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Anarel [89]
3 years ago
13

Electronegativity increases as the size of the atom increases. a. True b. False

Chemistry
1 answer:
stiv31 [10]3 years ago
5 0
B. False. Electronegativity is the property of an atom or element to attract electrons during a chemical reaction. If you look at the periodic trend for electronegativity you see Flourine has the highest electronegativity and Francium is the lowest. However, Francium has the largest atomic radius and Fluorine has the smallest (aside from Hydrogen and Helium). So your statement is false. The relationship between atomic radius and electronegativity is inversely proportional so a true statement would be: As atomic radius increases, electronegativity decreases.
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Depicted below you will see the molecules COH2 and SO3 . Why is COH2 polar while SO3 isn’t?
tia_tia [17]

Answer:

See explanation

Explanation:

When we look at SO3 we will notice that the compound is trigonal planar and it is symmetric. This means that it has equal charge distribution hence its dipoles cancel out resulting in a zero net dipole moment.

However, COH2 is also trigonal planar but is non-symmetric. Hence, its dipole moments do not cancel out, hence the molecule has a resultant dipole moment and is a polar molecule

7 0
2 years ago
What family are the elements carbon, germanium and lead in?
grin007 [14]

Answer:

Damian here!! :))

The carbon family consists of the elements carbon (C), silicon (Si), germanium (Ge), tin (Sn), lead (Pb), and flerovium (Fl). Atoms of elements in this group have four valence electrons. The carbon family is also known as the carbon group, group 14, or the tetrels. Elements in this family are of key importance for semiconductor technology.

Explanation:

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5 0
3 years ago
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sveticcg [70]
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3 years ago
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You have 17 liters of gas at STP. If the temperature rises to 94C and while the volume decreases to 12 liters, what will the ne
fenix001 [56]

Answer:

P_2=1.90atm

Explanation:

Hello!

In this case, according to the ideal gas equation ratio for two states:

\frac{P_1V_1}{P_2V_2} =\frac{n_1RT_1}{n_2RT_2}

Whereas both n and R are cancelled out as they don't change, we obtain:

\frac{P_1V_1}{P_2V_2} =\frac{T_1}{T_2}

Thus, by solving for the final pressure, we obtain:

\frac{P_2V_2}{P_1V_1} =\frac{T_2}{T_1}\\\\P_2=\frac{T_2P_1V_1}{V_2T_1}

Now, since initial conditions are 1.00 atm, 273.15 K and 17 L and final temperature and volume are 94 + 273 = 367 K and 12 L respectively, the resulting pressure turns out to be:

P_2=\frac{367K*1.00atm*17L}{12L*273.15K}\\\\P_2=1.90atm

Best regards!

7 0
3 years ago
After passing, it is important to always follow through. T/F
VMariaS [17]

Answer:

t

Explanation:

8 0
3 years ago
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