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2 years ago

Which electrons participate in chemical bonding?

Chemistry

2 answers:

OverLord2011 [107]2 years ago

7 0

B.) Valence Electrons. The nucleus of the valence electrons attracts and pulls atoms together.

VMariaS [17]2 years ago

4 0

Answer:

Explanation:

I took the test

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When a gas changes into a liquid , this is known as

Hitman42 [59]

Condensation, is the process of a gas changing into a liquid.

5 0

3 years ago

The height of a column of mercury in a closed-end manometer is 13.2 cm. What is the pressure of the gas in torr

AlladinOne [14]

Answer:

$P_{gas}=131.96torr$

Explanation:

Helo,

In this case, the pressure must be computed as follows:

$P_{gas}=gh\rho _{Hg}$

Which is using the density of mercury (13.6 g/mL) and its height, thus, we obtain (using the proper units):

$P_{gas}=9.8\frac{m}{s^2}*(13.2cm *\frac{1m}{100cm} )*(13.6\frac{g}{cm^3}*\frac{1kg}{1000g}*\frac{1x10^6cm^3}{1m^3} )\\\\P_{gas}=17592.96Pa*\frac{760torr}{101325Pa} \\\\P_{gas}=131.96torr$

Best regards.

5 0

2 years ago

B. Determine the percent composition of each element in NaCl. (1 point) Must show work to

andrey2020 [161]

Answer:

Sodium (Na): (.5 point)

22.99+35.453=58.433. 22.99/58.433= 39%

Chlorine (Cl): (.5 point)

22.99+35.453=58.433. 35.453/58.433= 61%

Explanation:

8 0

2 years ago

How much heat is released when 105 g of steam at 100.0°C is cooled to ice at -15.0°C? The enthalpy of vaporization of water is 4

Yuki888 [10]

1) (Hvap)(moles of water)=236.9783574kJ
(40.67)(105/18.02)
2) (change in temperature)(mass)(Cliquid)=43.9345172kJ
(100)(105/18.02)(75.4)/1000
3) (Hfus)(moles of water)=35.01942286kJ
(6.01)(105/18.02)
4) (change in temperature)(mass)(Csolid)=3.181465039kJ
(15)(105/18.02)(36.4)/1000
Total released=319.1137625kJ

5 0

2 years ago

The activation energy of a certain uncatalyzed reaction is 64 kJ/mol. In the presence of a catalyst, the Ea is 55 kJ/mol. How ma

Ksivusya [100]

Answer:

About 5 times faster.

Explanation:

Hello,

In this case, since the Arrhenius equation is considered for both the catalyzed reaction (1) and the uncatalized reaction (2), one determines the relationship between them as follows:

$\frac{k_1}{k_2}=\frac{Aexp(-\frac{Ea_1}{RT} )}{Aexp(-\frac{Ea_2}{RT})} \\\frac{k_1}{k_2}=\frac{exp(-\frac{Ea_1}{RT} )}{exp(-\frac{Ea_2}{RT})}$

By replacing the corresponding values we obtain:

$\frac{k_1}{k_2}=\frac{exp(-\frac{55000J/mol}{8.314J/molK*673.15K} )}{exp(-\frac{64000J/mol}{8.314J/molK*673.15K} )} =4.8$

Such result means that the catalyzed reaction is about five times faster than the uncatalyzed reaction.

Best regards.

4 0

3 years ago