Copper and iron(III) nitrate
Does it have a reaction?

Answer: yes it is false

Explanation:

The statement is false. A synthetic process with a lower E-factor produces less waste than a process with a higher E-factor.

The E-factor is a measure of the waste generated during a manufacturing process. It is calculated by dividing the total mass of waste produced by the mass of the desired product. A lower E-factor indicates that less waste is generated per unit of product.

In this case, the synthetic process with an E-factor of 3.0 produces less waste than the process with an E-factor of 17.4. This means that the process with an E-factor of 3.0 is more efficient in terms of waste reduction.

Answer:

See explanation below

Explanation:

In both cases the central atoms, C in CHCl₃ and O in H₂O, are sp³ hybridized .

Since they are sp³ hybridized we predict an angle  between the  H-C-Cl and H-O-H of 109.5 º ( tetrahedral ), but two of the sp³ orbitals in water are occupied by lone pairs.

These lone pairs do excercise more repulsion ( need more room ) than the bonds oxygen is making with hydrogen.

As a consequence of this repulsion the angles H-O-H are less than the predicted 109.5º in tetrahedra. ( Actually is 104.5 º)

Answer:

The ballance half reactions are:

Mg²⁺  + 2e⁻ → Mg

6OH⁻ + Si  → SiO₃²⁻ + 4e⁻ + 3 H₂O

Coefficients for H2O and OH– are 3 for H₂O (in products side) and 6 for OH⁻ (in reactants side)

Explanation:

Si (s) + Mg(OH)₂ (s) → Mg (s) + SiO₃²⁻ (aq)

Let's see the oxidations number.

As any element in ground state, we know that oxidation state is 0, so Si in reactants and Mg in products, have 0.

Mg in reactants, acts with +2, so the oxidation number has decreased.

This is the reduction, so it has gained electrons.

Si in reactants acts with 0 so in products we find it with +4. The oxidation number increased it, so this is oxidation. The element has lost electrons.

Let's take a look to half reactions:

Mg²⁺  + 2e⁻ → Mg

Si  → SiO₃²⁻ + 4e⁻

In basic medium, we have to add water, as the same amount of oxygen we have, IN THE SAME SIDE. We have 3 oxygens in products, so we add 3 H₂O and in the opposite site we can add OH⁻, to balance the hydrogen. The half reaciton will be:

6OH⁻ + Si  → SiO₃²⁻ + 4e⁻ + 3 H₂O

If we want to ballance the main reaction we have to multiply (x2) the half reaction of oxidation. So the electrons can be ballanced.

2Mg²⁺  + 4e⁻ → 2Mg

Now, that they are ballanced we can sum the half reactions:

2Mg²⁺  + 4e⁻ → 2Mg

6OH⁻ + Si  → SiO₃²⁻ + 4e⁻ + 3 H₂O

2Mg²⁺  + 4e⁻  + 6OH⁻ + Si  → 2Mg  +  SiO₃²⁻ + 4e⁻ + 3 H₂O

Given:

Radiation emission in Cs atom = 9,192,632,770 cycles

To determine:

The wavelength of the above radiation

Explanation:

It is given that :-

1 sec equivalent to 9,192, 631, 770

Now, frequency (ν) = cycles /sec = 9,192, 631, 770/sec

Wavelength of a radiation is given as:

λ = c/ν

where c = speed of light = 3*10⁸ m/s

λ = 3*10⁸ ms⁻¹/9,192, 631, 770 s⁻¹ = 0.0326 m

Ans: Thus the wavelength of this radiation is 0.033 m

Answer:

The chemical formula for the ionic compound formed by Au3+ and

HSO3-compound  is Au(HSO3)3

Explanation:

The charge on Au ion is

And the charge on HSO3- is

Thus, the number of atoms required by HSO3- to complete its octate is 1. On the other hand Au has 3 excess ions and hence it is to be released to reach the stable state.

So three molecules of HSO3- will combine with one atom of Au 3+

Thus, the compound formed by these two is Au(HSO3)3

Final answer:

The chemical formula for the ionic compound formed by Au3+ and HSO3- is Au(HSO3)3, as ionic compounds are always neutral.

Explanation:

The ionic compound formed by Au3+ (Gold ion) and HSO3- (Bisulfite ion) must have a net charge of zero since ionic compounds are neutral. Hence, we need 3 bisulfite ions to balance out one gold ion, which gives us the chemical formula as Au(HSO3)3.

Indeed, the formation of ionic compounds is a fascinating process. It involves the transfer of electrons from one atom (usually a metal) to another (usually a nonmetal), resulting in the formation of ions. These ions are then attracted to each other due to their opposite charges, forming an ionic compound. In this case, the gold ion (Au3+) donates three electrons, which are accepted by three bisulfite ions (HSO3-). This results in a neutral compound, as the positive and negative charges balance each other out. The resulting compound, Au(HSO3)3, is an example of how elements can combine in specific ratios to form neutral compounds.

Learn more about Chemical Formula here:

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Answer:

16 liters of the solution with 40% concentration must be mixed with 62 liters of the solution with 25% concentration in order to obtain 78 liters, 25% concentration solution.

Explanation:

Let the required volume of solution 1 be represented by x.

The required volume of solution 2 would then be 78-x.

The number of moles of solution 1 that would be required = 0.4x

The number of moles of solution 2 that would be required = 0.25(78-x)

The number of moles of the final mixture = 78 x 0.28 = 21.84

moles of solution 1 + moles of solution 2 = moles of final mixture

0.4x + 0.25(78 - x) = 21.84

  0.4x + 19.5 - 0.25x = 21.84

     0.4x - 0.25x = 21.84 - 19.5

          0.15x = 2.34

            x = 15.6 liters

To the nearest tenth = 16 liters

Liters of 40% solution needed = 16 liters

Liters of 25% solution needed = 78 - 16 = 62 liters.

Hence, 16 liters of the solution with 40% concentration must be mixed with 62 liters of the solution with 25% concentration in order to obtain 78 liters, 25% concentration solution.