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MrRa [10]
3 years ago
8

Is the enthalpy change when 1 mole of a substance undergoes complete

Chemistry
1 answer:
ipn [44]3 years ago
7 0

Answer:

Explanation:

Standard enthalpy of combustion (ΔH∘C) is the enthalpy change when 1 mole of a substance burns under standard state conditions;

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How many moles of sodium acetate (nach3coo) must be added to 1.000 liter of a 0.500 m solution of acetic acid (ch3cooh) to produ
Alenkinab [10]
When PH = -㏒[H3O+] 
and we have PH = 5.061 
by substitution:
∴ [ H3O+] = 10^-5.061 = 8.7x10^-6
when we have Ka = [CH3COO-][H3O+] / [CH3COOH]
we have Ka = 1.8x10^-5 & [H3O+] = 8.7x10^-6m & [CH3COOH] = 0.5 m 
by substitution in Ka formula:
1.8x10^-5 = [CH3COO-]*(8.7x10^-6) / 0.5
∴[CH3COO-] = 1.034
∴we need  to add 1.034 mol of sodium acetate
4 0
3 years ago
What are the most of the elements in the periodic table
vaieri [72.5K]

Answer:

Most of the elements in the periodic table are <u><em>metals</em></u>.

Hope that helps. x

5 0
2 years ago
Read 2 more answers
What is the oxidation number for zinc? How does zinc's electron configuration account for this oxidation number?
IRINA_888 [86]
+2 is the oxidation number for Zinc
<span>Oxidation Number for a Transition metal = Number of Unpaired d-electrons + Two s-electrons  
</span>[Ar]4s2 3d10 is the electron configuration for zinc. The d-block is full with 10 electrons meaning there are no unpaired d-electrons, so the oxidation number is 0 unpaired d-electrons + 2 s-electrons = +2
8 0
3 years ago
Practice Problem #1: __TiO2 + __Cl2 + __C __TiCl4 + __CO2 + __CO
ICE Princess25 [194]

Answer:

Explanation:

1)

Given data:

Moles of Cl₂ react = ?

Moles of carbon = 4.55 mol

Solution:

Chemical equation:

2TiO₂ + 4Cl₂ + 3C  → 2TiCl₄ + CO₂ + 2CO

Now we will compare the moles of Cl₂ with C.

                         C             :            Cl₂

                         3             :              4

                       4.55         :            4/3×4.55 = 6.1 mol

6.1 moles of chlorine will react with 4.55 moles of C.

2)

Given data:

Mass of TiO₂ react = ?

Moles of carbon = 4.55 mol

Solution:

Chemical equation:

2TiO₂ + 4Cl₂ + 3C  → 2TiCl₄ + CO₂ + 2CO

Now we will compare the moles of Cl₂ with C.

                         C             :            TiO₂

                         3             :              2

                       4.55         :            2/3×4.55 = 3 mol

Mass of TiO₂:

Mass = number of moles × molar mass

Mass = 3 mol × 79.87 g/mol

Mass =  239.61 g

3)

Given data:

Molecules of TiCl₄ formed = ?

Moles of TiO₂ react = 115 g

Solution:

Chemical equation:

2TiO₂ + 4Cl₂ + 3C  → 2TiCl₄ + CO₂ + 2CO

Number of moles of TiO₂:

Number of moles = mass/ molar mass

Number of moles = 115 g/ 79.87 g/mol

Number of moles = 1.44 mol

Now we will compare the moles of Cl₂ with C.

                      TiO₂           :            TiCl₄

                         2             :              2

                       1.44           :            1.44

Molecules of TiCl₄:

1 mole = 6.022 × 10²³ molecules

1.44 ×6.022 × 10²³ molecules

8.672× 10²³ molecules

8 0
3 years ago
Suppose 231.8 mgmg of PbCl2PbCl2 was added to 15.0 mLmL of water in a flask, and the solution was allowed to reach equilibrium a
vovangra [49]

Answer:

ksp = 2.2 x ⁻⁴

Explanation:

The equilibrium here is:

PbCl₂ (s)     ⇄ Pb²⁺ + 2 Cl⁻

we can recognize it as a product solubilty equilibrium once we are told that some undissolved PbCl₂ remained.

The equilibrium constant, Ksp is given by the equation

Ksp = [Pb²⁺][Cl⁻]²

where [Pb²⁺] and [Cl⁻]² are the concentrations (M) of Pb²⁺ and Cl⁻ in solution.

we have the mass of solid PbCl₂ placed in solution, so we can determine the number of moles it represents, and if  we  substract the moles of undissolved PbCl₂ we will know the moles of Pb²⁺ and Cl⁻ which went into solution.

From there we can calculate the molarity (M= moles/L solution) and finally plug the values into our expression for Ksp to answer this question.

molar mas PbCl₂ = 278.1 g/mol

1 milligram = 1 x 10⁻³ g

mol PbCl₂ initially = 231.8 x 10⁻³ g / 278.1 mol = 8.3 x 10⁻⁴ mol

Volume solution = 15 mL x 1L / 1000 mL = 0.015 L

mol undissolved PbCl₂ = 74 x 10⁻³ g / 278.1 g/mol = 2.7 x 10⁻⁴ mol

mol PbCl dissolved =   8.3 x 10⁻⁴ mol -  2.7 x 10⁻⁴ mol = 5.7 x 10⁻⁴ mol

Concentration of Pb²⁺ in solution = 5.7 x 10⁻⁴ mol / 0.015 L = 3.8 x 10⁻² M

Concentration of Cl⁻ in solution = 2 x 3.8 x 10⁻² M = 7.6 x 10⁻² M

(Note from the formula we we get 2 mol Cl⁻ per mol PbCl₂)

Plugging these values into the expression for Ksp we have

Ksp = 3.8 x 10⁻² x (7.6 x 10⁻²)² = 2.2 x 10⁻⁴

8 0
3 years ago
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