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3 years ago

When determining rock composition, describe the test that can be used to determine whether it contains the mineral calcite.

Chemistry

2 answers:

elena55 [62]3 years ago

7 0

Answer: ACID TEST

Explanation: Acid test is a test carried out to determine the presence of acid in a substance or a liquid. Acid tests are used to tests the composition of a ROCK,the test makes use of DILUTE HYDROCHLORIC ACID to test and classify rocks,it is capable of detecting CARBONATE MINERALS or other materials.

The process involves the ADDITION OF 5% TO 10% of HYDROCHLORIC ACID IN DROPS ON A Rock particle,the RELEASE OF BUBBLES OF CARBONDIOXIDE SIGNALS THE PRESENCE OF CARBONATE, DOLOMITE OR CALCITE MINERALS.

Alona [7]3 years ago

3 0

Answer:

Check explanation

Explanation:

There are two ways to test for calcite in rocks. Calcite is popularly known as calcium carbonate and it is found almost everywhere. There are many minerals that looks like calcite/calcium carbonates, so we there is a need for a good test for proper identification.

The tests are:

(1). ACID TEST: the calcite/calcium carbonate reacts vigorously with Hydrogen Chloride acid,HCl with the evolution of carbondioxide,CO2. The equation of reaction is given below;

CaCO3 + 2 HCl ---------> CaCl2 + CO2 + H2O.

(2). DOUBLE REFRACTION METHOD: There are other minerals in the rock that can react with Hydrochloric acid,HCl to evolve Carbondioxide,CO2(although their response to HCl differs). There is a need for a confirmatory test. This double refraction method is a confirmatory test. In this test, one will pass light through the calcite, if the light splits into two rays and reflected twice then, it is a calcite.

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CO2(g)+CCl4(g)⇌2COCl2(g) Calculate ΔG for this reaction at 25 ∘C under these conditions: PCO2PCCl4PCOCl2===0.140 atm0.185 atm0.7

padilas [110]

Answer: The $\Delta G$ for the reaction is 54.425 kJ/mol

Explanation:

For the given balanced chemical equation:

$CO_2(g)+CCl_4(g)\rightleftharpoons 2COCl_2(g)$

We are given:

$\Delta G^o_f_{CO_2}=-394.4kJ/mol\\\Delta G^o_f_{CCl_4}=-62.3kJ/mol\\\Delta G^o_f_{COCl_2}=-204.9kJ/mol$

To calculate $\Delta G^o_{rxn}$ for the reaction, we use the equation:

$\Delta G^o_{rxn}=\sum [n\times \Delta G_f(product)]-\sum [n\times \Delta G_f(reactant)]$

For the given equation:

$\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(COCl_2)})]-[(1\times \Delta G^o_f_{(CO_2)})+(1\times \Delta G^o_f_{(CCl_4)})]$

Putting values in above equation, we get:

$\Delta G^o_{rxn}=[(2\times (-204.9))-((1\times (-394.4))+(1\times (-62.3)))]\\\Delta G^o_{rxn}=46.9kJ=46900J$

Conversion factor used = 1 kJ = 1000 J

The expression of $K_p$ for the given reaction:

$K_p=\frac{(p_{COCl_2})^2}{p_{CO_2}\times p_{CCl_4}}$

We are given:

$p_{COCl_2}=0.735atm\\p_{CO_2}=0.140atm\\p_{CCl_4}=0.185atm$

Putting values in above equation, we get:

$K_p=\frac{(0.735)^2}{0.410\times 0.185}\\\\K_p=20.85$

To calculate the gibbs free energy of the reaction, we use the equation:

$\Delta G=\Delta G^o+RT\ln K_p$

where,

$\Delta G$ = Gibbs' free energy of the reaction = ?

$\Delta G^o$ = Standard gibbs' free energy change of the reaction = 46900 J

R = Gas constant = $8.314J/K mol$

T = Temperature = $25^oC=[25+273]K=298K$

$K_p$ = equilibrium constant in terms of partial pressure = 20.85

Putting values in above equation, we get:

$\Delta G=46900J+(8.314J/K.mol\times 298K\times \ln(20.85))\\\\\Delta G=54425.26J/mol=54.425kJ/mol$

Hence, the $\Delta G$ for the reaction is 54.425 kJ/mol

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Hey there!:

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absorbace = 0.85

log 100 = 2

- log transmitance = absorbace / log 100

0.85 / 2= 0.425

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