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sukhopar [10]
2 years ago
13

What is the empirical formula of a compound that contains 32.39 percent sodium, 22.53 percent sulfur, and 45.07 percent oxygen?

Chemistry
1 answer:
djverab [1.8K]2 years ago
5 0
To get the empirical formula of this compound, we take a basis of 100 grams which means each percentage is equivalent to 1 gram. Hence there is 32.39 grams sodium, 22. 53 grams sulfur and 45.07 grams oxygen. We convert each mass to their moles by dividing by their respective molar mass. Na: 1.408, S:0.704 and O:2.82. divide each with the lowest: Na: 2: S: 1 and O:4. Hence the formula is Na2SO4.
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What mass of agno3 can be dissolved in 250g of water at 20°c?
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Hello there,

You should know the <span>solubility of AgNO3 in water at 20°C equals to 2220 g/L.

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An equilibrium mixture contains 0.750 mol of each of the products (carbon dioxide and hydrogen gas) and 0.200 mol of each of the
sasho [114]

1.302  moles of carbon dioxide would have to be added

<h3>Further explanation</h3>

The equilibrium constant is the value of the product in the equilibrium state of the substance in the right (product) divided by the substance in the left (reactant) with the exponents of each reaction coefficient

The equilibrium constant is based on the concentration (Kc) in a reaction

pA + qB -----> mC + nD

\large {\boxed {\bold {Kc ~ = ~ \frac {[C] ^ m [D] ^ n} {[A] ^ p [B] ^ q}}}}

While the equilibrium constant is based on partial pressure

\large {\boxed {\bold {Kp ~ = ~ \frac {[pC] ^ m [pD] ^ n} {[pA] ^ p [pB] ^ q}}}}

The value of Kp and Kc can be linked to the formula '

\large {\boxed {\bold {Kp ~ = ~ Kc. (R.T) ^ {\Delta n}}}}

R = gas constant = 0.0821 L.atm / mol.K

=n = number of product coefficients-number of reactant coefficients

An equilibrium mixture: 0.750 moles of CO2 and H2, and 0.200 moles of CO and H2O

  • We determine Kc (constant concentration)

\displaystyle Kc=\frac{0.75.0.75}{0.2.0.2}\\\\Kc=14.1

  • the amount of carbon monoxide to 0.300 mol

Reaction :

                CO +H₂O ⇔ CO₂ + H₂

initially      0.2   0.2       0.75+x  0.75

reaction    0.1    0.1        0.1         0.1

product     0.3   0.3       0.65+x    0.65

\displaystyle Kc=\frac{(0.650+x)(0.65)}{0.3.0.3}\\\\14.1(0.09)=0.4225+0.65x\\\\1.269-0.4225=0.65x\\\\0.8465=0.65x\\\\x=1.302

<h3>Learn more</h3>

an equilibrium constant brainly.com/question/9173805

brainly.com/question/1109930

Calculate the value of the equilibrium constant, Kc

brainly.com/question/3612827

Concentration of hi at equilibrium

brainly.com/question/8962129

Keywords: constant equilibrium, Kc, concentration, product, reactant, reaction coefficient

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