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2 years ago

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Caustic soda, NaOH, can be prepared commercially by the reaction of Na2CO3 with slaked lime, Ca(OH)2. How many grams of NaOH can

be obtained by treating 1.350 kg of Na2CO3 with Ca(OH)2

1 answer:

torisob [31]2 years ago

8 0

Hey i dont have an answer but i need the points for finals today. Thank you

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a mixture containing 21.4 g of ice. (0.00 c) and 75.3 g of water (55.3 c) is placed in an insulatated container

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3 years ago

Using any data you can find in the ALEKS Data resource, calculate the equilibrium constant K at 30.0 °C for the following reacti

gayaneshka [121]

Answer : The value of $K$ for this reaction is, $2.6\times 10^{15}$

Explanation :

The given chemical reaction is:

$CH_3OH(g)+CO(g)\rightarrow HCH_3CO_2(g)$

Now we have to calculate value of $(\Delta G^o)$ .

$\Delta G^o=G_f_{product}-G_f_{reactant}$

$\Delta G^o=[n_{HCH_3CO_2(g)}\times \Delta G^0_{(HCH_3CO_2(g))}]-[n_{CH_3OH(g)}\times \Delta G^0_{(CH_3OH(g))}+n_{CO(g)}\times \Delta G^0_{(CO(g))}]$

where,

$\Delta G^o$ = Gibbs free energy of reaction = ?

n = number of moles

$\Delta G^0_{(HCH_3CO_2(g))}$ = -389.8 kJ/mol

$\Delta G^0_{(CH_3OH(g))}$ = -161.96 kJ/mol

$\Delta G^0_{(CO(g))}$ = -137.2 kJ/mol

Now put all the given values in this expression, we get:

$\Delta G^o=[1mole\times (-389.8kJ/mol)]-[1mole\times (-163.2kJ/mol)+1mole\times (-137.2kJ/mol)]$

$\Delta G^o=-89.4kJ/mol$

The relation between the equilibrium constant and standard Gibbs, free energy is:

$\Delta G^o=-RT\times \ln K$

where,

$\Delta G^o$ = standard Gibbs, free energy = -89.4 kJ/mol = -89400 J/mol

R = gas constant = 8.314 J/L.atm

T = temperature = $30.0^oC=273+30.0=303K$

$K$ = equilibrium constant = ?

Now put all the given values in this expression, we get:

$-89400J/mol=-(8.314J/L.atm)\times (303K)\times \ln K$

$K=2.6\times 10^{15}$

Thus, the value of $K$ for this reaction is, $2.6\times 10^{15}$

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3 years ago

A scientist, Dr. Sigma, states that a chemical taken internally once a day will prevent the common cold. However, no other scien

pshichka [43]

Answer:

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Explanation:

Expert judgment is a useful validation method to verify the reliability of an investigation that is defined as “an informed opinion of people with experience in the subject, who are recognized by others as qualified experts in it, and who can give information, evidence, judgments and assessments ”

After submitting an instrument for comparison to the consultation and expert opinion, it must meet two quality criteria: validity and reliability. The validity of content is often established based on two situations, one that concerns the design of a test and the other, the validation of an instrument subject to translation and standardization procedures to adapt it to different cultural meanings. It is here that the task of the expert becomes a fundamental task to eliminate irrelevant aspects, incorporate those that are essential and / or modify those that require it.

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A piece of metal that has a density of 5.2 g/cm3 and a mass of 100 g was placed in a full jar of water. How many mL of water was

Klio2033 [76]

Answer:

100g / (5.2g/cm3)

= 100g / (5.2g / 1cm3)

= 100g x 1cm3 / 5.2 g

= 19.2 cm3

Since 1 cm3 = 1 ml, your answer is 19.2 ml.

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3 years ago