{J}^{-1}}\\

E_{a} = \text{40 100 J}\cdot\text{mol}^{-1} = \textbf{40.1 kJ}\cdot \textbf{mol}^{-1}

3 0
3 years ago
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