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ratelena [41]
2 years ago
8

GUYS HELP ME PLEASE. MY TEACHER IS GONNA KILL ME IF I DONT SOLVE THIS RIGHT. Write the election configuration (H, Al, S,

Chemistry
1 answer:
ohaa [14]2 years ago
6 0

Explanation:

<em>1s 2s 2p</em>

<em>1s 2s 2p1s2 2s1 </em>

<em>1s 2s 2p1s2 2s1 2 1 1</em>

<em>1s 2s 2p1s2 2s1 2 1 11s2 2s2 </em>

You might be interested in
The total concentration of ions in a 0.75 M solution of HCl is
EastWind [94]
<h3>Answer:</h3>

The total concentration of ions in a 0.75 M solution of HCl is 1.5 M

That is; 0.75 M H⁺ and 0.75 M Cl⁻

<h3>Explanation:</h3>
  • Concentration or molarity is the number of moles of a compound or an ion contained in one liter of solution. It is measured in moles per liter (M).
  • The concentration of ions making a compound is determined by the ratio of moles of the compound and the constituents ions.
  • For instance, HCl dissociates to give H⁺ and Cl⁻

       HCl(aq) → H⁺(aq) + Cl⁻(aq)

  • Therefore, since the mole ratio between HCl and the constituent ions H⁺ and Cl⁻ is 1:1, then 0.75 M of HCl dissociates to give 0.75 M H⁺ and 0.75 m Cl⁻
  • Hence the total concentration of ions in a 0.75 M solution of HCl is 1.5 M (0.75 M H⁺ and 0.75 M Cl⁻)
4 0
3 years ago
Potassium iodide reacts with lead(II) nitrate in this precipitation reaction: 2 KI(aq) + Pb(NO3)2(aq) → 2 KNO3(aq) + PbI2(s) Wha
andrew11 [14]

Answer:

a. 174 mL

Explanation:

Let's consider the following reaction.

2 KI(aq) + Pb(NO₃)₂(aq) → 2 KNO₃(aq) + PbI₂(s)

We have 155.0 mL of a 0.112 M lead(II) nitrate solution. The moles of Pb(NO₃)₂ are:

0.1550 L × 0.112 mol/L = 0.0174 mol

The molar ratio of KI to Pb(NO₃)₂ is 2:1. The moles of KI are:

2 × 0.0174 mol = 0.0348 mol

The volume of a 0.200 M KI solution that contains 0.0348 moles is:

0.0348 mol × (1 L / 0.200 mol) = 0.174 L = 174 mL

5 0
2 years ago
When the following oxidation–reduction reaction in acidic solution is balanced, what is the lowest whole-number coefficient for
Svet_ta [14]

Answer:

c. 8, product side

Explanation:

In order to balance a redox reaction we use the ion-electron method, which has the following steps:

Step 1: identify oxidation and reduction half-reaction.

Oxidation: MnO₄⁻(aq) → Mn²⁺(aq)

Reduction: Br⁻(aq) → Br₂(l)

Step 2: perform the mass balance adding H⁺ and H₂O where necessary

8 H⁺(aq) + MnO₄⁻(aq) → Mn²⁺(aq) + 4 H₂O(l)

2 Br⁻(aq) → Br₂(l)

Step 3: perform the electrical balance adding electrons where necessary.

8 H⁺(aq) + MnO₄⁻(aq) + 5 e⁻ → Mn²⁺(aq) + 4 H₂O(l)

2 Br⁻(aq) → Br₂(l) + 2 e⁻

Step 4: multiply both half-reactions by numbers that secure that the number of electrons gained and lost are the same.

2 × (8 H⁺(aq) + MnO₄⁻(aq) + 5 e⁻ → Mn²⁺(aq) + 4 H₂O(l))

5 × (2 Br⁻(aq) → Br₂(l) + 2 e⁻)

Step 5: add both half-reactions side to side.

16 H⁺(aq) + 2 MnO₄⁻(aq) + 10 e⁻ + 10 Br⁻(aq) → 2 Mn²⁺(aq) + 8 H₂O(l) + 5 Br₂(l) + 10 e⁻

16 H⁺(aq) + 2 MnO₄⁻(aq) + 10 Br⁻(aq) → 2 Mn²⁺(aq) + 8 H₂O(l) + 5 Br₂(l)

3 0
2 years ago
Perform the following
vivado [14]
The correct number of significant figures and digits is 3
5 0
3 years ago
Combustion of hydrocarbons such as dodecane (C12H26) produces carbon dioxide, a "greenhouse gas." Greenhouse gases in the Earth'
miss Akunina [59]

Answer:

A. 2C12H26(l) + 37O2(g) —> 24CO2(g) + 26H2O(g)

B. 761.42 L

Explanation:

A. Step 1:

The equation for the reaction.

C12H26(l) + O2(g) —> CO2(g) + H2O(g)

A. Step 2:

Balancing the equation.

The equation can be balance as follow:

C12H26(l) + O2(g) —> CO2(g) + H2O(g)

There are 12 atoms of C on the left side and 1 atom on the right side. It can be balance by putting 12 in front of CO2 as illustrated below:

C12H26(l) + O2(g) —> 12CO2(g) + H2O(g)

There are 26 atoms of H on the left side and 2 atoms on the right side. It can be balance by putting 13 in front of H2O as illustrated below:

C12H26(l) + O2(g) —> 12CO2(g) + 13H2O(g)

Now, there are a total of 37 atoms of O2 on the right side and 2 atoms on the left. It can be balance by putting 37/2 in front of O2 as illustrated below:

C12H26(l) + 37/2O2(g) —> 12CO2(g) + 13H2O(g)

Multiply through by 2 to clear the fraction from the equation.

2C12H26(l) + 37O2(g) —> 24CO2(g) + 26H2O(g)

Now the equation is balanced

B. Step 1:

We'll by obtaining the number of mole of C12H26 in 0.450 kg of C12H26. This is illustrated below:

Molar Mass of C12H26 = (12x12) + (26x1) = 144 + 26 = 170g/mol

Mass of C12H26 = 0.450 kg = 0.450x1000 = 450g

Number of mole of C12H26 =?

Number of mole = Mass/Molar Mass

Number of mole of C12H26 = 450/170

Number of mole of C12H26 = 2.65 moles

B. Step 2:

Determination of the number of mole of CO2 produced by the reaction. This is illustrated below:

2C12H26(l) + 37O2(g) —> 24CO2(g) + 26H2O(g)

From the balanced equation above,

2 moles of C12H26 produced 24 moles of CO2.

Therefore, 2.65 moles of C12H26 will produce = (2.65x24)/2 = 31.8 moles of CO2.

B. Step 3:

Determination of the volume of CO2 produced by the reaction.

Pressure (P) = 1 atm

Temperature (T) = 19°C = 19°C + 273 = 292K

Gas constant (R) = 0.082atm.L/Kmol

Number of mole (n) = 31.8 moles

Volume (V) =?

The volume of CO2 produced by the reaction can b obtained by applying the ideal gas equation as follow:

PV = nRT

1 x V = 31.8 x 0.082 x 292

V = 761.42 L

Therefore, the volume of CO2 produced is 761.42 L

5 0
3 years ago
Read 2 more answers
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