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3 years ago

How can you identify the different elements found on the periodic table?

Chemistry

1 answer:

Firdavs [7]3 years ago

3 0

Answer:

In the typical periodic table, each element is listed by its element symbol and atomic number. For example, “H” denotes hydrogen, “Li” denotes lithium, and so on. Most elements are represented by the first letter or first two letters of their English name, but there are some exceptions.

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I am very confused, our teacher didn’t properly teach us this. How do you do it?

Fudgin [204]

Answer:

You multiply the 3 numbers together to get your volume, in this case it would be 4058.488 cm^3 (cm cubed)

V: 4058.488cm^3 ( round up to 4058.5 for convenience)

M: 27579

D: ?

So we divide mass by the volume to get density, which is

27579 / 4058.5 = ~6.79536 (can round up to 7 or 6.8)

This graph can help a lot, so maybe try and memorize it, hope I helped.

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3 years ago

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What is the symbol for Chromium?

kaheart [24]

The symbol for Chromium is Cr

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2 years ago

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How much thermal energy is needed to change the temperature of kg of water from 25 degrees Celsius to boiling (Hint : boiling in

uysha [10]

∆H = mc∆T

c = specific heat capacity

c of water = 4200 J/kg °C

∆H = 1 * 4200 * (100°C - 25°C)

∆H = 315000J

∆H = 315 KJ

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3 years ago

What is the frequency of radiation emitted by a photon of light if the energy released during its transition to ground state is

notka56 [123]

I cant see the units properly. You can try using E = hf, where h is the Planck's constant and f is the frequency that you need and E is the energy released in the transition.

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3 years ago

Magicians often use ‘flash powder’ in their acts. Flash powder contains 69.0% potassium chlorate (31.9% K, 28.9% Cl, 39.2% O) an

mafiozo [28]

Answer:

$\rm K$ : approximately $22.0\%$ .

$\rm Cl$ : approximately $19.9\%$ .

$\rm O$ : approximately $27.0\%$ .

$\rm Al$ : $31.0\%$ .

Explanation:

Consider a $100\; \rm g$ sample. There would be $69.0\% \times 100\; \rm g = 69.0\; \rm g$ of $\rm KClO_{3}$ and $100.0\; \rm g - 69.0\; \rm g = 31.0\; \rm g$ of $\rm Al$ .

Out of that $69.0\; \rm g$ of $\rm KClO_{3}$ :

$31.9\%$ (by mass) would be $\rm K$ : $31.9\% \times 69.0\; \rm g \approx 22.0\; \rm g$ .

$28.9\%$ (by mass) would be $\rm Cl$ : $28.9\% \times 69.0\; \rm g \approx 19.9\; \rm g$ .

$39.2\%$ (by mass) would be $\rm O$ : $39.2\% \times 69.0\; \rm g \approx 27.0\; \rm g$ .

Overall, the composition (by mass) of each element in this mixture would be:

$\rm K$ :

$\begin{aligned} & \frac{31.9\% \times 69.0\; \rm g}{100\; \rm g} \\ =\; & 31.9\% \times 69.0\% \\ \approx \; & 22.0\%\end{aligned}$ .

$\rm Cl$ :

$\begin{aligned} & \frac{28.9\% \times 69.0\; \rm g}{100\; \rm g} \\ =\; & 28.9\% \times 69.0\% \\ \approx \; & 19.9\%\end{aligned}$ .

$\rm O$ :

$\begin{aligned} & \frac{39.2\% \times 69.0\; \rm g}{100\; \rm g} \\ =\; & 39.2\% \times 69.0\% \\ \approx \; & 27.0\%\end{aligned}$ .

$\rm Al$ :

$\begin{aligned}& \frac{100\; \rm g - 69.0\; \rm g}{100\; \rm g} \\ =\; & 100\% - 69.0\% \\ =\; & 31.0\%\end{aligned}$ .

7 0

2 years ago