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Alexandra [31]
4 years ago
13

Magnesium is an alkaline earth metal with an atomic number of 12. It’s in the third row in the periodic table. Which expression

shows one way to find the number of valence electrons in one atom of magnesium?
Chemistry
1 answer:
Leviafan [203]4 years ago
7 0

Answer:

group # = # valence electrons

group 2= 2 valence electrons

Explanation:

Mg is in group #2 and subsequently has 2 valence electrons (electrons on the outermost shell)

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The relative atomic mass of Chlorine is 35.45. Calculate the percentage abundance of the two isotopes of Chlorine, 35Cl and 37Cl
nata0808 [166]

Answer:

35Cl = 75.9 %

37Cl = 24.1 %

Explanation:

Step 1: Data given

The relative atomic mass of Chlorine = 35.45 amu

Mass of the isotopes:

35Cl = 34.96885269 amu

37Cl = 36.96590258 amu

Step 2: Calculate percentage abundance

35.45 = x*34.96885269 + y*36.96590258

x+y = 1  x = 1-y

35.45 = (1-y)*34.96885269 + y*36.96590258

35.45 = 34.96885269 - 34.96885269y +36.96590258y

0.48114731 = 1,99704989‬y

y = 0.241 = 24.1 %

35Cl = 34.96885269 amu = 75.9 %

37Cl = 36.96590258 amu = 24.1 %

3 0
3 years ago
If you leave a glass of water and ice on the kitchen counter long enough, after a while you will have just a glass of water. why
irakobra [83]
The heat of the water is shared with the ice and there for you drink will get cold but your ice will melt away from the heat going in the ice
8 0
3 years ago
Which element has a larger atomic radius than sulfur? chlorine cadmium fluorine oxygen
zhuklara [117]

Answer:

  • <u>Cadmium has larger atomic radius than sulfur.</u>

Explanation:

Down a period, atomic radii decrease from left to right due to the increase in the number of protons and electrons across a period: when a proton is added the pull of the electrons towards the nucleus is larger, so the size of the atom decreases.

Hence, you can compare the elements that belong to a same period and predict that the atom with lower atomic number (number of protons) will haver larger atomic radius. With that:

  • Oxygen and fluorine are in the period 3, being oxygen to the left of fluorine, so oxygen is larger than fluorine.

  • Sulfur and chlorine are in the period 4, being sulfur to the left of chlorine, so sulfur is larger than chlorine.

Now see whan happens down a group. Atomic radius increases from top to bottom within a group due to electron shielding. That permits you to compare the size of the elements in a group:

  • Fluorine and chlorine are in the same group (17), with chlorine directly below fluorine, so the atomic radius of chlorine is larger than the atomic radius of fluorine.

  • Sulfur and oxygen are in the same group (16), with sulfur directlly below oxygen, so sulfur the atomic radius of sulfur is larger than the atocmi radius of oxygen.

So far, you can rank the atomic radius of sulfur, chlorine, fluorine, and oxygen, in increasing order as:

  • O < F < Cl < S, concluding that O, F, and Cl have smaller atomic radius than S.

Cadmiun, Cd, is to the left and below sulfur, so both electron shielding (down a group) and increase of the number of protons (down a period) lead to predict the cadmium has a larger atomic radius than sulfur.

8 0
3 years ago
A student shines a flashlight on a mirror.
irga5000 [103]

Answer:

a

Explanation:

mirrors reflect light

7 0
3 years ago
Read 2 more answers
A chemist mixes 75.0 g of an unknown substance at 96.5°C with 1,150 g of water at 25.0°C. If the final temperature of the system
AleksAgata [21]
To do this problem it is necessary to take into account that the heat given by the unknown substance is equal to the heat absorbed by the water, but considering the correct sign:

-m\cdot c_e\cdot \Delta T = m_w\cdot c_e_w\cdot \Delta T_w

Clearing the specific heat of the unknown substance:

c_e = \frac{m_w\cdot c_e_w\cdot \Delta T_w}{m\cdot \Delta T} = -\frac{1\ 150\ g\cdot 4.184\frac{J}{g\cdot ^\circ C}\cdot (37.1 - 25.0)^\circ C}{75\ g\cdot (37.1 - 96.5)^\circ C}

c_e = -\frac{1\ 150\ g\cdot 4.184\frac{J}{g\cdot ^\circ C}\cdot 12.1^\circ C}{75\ g\cdot (-59.4)^\circ C} = \bf 13.07\frac{J}{g\cdot ^\circ C}
6 0
3 years ago
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