Friction is generated when _____ interfere with each other on sliding surfaces. solids molecules forces fluids

32. In general, ionization energy increases from left to right across a given

miskamm [114]

Aluminum's outer electron in 3p-orbital is slightly further from the nucleus and protected by 3s-orbital. This causes it to need less energy. Magnesium's outer electron is closer and not protected by an outer orbital.
Hope This Helps and Ged Bless!

6 0

3 years ago

Hydrogen gas (a potential future fuel) can be formed by the reaction of methane with water according to the following equation:

dem82 [27]

Answer:

The percent yield of the reaction is 62.05 %

Explanation:

Step 1: Data given

Volume of methane = 25.5 L

Pressure of methane = 732 torr

Temperature = 25.0 °C = 298 K

Volume of water vapor = 22.0 L

Pressure of H2O = 704 torr

Temperature = 125 °C

The reaction produces 26.0 L of hydrogen gas measured at STP

Step 2: The balanced equation

CH4(g) + H2O(g) → CO(g) + 3H2(g)

Step 3: Calculate moles methane

p*V = n*R*T

⇒with p = the pressure of methane = 0.963158 atm

⇒with V = the volume of methane = 25.5 L

⇒with n = the moles of methane = TO BE DETERMINED

⇒with R = the gas constant = 0.08206 L*atm/mol*K

⇒with T = the temperature = 298 K

n = (p*V) / (R*T)

n = (0.963158 * 25.5 ) / ( 0.08206 * 298)

n = 1.0044 moles

Step 4: Calculate moles H2O

p*V = n*R*T

⇒with p = the pressure of methane = 0.926316 atm

⇒with V = the volume of methane = 22.0 L

⇒with n = the moles of methane = TO BE DETERMINED

⇒with R = the gas constant = 0.08206 L*atm/mol*K

⇒with T = the temperature = 398 K

n = (p*V) / (R*T)

n = (0.926316 * 22.0) / (0.08206 * 398)

n = 0.624 moles

Step 5: Calculate the limiting reactant

For 1 mol methane we need 1 mol H2O to produce 1 mol CO and 3 moles H2

H2O is the limiting reactant. It will completely be consumed (0.624 moles).

Methane is in excess. There will react 0.624 moles. There will remain 1.0044 - 0.624 moles = 0.3804 moles methane

Step 6: Calculate moles hydrogen gas

For 1 mol methane we need 1 mol H2O to produce 1 mol CO and 3 moles H2

For 0.624 moles H2O we'll have 3*0.624 = 1.872 moles

Step 9: Calculate volume of H2 at STP

1.0 mol at STP has a volume of 22.4 L

1.872 moles has a volume of 1.872 * 22.4 = 41.9 L

Step 10: Calculate the percent yield of the reaction

% yield = (actual yield / theoretical yield) * 100 %

% yield = ( 26.0 L / 41.9 L) *100 %

% yield = 62.05 %

The percent yield of the reaction is 62.05 %

6 0

3 years ago

Read 2 more answers

Which list of elements is arranged in order of increasing electronegativity?

muminat

K, ca, sc is the right answer. Take a look at table S of your chemistry reference table.

6 0

3 years ago

Read 2 more answers

_Ca + _H20 --> _Ca(OH)2 + _H2

Liono4ka [1.6K]

2Ca + 2H2O —-> Ca(OH)2 + H2

5 0

4 years ago

Consider the elements: Na, Mg, Al, Si, P.

Olegator [25]

Answer:

The elements mentioned in the series are Na, Mg, Al, Si, and P. It can be seen that all these elements are located in the same period. The atomic number of the mentioned elements are Na-11, Mg-12, Al-13, Si-14, and P-15.

a) There will be an increase in the ionization energy with the increase in the element's atomic number across a period. More energy is needed to withdraw an electron from a completely occupied shell in comparison to an incompletely occupied shell.

The atomic number of Na is 11. When one electron is withdrawn from Na, it gets converted into the inert gas configuration of Ne. Thus, it will require more energy to withdraw the second electron from Na. Hence, Na exhibits the lowest second ionization energy.

b) Across the period, with an increase in the element's atomic number, the atomic radii reduces from left to right. Thus, P exhibits the smallest atomic radius.

c) The metallic nature of the elements reduces from left to right across a period in the periodic table. Thus, P is the least metallic element.

d) Diamagnetic signifies towards the element that exhibits pair electrons in its sub-shells. The electronic configuration of Mg is,

1s2 2s2 2p6 3s2

In Mg, no unpaired electrons are present, while all the remaining elements mentioned exhibit unpaired electrons in their valence shell. Thus, Mg is the diamagnetic element.

8 0

3 years ago