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Write the name of the ionic compound Fe(NO2)2Answer
iron (III) nitrite
iron (II) nitrite
iron (1) nitrite

Answers

Answer 1
Answer:

Final answer:

The name of the ionic compound Fe(NO2)2 is iron (II) nitrite.


Explanation:

The name of the ionic compound Fe(NO2)2 is iron (II) nitrite.


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The Michael reaction is a conjugate addition process wherein a nucleophilic enolate anion (the donor) reacts with an α,β-unsaturated carbonyl compound (the acceptor). The best Michael reactions are those that take place when a particularly stable enolate anion is formed via treatment of the donor with a strong base. Alternatively, milder conditions can be used if an enamine is chosen as the donor, this variant is termed the Stork reaction. In the second step, the donor adds to the β-carbon of the acceptor in a conjugate addition, generating a new enolate. The enolate abstracts a proton from solvent or from a new donor molecule to give the conjugate addition product. Draw curved arrows to show the movement of electrons in this step of the mechanism.

Answers

Answer:

See the attached file for the structure

Explanation:

See the attached file

Four different methods are described for validating the results of a particular analysis. Indicate for each whether the method primarily checks the accuracy of the analysis or the precision of the analysis.a. Five aliquots of the same sample are injected for gas chromatographic analysis by one person on the same day.
i. Accuracy
ii. Precision
b. A known amount of analyte is added to an aliquot of the sample and analyzed with sample.
i. Accuracy
ii. Precision
c. Aliquots from a blood sample are sent to three separate laboratories for analysis using the same method.
i. Accuracy
ii. Precision
d. Identical standard are analyzed by two different methods.
i. Accuracy
ii. Precision

Answers

Answer:

a) Precision

b) Accuracy

c) Accuracy and precision

d) Accuracy

Explanation:

When an experiment is done more than once to determine if the results are statistically ok, two forms of the validations are possible the accuracy and precision. When the values of the various experiments are close to the known value, then they are accurate. When the values are close to each other they are precise. So, sometimes the results are precise but are not accurate, and vice-versa.

a) Here, the person wants to find if the 5 aliquots will have close results, so, he or she is looking for precision.

b) Here the amount of analyte is already known, and the person wants to identify if the value will be the same when analyzed together with another sample, thus he or she is looking for accuracy.

c) Here the three results will be compared with each other (precision) and with the standard value of the method (accuracy).

d) The methods will be tested, and the values will be compared with the standard known value, so the person is looking for accuracy.

The following data were obtained in a kinetics study of the hypothetical reaction A + B + C → products. [A]0 (M) [B]0 (M) [C]0 (M) Initial Rate (10–3 M/s) 0.4 0.4 0.2 160 0.2 0.4 0.4 80 0.6 0.1 0.2 15 0.2 0.1 0.2 5 0.2 0.2 0.4 20 Using the initial-rate method, what is the order of the reaction with respect to C? a. zero-order b. first-order c. third-order d. second-order e. impossible to tell from the data given

Answers

The dependence of the power of the reaction rate on the concentration is called the order of the reaction. The order of the reaction is the first order.

What is the initial-rate method?

The initial rate method is the estimation of the order of the reaction by the initial rates of the reactants and products and by performing the reaction several times by measuring the rate.

The reaction is given as,

\rm A + B + C \rightarrow Products

The rate of reaction can be given as:

\rm rate = k[A]^(x)[B]^(y)[C]^(z)

Here the variables x, y and z are orders respective to the reactant concentration and k is the rate constant.

Value of x with respect to A:

\begin{aligned} \rm \frac {Rate 3}{Rate 4} &= \rm [([A(3)])/([A(4)])]^(\rm x)\n\n(15)/(5) &= [([0.6])/([0.2])]^(\rm x)\n\n\rm x &= 1\end{aligned}

Value of y with respect to B:

\begin{aligned}\rm \frac {Rate 2}{Rate 5} &= \rm [([B(2)])/([B(5)])]^(\rm y)\n\n(80)/(20) &= [([0.4])/([0.2])]^(\rm y)\n\n\rm y &= 2\end{aligned}

Value of z  with respect to C:

\rm \frac {Rate 1}{Rate 2} &= [([A(1)])/([A(2)])]^(x) [([B(1)])/([B(2)])]^(y) [([C(1)])/([C(2)])]^(z)

Substituting value of x = 1 and y = 2 in the above equation:

\begin{aligned}(160)/(80) &= [([0.4])/([0.2])]^(1)[([0.4])/([0.4])]^(2) [([0.2])/([0.4])]^(\rm z)\n\n1 &= (0.5)^(\rm z)\n\n&= 1\end{aligned}

Therefore option b. with respect to C = 1, the order of the reaction is first-order.

