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3 years ago

How many bonding electrons are shown below? H-C=C-H

Chemistry

1 answer:

padilas [110]3 years ago

7 0

Answer:

There are total 8 bonding electrons 6 frm the both Carbons and 2 from both hydrogens.

Explanation:

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Where are non-metals located on the periodic table on the right of the periodic table after the staircase where would you find t

djyliett [7]

Did you mean why? not where?

3 0

3 years ago

Chemist preformed Reaction of solid magnesium burning and found that the mass of the magnesium oxide what is greater than the ma

tankabanditka [31]

Answer:

The magnesium reacted with the oxygen in the air.

Explanation:

For argument’s sake, let’s say that the mass of magnesium oxide was 3 g and that of the oxide was 5 g.

The reaction was

magnesium + oxygen ⟶ magnesium oxide

Mass: 3 g 5 g

Mass of oxygen = 5 g – 3 g = 2 g

The 3 g of magnesium must have combined with 2 g of oxygen to form 5 g of magnesium oxide.

4 0

3 years ago

Consider the reaction: 2HI(g) ⇄ H2(g) + I2(g). It is found that, when equilibrium is reached at a certain temperature, HI is 35.

Sliva [168]

Answer:

Kc = 0.075

Explanation:

The dissociation (α) is the initial quantity that ionized divided by the total dissolved. So, let's calling x the ionized quantity, and M the initial one:

α = x/M

x = M*α

x = 0.354M

For the stoichiometry of the reaction (2:1:1), the concentration of H₂ and I₂ must be half of the acid. So the equilibrium table must be:

2HI(g) ⇄ H₂(g) + I₂(g)

M 0 0 Initial

-0.354M +0.177M +0.177M Reacts

0.646M 0.177M 0.177M Equilibrium

The equilibrium constant Kc is the multiplication of the products' concentrations (elevated by their coefficients) divided by the multiplication of the reactants' concentrations (elevated by their coefficients):

$Kc = \frac{[H2]*[I2]}{[HI]^2}$

$Kc = \frac{0.177M*0.177M}{(0.646M)^2}$

$Kc = \frac{0.03133M^2}{0.41732M^2}$

Kc = 0.075

5 0

3 years ago

Calculate the net change in enthalpy for the formation of one mole of acrylic acid from calcium carbide, water and carbon dioxid

Zielflug [23.3K]

Check the attached file for the answer.

5 0

2 years ago

How much heat must be removed from 456 g of water at 25.0°c to change it into ice at?

N76 [4]

This question can be simply solved by using heat formula,
Q = mCΔT

Q = heat energy (J)
m = Mass (kg)
C = Specific heat capacity (J / kg K)
ΔT = Temperature change (K)

when water freezes, it produces ice at 0°C (273 K)
hence the temperature change is 25 K (298 K - 273 K)
C for water is 4186 J / kg K or 4.186 J / g K
By applying the equation,
Q = 456 g x 4.186 J / g K x 25 K
= 47720.4 J
= 47.72 kJ

hence 47.72 kJ of heat energy should be removed.

5 0

2 years ago