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An element x is found to have a mass number of 31 and atomic number of 17. idenrify the group and the period to which it belongs?​

Answers

Answer 1
Answer:

Answer:

Well atomic number 17 is Chlorine, which is most commonly found as a gas, and is period 7.

Explanation:

elements found on period 7 are some of the most unstable elements.

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Of the following elements, which one has the lowest first ionization energy?boron carbon aluminum silicon

Answers

Answer:

Boron

Explanation:

Because it has a complete 2s orbital and therefore, an increased shielding of the 2s orbital will reduce the ionisation energy.

Final answer:

Among boron, carbon, aluminum, and silicon, aluminum has the lowest first ionization energy due to its position on the periodic table, which is further to the left and in a higher period than the other elements.

Explanation:

Ionization energy refers to the energy necessary to remove an electron from an atom in its gaseous state. The element with the lowest first ionization energy among boron, carbon, aluminum, and silicon is aluminum. Ionization energy increases from left to right across a period in the periodic table and from bottom to top in a group. Thus, aluminum, being to the left of boron, carbon, and silicon, has the lowest first ionization energy. Furthermore, aluminum is in the third period, which is below boron and carbon's second period.

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Which of the following is true for all exergonic reactions? The reaction releases energy. A net input of energy from the surroundings is required for the reactions to proceed. The reactions are rapid. The products have more total energy than the reactants. The reaction goes only in a forward direction: all reactants will be converted to products, but no products will be converted to reactants.

Answers

Answer:

The reaction releases energy

Explanation:

The products of an exergonic reaction have a lower energy state (Delta-G) compared to the reactants. Therefore there is a negative delta –G between products and reactants after the reactions. This means some energy is lost into the environment usually through light or heat.

Final answer:

Exergonic reactions are characterized by a net release of energy but they still require a small initial energy input to start, referred to as the 'activation energy'. The speed or direction of the reaction is not determined by whether it's exergonic.

Explanation:

In the context of chemical reactions, the true statement for all exergonic reactions is that such reactions result in a net release of energy. However, even exergonic reactions, which are characterized by energy release, require a small initial input of energy to get started. This initial energy demand is referred to as the 'activation energy'. Also, it's important to note that the speed of the reaction or its directionality (whether it proceeds only in a forward direction) are not inherently determined by whether a reaction is exergonic. These aspects depend on other reaction conditions and catalysis.

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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

The standard entropy of Pb(s) at 298.15 K is 64.80 J K–1 mol–1. Assume that the heat capacity of Pb(s) is given by: CP,m(Pb, s) J K−1mol−1 = 22.13 + 0.01172 T K + 1.00 x 10−5 T 2 K2 The melting point is 327.4 ℃ and the heat of fusion is 4770 J mol-1. Assume that the heat capacity of Pb(l) is given by: CP,m(Pb, l) J K−1mol−1 = 32.51 − 0.00301 T K Calculate the standard entropy of Pb(l) at 500 ℃

Answers

Answer:

s_(Pb(l),500\°C)=100.83(J)/(mol*K)

Explanation:

Hello,

In this case, for the calculation of the standard entropy of liquid lead at 500 °C (773.15 K), starting by solid lead 298.15 K we need to consider three processes:

1. Heating of solid lead at 298.15 K to 600.55 K (melting point).

2. Melting of solid lead to liquid lead.

3. Heating of liquid lead at 600.55 K (melting point) to 773.15 K.

Which can be written in terms of entropy by:

s_(Pb(l),500\°C)=s_(Pb(s),298.15K)+s_1+s_2+s_3

Whereas each entropy is computed as follows:

s_1=\int\limits^(600.55K)_(298.15K) {(22.13 + 0.01172 T + 1.00 x 10^(-5) T^2)/(T) } \, dT =20.4(J)/(mol*K)\n\n\ns_2=(4770(J)/(mol) )/(600.55K)= 7.94(J)/(mol*K)\n\n\ns_3=\int\limits^(773.15K)_(600.55K) {(32.51-0.00301T)/(T) } \, dT=7.69(J)/(mol*K)

Therefore, the standard entropy of liquid lead at 500 °C turns out:

s_(Pb(l),500\°C)=64.80+20.4+7.94+7.69\n\ns_(Pb(l),500\°C)=100.83(J)/(mol*K)

Best regards.

You just measured a metal cylinder and obtained the following information: mass - 3.543 g diameter -0.53 cm height = 4.40 cm. 265 • Determine the volume (V). (V=nrºh, where r = radius, h = height, T - 3.14) V= (3.14 36205² 4.402.17cm? • Determine density of the cylinder D. 3. SMS-365923 197

Answers

The density of the cylinder would be 3.652 gram/ cm³

.

What is density?

It can be defined as the mass of any object or body per unit volume of the particular object or body. Generally, it is expressed as in gram per cm³ or kilogram per meter³.

As given in the problem, you just measured a metal cylinder and obtained the following information: mass - 3.543 grams, diameter 0.53 cm, height = 4.40 cm , and we have to calculate the density of the cylinder,

mass of the cylinder = 3.543 grams

the volume of the cylinder = πr²h

                                            = 3.14 ×.265²×4.4

                                            =0.97 cm³

By using the above formula for density

ρ = mass of the cylinder/volume of the cylinder

  = 3.543 grams/0.97 cm³

  =3.652 grams/ cm³

Thus,the density of the cylinder would be 3.652 grams/ cm³.

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Answer:

V cylinder = \pi r^2 h = \pi(0.53 cm/2)^24.4cm= 0.97 cm^3

note that r=diameter/2

density = m/V= 3.543/0.97= 3.65 g/cm^3

If the mass percentage composition of a compound is 72.1% Mn and 27.9% O, its empirical formula is

Answers

Answer:

MnO- Manganese Oxide

Explanation:

Empirical formula: This is the formula that shows the ratio of elements

present in a  

compound.

   

How to determine Empirical formula

1. First arrange the symbols of the elements present in the compound

alphabetically to  determine the real empirical formula. Although, there

are exceptions to this rule, E.g H2So4

2. Divide the percentage composition by the mass number.

3. Then divide through by the smallest number.

4. The resulting answer is the ratio attached to the elements present in

a compound.

           

                                                                              Mn                         O    

                         

% composition                                                      72.1                      27.9    

                       

Divide by mass number                                       54.94                     16  

                                 

                                                                               1.31                      1.74    

                       

Divide by the smallest number                         1.31                      1.31                          

                                                                               1                    1.3

                                                 

The resulting ratio is 1:1

 

Hence the Empirical formula is MnO, Manganese oxide
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