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Anna007 [38]
3 years ago
5

A sample of a compound contains only the elements sodium, sulfur, and oxygen. it is found by analysis to contain 0.979 g na, 1.3

65 g s, and 1.021 g o. determine its empirical formula.
Chemistry
2 answers:
Aloiza [94]3 years ago
4 0
To determine the empirical formula for the compound that contains <span>0.979 g Na, 1.365 g S, and 1.021 g O, we convert these to mole units. The molar masses to be used are:

Molar mass of Na = 23 g/mol
</span>Molar mass of S = 32 g/mol
Molar mass of O = 16 g/ mol

The number of moles is obtained using the molar mass for each element.

moles Na = 0.979 g Na/ 23 g/mol Na = 0.04256
moles S = 1.365 g Na/ 32 g/mol Na = 0.04265
moles O = 1.021 g O/ 16 g/mol Na = 0.06326

We then divide each with the smallest number of moles obtained. 

Na: 0.04256/ 0.04256 = 1 
S: 0.04265/ 0.04256 = 1.002 ≈ 1
O: 0.06326/ 0.04256 = 1.49 ≈ 1.5

We then have an empirical formula of NaSO₁.₅. However, chemical formulas must have only integers as subscripts, thus, we multiply each to 2. The empirical formula is then Na₂S₂O₃ also known as sodium thiosulfate.
Mumz [18]3 years ago
3 0
First, convert the masses into moles using the molar weights: 23 g Na/mol, 32.06 g S/mol and 16 g O/mol.

Mol Na: 0.979/23 = 0.04256522
Mol S: 1.365/32.06 = 0.04257642
Mol O: 1.021/16 = 0.0638125

The least value is mol Na. Thus, divide all mole compositions to that of mol Na:
Mol Na: 0.04256522/0.04256522 = 1
Mol S: 0.04257642/0.04256522 = 1
Mol O: 0.0638125/0.04256522 = 1.5

The ratios must be whole numbers. To make 1.5 a whole number, multiply all ratios with 2 to keep it uniform.
Mol Na: 1*2 = 2
Mol S: 1*2 = 2
Mol O: 1.5*2 = 3

<em>Thus, the empirical formula is Na₂S₂O₃</em>.
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maria [59]

Answer:

3.24

Explanation:

The dissociation equation for the carboxylic acid can be represented as follows:

RCOOH —-> RCOO- + H+

We can use an ICE table to get the value of the concentration of the hydrogen ion. ICE stands for initial, change and equilibrium.

RCOOH RCOO- H+

Initial 0.2 0.0. 0.0

Change -x +x. +x

Equilibrium 0.2-x. x. x

We can now find the value of x as follows:

Ka = [RCOO-][H+]/[RCOOH]

(1.66* 10^-6) = (x * x)/(0.2-x)

(1.66 * 10^-6) (0.2-x) = x^2

x^2 = (3.32* 10^-7) - (1.66*10^-6)x

x^2 + (1.66 * 10^-6)x - (3.32* 10^-7) = 0

Solving the quadratic equation to get x:

x = 0.0005753650094369094 or - 0.0005753650094369094

As concentration cannot be negative, we discard the negative answer

Hence [H+] = 0.0005753650094369094

By definition, pH = -log[H+]

pH = -log(0.0005753650094369094)

pH = 3.24

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3 years ago
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When 0.50 L of a 12 M solution is diluted to 1.0 L, what is the resulting molarity?
Lyrx [107]

Answer:

The resulting molarity is 6M.

Explanation:

A dilution consists of the decrease of concentration of a substance in a solution (the higher the volume of the solvent, the lower the concentration).

We use the formula for dilutions:

C1 x V1 = C2 x V2

12 M x 0,5L = C2 x 1,0 L

C2= (12 M x 0,5 L)/1,0 L

<em>C2= 6 M</em>

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How many grams of P4O10 (292.88 g/mol) form when phelpsphorous (P4, 125.52 g/mol) reacts with 16.2 L of O2 (33.472 g/mol) ) at s
nevsk [136]

Answer:

40.5 g of P₄O₁₀ are produced

Explanation:

We state the reaction:

P₄ + 5O₂ → P₄O₁₀

We do not have data from P₄ so we assume, it's the excess reactant.

We need to determine mass of oxygen and we only have volumne so we need to apply density.

Density = mass / volume, so Mass = density . volume

Denstiy of oxygen at STP is: 1.429 g/L

1.429 g/L . 16.2L = 23.15 g

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5 moles of O₂ can produce 1 mol of P₄O₁₀

Our 0.692 moles may produce (0.692 . 1)/ 5 = 0.138 moles

We determine the mass of product:

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