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2 years ago

Models may change as new discoveries are made. Please select the best answer from the choices provided T F

1 answer:

8 0

True.
For an example, look at the models of the atom. First, we have the plumb pudding model. Then that was replaced by the planetary model. That model was replaced by the quantum model.

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Write the concentration equilibrium constant expression for this reaction.

Akimi4 [234]

In a chemical reaction, the equilibrium constant refers to the value of its reaction quotient at chemical equilibrium, that is, a condition attained by a dynamic chemical system after adequate time has passed, and at which its composition has no measurable capacity to undergo any kind of further modification.

The given reaction is: HCN (aq) + OH⁻ = CN⁻ (aq) + H2O (l)

The equilibrium constant = product of concentration of products / product of concentration of reactants

(Here, H2O is not considered as its concentration is very high)

So, Keq = [CN⁻] / [HCN] [OH⁻]

8 0

2 years ago

Why water can be consider as an acid as well as base .explain briefly

faltersainse [42]

It depends on the pH level of the water.

4 0

2 years ago

Read 2 more answers

During the chemical reaction given below 21.71 grams of each reagent were allowed to react. Determine how many grams of the exce

swat32

Answer: 16.32 g of $O_2$ as excess reagent are left.

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$

$\text{Moles of} SO_2=\frac{21.71g}{64g/mol}=0.34mol$

$\text{Moles of} O_2=\frac{21.71g}{32g/mol}=0.68mol$

$2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)$

According to stoichiometry :

2 moles of $SO_2$ require = 1 mole of $O_2$

Thus 0.34 moles of $SO_2$ will require= $\frac{1}{2}\times 0.34=0.17moles$ of $O_2$

Thus $SO_2$ is the limiting reagent as it limits the formation of product and $O_2$ is the excess reagent.

Moles of $O_2$ left = (0.68-0.17) mol = 0.51 mol

Mass of $O_2=moles\times {\text {Molar mass}}=0.51moles\times 32g/mol=16.32g$

Thus 16.32 g of $O_2$ as excess reagent are left.

3 0

2 years ago

3.0 cm x 4.0 cm x 1.0 cm [?]cm^3

CaHeK987 [17]

Explanation:

Hi there!!

you asked to multiply these all right,

you can simply multiply it ,

=3cm × 4 cm × 1cm

= 12cm^2×1cm (4×3=12)

= 12cm^3 (12×1=12)

Therefore, theanswer is 12 cm^3.

Hope it helps..

4 0

2 years ago

One mole of calcium ions has the same number of representative particles as which of the following? one mole of calcium nuclei o

Kruka [31]

One mole is always the same number: 6.02 * 10^ 23.

So, one mole of cars = 6.02 * 10 ^23 cars; one mole of pencils = 6.02 * 10^23 pencils; one mole of atoms = 6.02 * 10^23 atoms; one mole of molecules = 6.02 * 10^23 molecules.

So, all the options are correct: one mole of calcium ions has 6.02 * 10^23 representative particles, such as one mole of calcium nuclei and one of calcium atoms.

3 0

2 years ago