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3 years ago

The specific heat of a given substance

1 answer:

7 0

The specific heat is the amount of heat per unit mass required to raise the temperature to 1 degree Celsius. (This is from google)

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Calculate molecules in 1dm^3 of oxygen

Archy [21]

Number of molecules in 1 dm³ Oxygen = 2.71 x 10²²

<h3>Further explanation</h3>

Conditions at T 0 ° C and P 1 atm are stated by STP (Standard Temperature and Pressure). At STP, Vm is 22.4 liters/mol.

The mole is the number of particles contained in a substance

1 mol = 6.02.10²³

1 dm³ of oxygen = 1 L Oxygen

mol Oxygen :

$\tt \dfrac{1}{22.4}=0.045`mol$

molecules of Oxygen :

n=mol=0.045

No = 6.02.10²³

$\tt N=n\times No\\\\N=0.045\times 6.02\times 10^{23}\\\\N=2.71\times 10^{22}$

5 0

3 years ago

Lavoisier's discovery that air was not a substance but rather a mixture of nitrogen and oxygen is described as history-making. w

Taya2010 [7]

Antoine Lavoisier was part of a wealthy family in Paris. He pursued to study science upon realizing that chemistry or the study of the elements was not a well-studied field. His discovery of that air was a mixture of nitrogen and oxygen gave rise to the concept of COMBUSTION after repeating the experiments made by Priestly using mercury and other metal oxides.

The event was such a history-making because it disproved that concept that air was a pure substance along with 3 others: earth, fire, and water.

6 0

3 years ago

Help one number 46 please

Juli2301 [7.4K]

Say the dereference between the two is that "Exo" means "outside" and "Endo" means "inside" meaning one pulls heat inside (endothermic) and the other expels heat (exothermic)

3 0

3 years ago

Calculate the concentration of oxalate ion (c2o42–) in a 0.175 m solution of oxalic acid (c2h2o4). [for oxalic acid, ka1 = 6.5 ×

Darina [25.2K]

C2H2O4 C2HO4- + H+

0.175 - x x x + y

C2HO4- c2o42– + H+

x - y y x + y

K2 = (y) (x +y) / (x-y)

As, y << x

So, K2 = (y) (x) / (x)

K2 = y =6.1 × 10^–5

Hence, concentration of (c2o42–) 6.1 × 10^–5 M

7 0

4 years ago

PbSO4 has a Ksp = 1.3 * 10-8 (mol/L)2.

Oduvanchick [21]

i. The dissolution of PbSO₄ in water entails its ionizing into its constituent ions:

$\mathrm{PbSO_{4}}(aq) \rightleftharpoons \mathrm{Pb^{2+}}(aq)+\mathrm{SO_4^{2-}}(aq).$

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ii. Given the dissolution of some substance

$xA{(s)} \rightleftharpoons yB{(aq)} + zC{(aq)}$ ,

the Ksp, or the solubility product constant, of the preceding equation takes the general form

$K_{sp} = [B]^y [C]^z$ .

The concentrations of pure solids (like substance A) and liquids are excluded from the equilibrium expression.

So, given our dissociation equation in question i., our Ksp expression would be written as:

$K_{sp} = \mathrm{[Pb^{2+}] [SO_4^{2-}]}$ .

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iii. Presumably, what we're being asked for here is the <em>molar </em>solubility of PbSO4 (at the standard 25 °C, as Ksp is temperature dependent). We have all the information needed to calculate the molar solubility. Since the Ksp tells us the ratio of equilibrium concentrations of PbSO4 in solution, we can consider either [Pb2+] or [SO4^2-] as equivalent to our molar solubility (since the concentration of either ion is the extent to which solid PbSO4 will dissociate or dissolve in water).

We know that Ksp = [Pb2+][SO4^2-], and we are given the value of the Ksp of for PbSO4 as 1.3 × 10⁻⁸. Since the molar ratio between the two ions are the same, we can use an equivalent variable to represent both:

$1.3 \times 10^{-8} = s \times s = s^2 \\s = \sqrt{1.3 \times 10^{-8}} = 1.14 \times 10^{-4} \text{ mol/L}.$

So, the molar solubility of PbSO4 is 1.1 × 10⁻⁴ mol/L. The answer is given to two significant figures since the Ksp is given to two significant figures.

8 0

3 years ago