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Answer:

B. First order, Order with respect to C = 1

Explanation:

The given kinetic data is as follows:

A + B + C → Products

     [A]₀     [B]₀    [C]₀       Initial Rate (10⁻³ M/s)

1.   0.4      0.4     0.2       160

2.  0.2      0.4      0.4       80

3.   0.6     0.1       0.2       15

4.   0.2     0.1       0.2        5

5.   0.2     0.2      0.4       20

The rate of the above reaction is given as:

Rate = k[A]^(x)[B]^(y)[C]^(z)

where x, y and z are the order with respect to A, B and C respectively.

k = rate constant

[A], [B], [C] are the concentrations

In the method of initial rates, the given reaction is run multiple times. The order with respect to a particular reactant is deduced by keeping the concentrations of the remaining reactants constant and measuring the rates. The ratio of the rates from the two runs gives the order relative to that reactant.

Order w.r.t A : Use trials 3 and 4

(Rate3)/(Rate4)= [([A(3)])/([A(4)])]^(x)

(15)/(5)= [([0.6])/([0.2])]^(x)

3 = 3^(x) \n\nx =1

Order w.r.t B : Use trials 2 and 5

(Rate2)/(Rate5)= [([B(2)])/([B(5)])]^(y)

(80)/(20)= [([0.4])/([0.2])]^(y)

4 = 2^(y) \n\ny =2

Order w.r.t C : Use trials 1 and 2

(Rate1)/(Rate2)= [([A(1)])/([A(2)])]^(x)[([B(1)])/([B(2)])]^(y)[([C(1)])/([C(2)])]^(z)

we know that x = 1 and y = 2, substituting the appropriate values in the above equation gives:

(160)/(80)= [([0.4])/([0.2])]^(1)[([0.4])/([0.4])]^(2)[([0.2])/([0.4])]^(z)

1 = (0.5)^(z)

z = 1

Therefore, order w.r.t C = 1

A eraser has a mass of 4g and a volume of 2cm3 what is it’s density

Answers

Answer:

The answer is 2.0 g/cm³

Explanation:

The density of a substance can be found by using the formula

density = (mass)/(volume) \n

From the question

mass = 4 g

volume = 2 cm³

We have

density = (4)/(2) \n

We have the final answer as

2.0 g/cm³

Hope this helps you

Identify two ions that have the following ground-state electron configurations Part B
[Ar]3d^5
Check all that apply.
A- Fe2+
B- Fe3+
C- Mn2+
D- V+
E- Sc2+

Answers

Answer: Fe^(2+):24:[Ar]3d^5

Mn^(2+):23:[Ar]3d^5

Explanation:

Electronic configuration represents the total number of electrons that a neutral element contains. We add all the superscripts to know the number of electrons in an atom.  The electrons are filled according to Afbau's rule in order of increasing energies.

The electronic configuration for given elements is as follows:

Fe:26:[Ar]3d^64s^2

Fe^(2+):24:[Ar]3d^5

Fe^(3+):23:[Ar]3d^4

Mn:25:[Ar]3d^54s^2

Mn^(2+):23:[Ar]3d^5

V:23:[Ar]3d^34s^2

V^+:22:[Ar]3d^34s^1

Sc:21:[Ar]3d^14s^2

Sc^(2+):19:[Ar]3d^1

Final answer:

The ions Fe2+ and Mn2+ have the ground-state electron configuration [Ar]3d^5.

Explanation:

The ground-state electron configuration [Ar]3d^5 indicates a level of electrons in 3d subshell after the Argon core electron configuration. Now, iron (Fe) has a base atomic configuration of [Ar]3d^6 4s2. When it loses 2 electrons (to form Fe2+), it tends to lose from both the 3d and the 4s sublevels, giving [Ar]3d^5 (which is our required configuration).

However, it's also important to consider Manganese (Mn), which has a base configuration of [Ar]3d^5 4s2. It usually loses 2 electrons from the 4s sublevel first when it forms Mn2+ which results in a configuration [Ar]3d^5.

So, the two ions with the electron configuration [Ar]3d^5 are Fe2+ and Mn2+.

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Construction crews sometimes use this reaction for welding underwater structures:Fe2O3 + 2Al → Al2O3 + 2Fe

How many moles of iron (Fe) would be produced if 2.50 mol Fe2O3 react? Make sure to use the correct number of significant figures in your answer.

2.50 mol Fe2O3 =

Answers

Answer:

5 moles of iron formed

Explanation:

Given data:

Moles of iron formed = ?

Moles of iron oxide react = 2.50 mol

Solution:

Chemical equation:

Fe₂O₃ + 2Al     →     Al₂O₃ + 2Fe

Now we will compare the moles of iron with iron oxide.

                  Fe₂O₃        :           Fe

                      1             ;            2

                  2.50          :          2×2.50 = 5 mol    
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