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2 years ago

Use the periodic to fill in the numbers in the electron configurations shown below. B: 1s2 2sA2pB

Chemistry

2 answers:

Inga [223]2 years ago

8 0

Explanation:

Atomic number of boron is 5 and its electronic configuration is $1s^{2}2s^{2}2p^{1}$ . Boron is a non-metal and belongs to group 13.

That is, 2s shell of boron atom contains 2 electrons and 2p shell of the atom has 1 electron.

Therefore, we can conclude that the value of A equals 2 and value of B equals 1 in B: 1s2 2sA2pB.

Bezzdna [24]2 years ago

6 0

The answer is

A= 2

B= 1

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Sulfurous acid, h2so3, breaks down into water (h2o) and sulfur dioxide (so2). if only one molecule of sulfurous acid was involve

devlian [24]

Explanation:

The given reaction equation will be as follows.

$H_{2}SO_{3} \rightarrow H_{2}O + SO_{2}$

Now, number of atoms on reactant side are as follows.

H = 2

S = 1

O = 3

Number of atoms on product side are as follows.

H = 2

S = 1

O = 3

Therefore, this equation is balanced since atoms on both reactant and product sides are equal.

Thus, we can conclude that there is one sulfur atom in the products.

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2 years ago

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A. The measured pH of a 0.100 M HCl solution at 25 degrees Celsius is 1.092. From this information, calculate the activity coeff

ra1l [238]

Answer:

activity coefficient $\mathbf{\gamma =0.809}$

activity coefficient $\mathbf{\gamma = 0.791}$

The change in pH in part A = 0.092

The change in pH in part B = 0.102

Explanation:

From the given information:

pH of HCl solution = 1.092

Activity of the pH solution [a] = $10^{-1.092}$

[a] = 0.0809 M

Recall that [a] = $\gamma$ × C

where;

$\gamma$ = activity coefficient

C = concentration

Making the activity coefficient the subject of the formula, we have:

$\gamma = \dfrac{[a]}{C}$

$\gamma = \dfrac{0.0809 \ M}{0.100 \ M}$

$\mathbf{\gamma =0.809}$

The pH of a solution of HCl and KCl = 2.102

[a] = $10^{-2.102}$

[a] = 0.00791 M

activity coefficient $\gamma = \dfrac{0.00791 \ M}{0.01 \ M}$

$\mathbf{\gamma = 0.791}$

C. The change in pH in part A = 1.091 - 1.0 = 0.092

The change in pH in part B = 2.102 -2.00 = 0.102

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2 years ago

Determine Z and V for steam at 250°C and 1800 kPa by the following: (a) The truncated virial equation [Eq. (3.38)] with the foll

makvit [3.9K]

Answer:

Explanation:

Given that:

the temperature $T_1$ = 250 °C= ( 250+ 273.15 ) K = 523.15 K

Pressure = 1800 kPa

The truncated viral equation is expressed as:

$\frac{PV}{RT} = 1 + \frac{B}{V} + \frac{C}{V^2}$

where; B = - $152.5 \ cm^3 /mol$ C = -5800 $cm^6/mol^2$

R = 8.314 × 10³ cm³ kPa. K⁻¹.mol⁻¹

Plugging all our values; we have

$\frac{1800*V}{8.314*10^3*523.15} = 1+ \frac{-152.5}{V} + \frac{-5800}{V^2}$

$4.138*10^{-4} \ V= 1+ \frac{-152.5}{V} + \frac{-5800}{V^2}$

Multiplying through with V² ; we have

$4.138*10^4 \ V ^3 = V^2 - 152.5 V - 5800 = 0$

$4.138*10^4 \ V ^3 - V^2 + 152.5 V + 5800 = 0$

V = 2250.06 cm³ mol⁻¹

Z = $\frac{PV}{RT}$

Z = $\frac{1800*2250.06}{8.314*10^3*523.15}$

Z = 0.931

b) The truncated virial equation [Eq. (3.36)], with a value of B from the generalized Pitzer correlation [Eqs. (3.58)–(3.62)].

