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Veseljchak [2.6K]
2 years ago
14

a compound containing only sulfur and nitrogen is by mass; the molar mass is g/mol. what are the empirical and molecular formula

s of the compound?'
Chemistry
1 answer:
kolezko [41]2 years ago
3 0

The chemical compound's empirical formula is NS.

The chemical compound's molecular formula is N4S4.

<h3>What does a chemical empirical formula look like?</h3>
  • The empirical formula of a compound that gives the proportion (ratios) of the elements in the complex but not the precise number or arrangement of atoms is known as an empirical formula.
  • This would be the compound's element to whole number ratio with the lowest value.
<h3>What sort of empirical formula would that be?</h3>
  • The chemical structure of glucose is C6H12O6. Every mole of carbon and oxygen is accompanied by two moles of hydrogen.
  • Glucose has the empirical formula CH2O.
  • Ribose has the chemical formula C5H10O5, which can be simplified to the empirical formula CH2O.

learn more about empirical formula here

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the question you are looking for is

A compound containing only sulfur and nitrogen is 69.6% S by mass; the molar mass is 184 g/mol. What are the empirical and molecular formulas of the compound?

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What is the maximum amount in moles of P2O5P2O5 that can theoretically be made from 112 gg of O2O2 and excess phosphorus
Nastasia [14]

Answer:

n_{P_2O_5}^{max}=1.4molP_2O_5

Explanation:

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In this case, since the reaction between phosphorous and oxygen to form diphosphorous pentoxide is:

2P+\frac{5}{2}O_2\rightarrow P_2O_5

Thus, since phosphorous is in excess and oxygen and diphosphorous pentoxide are in a 5/2:1 mole ratio, we can compute the maximum moles of product as shown below:

n_{P_2O_5}^{max}=112 gO_2*\frac{1molO_2}{32.00gO_2}*\frac{1molP_2O_5}{5/2molO_2}\\\\  n_{P_2O_5}^{max}=1.4molP_2O_5

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The following data were obtained in a kinetics study of the hypothetical reaction A + B + C → products. [A]0 (M) [B]0 (M) [C]0 (
Vladimir [108]

Answer:

B. First order, Order with respect to C = 1

Explanation:

The given kinetic data is as follows:

A + B + C → Products

     [A]₀     [B]₀    [C]₀       Initial Rate (10⁻³ M/s)

1.   0.4      0.4     0.2       160

2.  0.2      0.4      0.4       80

3.   0.6     0.1       0.2       15

4.   0.2     0.1       0.2        5

5.   0.2     0.2      0.4       20

The rate of the above reaction is given as:

Rate = k[A]^{x}[B]^{y}[C]^{z}

where x, y and z are the order with respect to A, B and C respectively.

k = rate constant

[A], [B], [C] are the concentrations

In the method of initial rates, the given reaction is run multiple times. The order with respect to a particular reactant is deduced by keeping the concentrations of the remaining reactants constant and measuring the rates. The ratio of the rates from the two runs gives the order relative to that reactant.

Order w.r.t A : Use trials 3 and 4

\frac{Rate3}{Rate4}= [\frac{[A(3)]}{[A(4)]}]^{x}

\frac{15}{5}= [\frac{[0.6]}{[0.2]}]^{x}

3 = 3^{x} \\\\x =1

Order w.r.t B : Use trials 2 and 5

\frac{Rate2}{Rate5}= [\frac{[B(2)]}{[B(5)]}]^{y}

\frac{80}{20}= [\frac{[0.4]}{[0.2]}]^{y}

4 = 2^{y} \\\\y =2

Order w.r.t C : Use trials 1 and 2

\frac{Rate1}{Rate2}= [\frac{[A(1)]}{[A(2)]}]^{x}[\frac{[B(1)]}{[B(2)]}]^{y}[\frac{[C(1)]}{[C(2)]}]^{z}

we know that x = 1 and y = 2, substituting the appropriate values in the above equation gives:

\frac{160}{80}= [\frac{[0.4]}{[0.2]}]^{1}[\frac{[0.4]}{[0.4]}]^{2}[\frac{[0.2]}{[0.4]}]^{z}

1 = (0.5)^{z}

z = 1

Therefore, order w.r.t C = 1

8 0
4 years ago
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