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son4ous [18]
3 years ago
14

¿Que hace diferente ala química de la alquimia?

Chemistry
1 answer:
viva [34]3 years ago
8 0

Answer:

La distinción entre la química teóricamente teórica y la química avanzada es la química teórica teórica que se basa en un espiritualista, potente que percibe visualmente la autenticidad, mientras que la química espera que la autenticidad sea fundamentalmente mundana. Eso engendra una tremenda distinción, y la química nunca se hubiera alejado excepcionalmente de la remota posibilidad de que se hubiera quedado con el trascendentalismo antediluviano.

Explanation:

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maxonik [38]

len toExplanation:tal esta echpo de los materia

4 0
3 years ago
What is the entropy change for freezing 2.71 g of C2H5OH at 158.7 K? ∆H = −4600 J/mol. Answer in units of J/K.
Ira Lisetskai [31]

Answer:

-1.71 J/K

Explanation:

To solve this problem we use the formula

ΔS = n*ΔH/T

Where n is mol, ΔH is enthalpy and T is temperature.

ΔH and T are already given by the problem, so now we calculate n:

Molar Mass C₂H₅OH = 46 g/mol

2.71 g C₂H₅OH ÷ 46g/mol = 0.0589 mol

Now we calculate ΔS:

ΔS = 0.0589 mol * −4600 J/mol / 158.7 K

ΔS = -1.71 J/K

4 0
3 years ago
How many grams of a nonelectrolyte (78.2 g/mol) must be dissolved in 1 kg of solvent to obtain a freezing point of solution of 1
Dimas [21]

Answer Tu mama por si acaso

Explanation:

jkand

8 0
3 years ago
Two aqueous sulfuric acid solutions containing 20.0 wt% H2SO4
kozerog [31]

Answer:

The feed ratio (liters 20%  solution/liter 60% solution) is 3,08

Explanation:

In this problem you have a 20,0 wt% H₂SO₄ and a 60,0 wt% H₂SO₄ solutions.

100 kg of 20% solution are 100kg/1,139 kg/L = <em>87,8 L </em>

100kg×20wt% = 20 kg H₂SO₄. In moles:

20 kg H₂SO₄ × (1 kmol/98,08 kg) = 0,2039 kmol H₂SO₄≡ <em>203,9 mol</em>

The final molarity 4,00M comes from:

\frac{203,9 moles+ Xmoles}{87,8L + Yliters} <em>(1)</em>

Where X moles and Y liters comes from 60,0 wt% H₂SO₄

100 L of 60,0 wt% H₂SO₄ are:

100L×\frac{1,498 kgsolution}{L}×\frac{60 kg H_{2}SO_{4}}{100kgSolution}×\frac{1kmol}{98,08kg H_{2}SO_{4}} = <em>0,9164 kmolH₂SO₄ ≡ 916,4 moles</em>

That means:

X/Y = 916,4/100 = 9,164 <em>(2)</em>

Replacing (2) in (1):

Y(liters of 60,0 wt% H₂SO₄) = <em>28,52 L</em>

Thus, feed ratio (liters 20%  solution/liter 60% solution):

87,8L/28,52L = <em>3,08</em>

I hope it helps!

8 0
3 years ago
Which of the following is an accurate statement
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What are the statements?
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