How many electrons can be ach orbital hold?

If the same large amount of heat is added to a 250 g piece of aluminum and a 150 g piece of aluminum, what will happen? The 150

Inga [223]

Answer:

The 150 g Al will reach a higher temperature.

Explanation:

The amount of heat added to a substance (Q) can be calculated from the relation:

Q = m.c.ΔT.

where, Q is the amount of heat added,

m is the mass of the substance,

c is the specific heat of the substance,

ΔT is the temperature difference (final T - initial T).

Since, Q and c is constant, ΔT will depend only on the mass of the substance (m).

∵ ΔT is inversely proportional to the mass of the substance.

∴ The piece with the lowest mass (150.0 g) will reach a higher temperature than that of a higher mass (250.0 g).

So, the right choice is: The 150 g Al will reach a higher temperature.

3 0

3 years ago

What is the total number of electrons shared in a double covalent bond?

SCORPION-xisa [38]

The correct answer is option 4. In a double covalent bond, two pairs of electrons are bonded or there is a total of four electrons being shared in a bond. For example, we have oxygen. Two atoms of oxygen share two pairs of electrons to have a stable structure.

3 0

4 years ago

What is the pH of this solution?

Vesnalui [34]

Answer:

pH = 11.216.

Explanation:

Hello there!

In this case, according to the ionization of ammonia in aqueous solution:

$NH_3+H_2O\rightleftharpoons NH_4^++OH^-$

We can set up its equilibrium expression in terms of x as the reaction extent equal to the concentration of each product at equilibrium:

$Kb=\frac{[NH_4^+][OH^-]}{[NH_3]} \\\\1.80x10^{-5}=\frac{x*x}{0.150-x}$

However, since Kb<<<1 we can neglect the x on bottom and easily compute it via:

$1.80x10^{-5}=\frac{x*x}{0.150}\\\\x=\sqrt{1.80x10^{-5}*0.150}=1.643x10^{-3}M$

Which is also:

$[OH^-]=1.643x10^{-3}M$

Thereafter we can compute the pOH first:

$pOH=-log(1.643x10^{-3}M)\\\\pOH=2.784$

Finally, the pH turns out:

$pH=14-2.784\\\\pH=11.216$

Regards!

5 0

3 years ago

A compound with a molar mass of 60g/mol is 40.4% carbon, 6.7% hydrogen and 53.3% oxygen (by mass). determine the emperical and m

Fittoniya [83]

A compound is found to be 40.0% carbon, 6.7% hydrogen and 53.5% oxygen. Its molecular mass is 60. g/mol.
Q1)
Empirical formula is the simplest ratio of whole numbers of components making up a compound.
the percentages have been given, therefore we can calculate for 100 g of the compound.

C H O
Mass in 100 g 40.0 g 6.7 g 53.5 g
Molar mass 12 g/mol 1 g/mol 16 g/mol
Number of moles 40.0/12= 3.33 6.7/1 = 6.7 53.5/16 = 3.34
Divide by the least number of moles
3.33/3.33 = 1 6.7/3.33 = 2.01 3.34/3.33 = 1.00
after rounding off
C - 1
H - 2
O - 1

Empirical formula - CH₂O

Q2)
Molecular formula is the actual number of components making up the compound.
To find the number of empirical units we have to find the mass of one empirical unit.
Mass of one empirical unit = CH₂O - 12 + (1x2) + 16 = 30 g
Mass of one mole of compound = 60 g
Number of empirical units = 60 g / 30 g = 2
Therefore molecular formula - 2(CH₂O)
Molecular formula - C₂H₄O₂

4 0

3 years ago

What factors affect the dynamic state of equilibrium in a chemical reaction and how?

yanalaym [24]

Answer:

Only changes in temperature will influence the equilibrium constant $K_c$ . The system will shift in response to certain external shocks. At the new equilibrium $Q$ will still be equal to $K_c$ , but the final concentrations will be different.

The question is asking for sources of the shocks that will influence the value of $Q$ . For most reversible reactions:

External changes in the relative concentration of the products and reactants.

For some reversible reactions that involve gases:

Changes in pressure due to volume changes.

Catalysts do not influence the value of $Q$ . See explanation.

Explanation:

$\displaystyle K_c = {e}^{\Delta G/(R\cdot T)}$ .

Similar to the rate constant, the equilibrium constant $K_c$ depends only on:

$\Delta G$ the standard Gibbs energy change of the reaction, and
$T$ the absolute temperature (in degrees Kelvins.)

The reversible reaction is in a dynamic equilibrium when the rate of the forward reaction is equal to the rate of the backward reaction. Reactants are constantly converted to products; products are constantly converted back to reactants. However, at equilibrium $Q = K_c$ the two processes balance each other. The concentration of each species will stay the same.

Factors that alter the rate of one reaction more than the other will disrupt the equilibrium. These factors shall change the rate of successful collisions and hence the reaction rate.

Changes in concentration influence the number of particles per unit space.
Changes in temperature influence both the rate of collision and the percentage of particles with sufficient energy of reaction.

For reactions that involve gases,

Changing the volume of the container will change the concentration of gases and change the reaction rate.

However, there are cases where the number of gases particles on the reactant side and the product side are equal. Rates of the forward and backward reaction will change by the same extent. In such cases, there will not be a change in the final concentrations. Similarly, catalysts change the two rates by the same extent and will not change the final concentrations. Adding noble gases will also change the pressure. However, concentrations stay the same and the equilibrium position will not change.

8 0

3 years ago