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3 years ago

Which statements are true of precipitation reactions?

2 answers:

7 0

In chemistry, a precipitation reaction is a double replacement reaction. It involves two reactants and two products. It means that the cations and anions of the compounds interchange with one another to yield two products. But what makes precipitation reactions different is that when two liquid reactants are allowed to react together, a solid substance called precipitate is formed. The solid appears because it is insoluble to the other product in aqueous state.

Therefore, basing on the choices given, precipitation reactions apply to letters C and D.

azamat3 years ago

7 0

the correct answer to this question would be B and C.

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photoshop1234 [79]

If the element have 39 protons, that means that his atomic number is 39, being Yttrium the element. And, the number of protons is the same of the number of electrons, so, 39 electrons.

6 0

3 years ago

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The Balmer series consists of the spectral lines from hydrogen for an electron making a transition from an excited state to the

MaRussiya [10]

Answer:

Wavelength of the first four spectral line:

$\lambda_1=1216{\AA}$

$\lambda_2=1026{\AA}$

$\lambda_3=972.8{\AA}$

$\lambda_4=950{\AA}$

Explanation:

Lman series when electron from higher shell number and jumps to 1st shell then all the possible lines is called as lyman series.

to calculate wavelength of first four spectral line:

For hydrogen Z=1;

by using rydberg equation

$\frac{1}{\lambda} =RZ^2[\frac{1}{n_1^2} -\frac{1}{n_2^2} ]$

1. n=2 to n=1

$\frac{1}{\lambda} =RZ^2[\frac{1}{n_1^2} -\frac{1}{n_2^2} ]$

$\frac{1}{R} =912{\AA}$ =rydberg constant

$\lambda=\frac{4}{3R}$

$\lambda_1=1216{\AA}$

2. n=3 to n=1

$\lambda=\frac{9}{8R}$

$\lambda_2=1026{\AA}$

3. n=4 to n=1

$\lambda=\frac{16}{15R}$

$\lambda_3=972.8{\AA}$

4. n=5 to n=1

$\lambda=\frac{25}{24R}$

$\lambda_4=950{\AA}$

5 0

3 years ago

. The net ionic equation for the reaction that occurs during the titration of nitrous acid with sodium hydroxide is (a) HNO2 Na

hjlf

Answer:

$H^+_{(aq)}+OH^{-}_{(aq)}\rightarrow H_2O_{(l)}$

Explanation:

Complete ionic equation : In complete ionic equation, all the substance that are strong electrolyte and present in an aqueous are represented in the form of ions.

Net ionic equation : In the net ionic equations, we are not include the spectator ions in the equations.

Only the species which are present in aqueous state dissociate.

Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.

(a)

The balanced molecular equation will be,

$HNO_2_{(aq)}+NaOH_{(aq)}\rightarrow H_2O_{(l)}+NaNO_2_{(aq)}$

The complete ionic equation in separated aqueous solution will be,

$H^+_{(aq)}+NO_2^-_{(aq)}+Na^+_{(aq)}+OH^{-}_{(aq)}\rightarrow H_2O_{(l)}+Na^+_{(aq)}+NO_2^-_{(aq)}$

In this equation the species present are, $Na^+\text{ and }NO_2^-$ are the spectator ions.

Hence, the net ionic equation contains specie is

$H^+_{(aq)}+OH^{-}_{(aq)}\rightarrow H_2O_{(l)}$

6 0

3 years ago

Consider the following reaction:C2H4(g) + F2(g) -----------> C2H4F2(g) Delta H = -549 kJEstimate the carbon-fluorine bond ene

frutty [35]

Answer:

Bond energy of carbon-fluorine bond is 485 kJ/mol

Explanation:

Enthalpy change for a reaction, is given as:

$\Delta H_{rxn}=\sum [n_{i}\times (E_{bond})_{i}]-\sum [n_{j}\times (E_{bond})_{j}]$

Where $(E_{bond})_{i}$ and $(E_{bond})_{j}$ represents average bond energy in breaking "i" th bond and forming "j" th bond respectively. $n_{i}$ and $n_{j}$ are number of moles of bond break and form respectively.

In this reaction, one mol of C=C, four moles of C-H and one mol of F-F bonds are broken. One mol of C-C bond, four moles of C-H bonds and two moles of C-F bonds are formed

So, $-549kJ=(1mol\times 614kJ/mo)+(4mol\times E_{C-H})+(1mol\times 154kJ/mol)-(1mol\times 347kJ/mol)-(4mol\times E_{C-H})-(2mol\times E_{C-F})$

or, $-549kJ=(1mol\times 614kJ/mo)+(1mol\times 154kJ/mol)-(1mol\times 347kJ/mol)-(2mol\times E_{C-F})$

or, $E_{C-F}=485kJ/mol$

So bond energy of carbon-fluorine bond is 485 kJ/mol

8 0

3 years ago

Which of the following pairs of elements could possibly be in the same group? X has a 1+ ion; Y has a 1- ion. X tends to form a

exis [7]

Let's eliminate these one by one.
The first pair would not be the same, as X would most likely be in group IA, and Y would be in group VIIA, because of their tendency to gain and lose electrons.
The second pair would also violate the same rule, but X would most likely be in group IIA, and Y would most likely be in group VIA.
The third pair would not be the same, as X is most likely in group VIIA, and since Y has eight valence electrons, it is most likely a noble gas.
The final pair has X with atomic number 15, making it phosphorous. Phosphorous wants to gain 3 electrons to have a full octet of 8 outer "valence" electrons, and Y would also like to gain 3 electrons. This means it is possible that the final pair would be in the same group.

5 0

3 years ago

Read 2 more answers