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dusya [7]
3 years ago
12

Calcule el peso de cromato sodico, Na2CrO4, necesario para preparar 200g de una disolución al 20% de Na2CrO4

Chemistry
1 answer:
olga55 [171]3 years ago
6 0

Answer:

40g de cromato sódico serán necesarios

Explanation:

Una solución 20% de Na2CrO4 contiene 20g de Na2CrO4 por cada 100g de solución (Esto es, soluto + solvente). Para preparar 200g de la disolución en mención serán necesarios:

200g solucion * (20g Na2CrO4 / 100g Solución) =

<h3>40g de cromato sódico serán necesarios</h3>
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Answer:

Explanation:

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3 0
3 years ago
Suppose 0.410 kg of hexane are burned in air at a pressure of exactly 1 atm and a temperature of 13.0 °C. Calculate the volume o
3241004551 [841]

Answer:

The answer is 671 litres of carbon dioxide is produced from 0.410 kg of hexane

Explanation:

We first write a balanced reaction for the complete combustion of hexane thus

The stoichiometry of the cumbustion of hexane in air is

2C6H14(g)+18O2(g)→12CO2(g)+14H2O(l) or

C6H14(g)+9O2(g)→6CO2(g)+7H2O(l)

From the above reaction it is observed that one mole of hexane burns completely in the presence of oxygen to produce 6 moles of carbon dioxide

Therefore we calculate the nuber of moles of hexane present in the sample thus

Mass hexane of sample = 0.41 kg

Molar nass of hexane = 86.18 g/mol

number of moles of hexane = (mass of hexane)/(molar mass of hexane) = (0.41×1000)/86.16 = 410/86.16 = 4.76 moles

As we have seen from the chemical reaction, 1 mole of H6H14 produces 6 moles of CO2 hence 4.76 moles of Hexane produces

4.76×6 moles of CO2 which is 28.55 moles of CO2

From the question we have the temperature and the pressure of the production of CO2 as

Temperature of reaction = 13° C converting to kelving gives= 13+273.15 = 286.15 K

and pressure = 1 atmosphere or 101325 Pa

13.0∘C=13.0∘C+273.15=286.15 K

The volume of the produced CO2 can be calculated using the combined ideal gas equation given by

P×V=n×R×T where

Here

P = Gas pressure (of CO2 )

V = Volume (of the CO2)

n = number of moles of gas (CO2) present

R = universal gas constant, equal to 0.0821 atm× L/(mol× K )

T = absolute temperature in Kelvin

Thus we have

1×V = 28.55×0.0821×286.15  or V  = 670.76L

Rounding up the answer to 3 significant digits we have

670.76L ≅ 671L

671 litres of carbon dioxide is produced from 0.410 kg of hexane

8 0
4 years ago
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almond37 [142]

Answer:

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Explanation:

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3 years ago
1. True or false, phosphorus-33 is an isotope
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Answer:

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6 0
3 years ago
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