The generalized Pitzer correlation is :

$T_c = 647.1 \ K \\ \\ P_c = 22055 \ kPa \\ \\ \omega = 0.345$

$T__{\gamma}} = \frac{T}{T_c}$

$T__{\gamma}} = \frac{523.15}{647.1}$

$T__{\gamma}} = 0.808$

$P__{\gamma}} = \frac{P}{P_c}$

$P__{\gamma}} = \frac{1800}{22055}$

$P__{\gamma}} = 0.0816$

$B_o = 0.083 - \frac{0.422}{T__{\gamma}}^{1.6}}$

$B_o = 0.083 - \frac{0.422}{0.808^{1.6}}$

$B_o = 0.51$

$B_1 = 0.139 - \frac{0.172}{T__{\gamma}}^{ \ 4.2}}$

$B_1 = -0.282$

The compressibility is calculated as:

$Z = 1+ (B_o + \omega B_1 ) \frac{P__{\gamma}}{T__{\gamma}}$

$Z = 1+ (-0.51 +(0.345* - 0.282) ) \frac{0.0816}{0.808}$

Z = 0.9386

$V= \frac{ZRT}{P}$

$V= \frac{0.9386*8.314*10^3*523.15}{1800}$

V = 2268.01 cm³ mol⁻¹

c) From the steam tables (App. E).

At $T_1 = 523.15 \ K \ and \ P = 1800 \ k Pa$

V = 0.1249 m³/ kg

M (molecular weight) = 18.015 gm/mol

V = 0.1249 × 10³ × 18.015

V = 2250.07 cm³/mol⁻¹

R = 729.77 J/kg.K

Z = $\frac{PV}{RT}$

Z = $\frac{1800*10^3 *0.1249}{729.77*523.15}$

Z = 0.588

3 0

2 years ago

A student adds solid KCl to water in a flask. The flask is sealed with a stopper and thoroughly shaken until no more solid KCl d

AlladinOne [14]

Answer:

Option (A) saturated and is at equilibrium with the solid KCl

Explanation:

A saturated solution is a solution which can not dissolve more solute in the solution.

From the question given above, we can see that the solution is saturated as it can not further dissolve any more KCl as some KCl is still visible in the flask.

Equilibrium is attained in a chemical reaction when there is no observable change in the reaction system with time. Now, observing the question given we can see that there is no change in flask as some KCl is still visible even after thorough shaking. This simply implies that the solution is in equilibrium with the KCl solid as no further dissolution occurs.

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2 years ago

How does 0.5 m sucrose (molecular mass 342) solution compare to 0.5 m glucose (molecular mass 180) solution?

mash [69]

Answer : Both solutions contain $3.011 X 10^{23}$ molecules.

Explanation : The number of molecules of 0.5 M of sucrose is equal to the number of molecules in 0.5 M of glucose. Both solutions contain $3.011 x 10^{23}$ molecules.

Avogadro's Number is $N_{A}$ = $6.022 X 10^{23}$ which represents particles per mole and particles may be typically molecules, atoms, ions, electrons, etc.

Here, only molarity values are given; where molarity is a measurement of concentration in terms of moles of the solute per liter of solvent.

Since each substance has the same concentration, 0.5 M, each will have the same number of molecules present per liter of solution.

Addition of molar mass for individual substance is not needed. As if both are considered in 1 Liter they would have same moles which is 0.5.

We can calculate the number of molecules for each;

Number of molecules = $N_{A} X M$ ;

∴ Number of molecules = $6.022 X 10^{23} X 0.5 mol/L X 1 L$ which will be = $3.011 X 10^{23}$

Thus, these solutions compare to each other in that they have not only the same concentration, but they will have the same number of solvated sugar molecules. But the mass of glucose dissolved will be less than the mass of sucrose.

7 0

2 years ago

Read 2 more answers