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Alisiya [41]
3 years ago
12

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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Answer:

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3 years ago
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Ulleksa [173]
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5 0
3 years ago
(Please explain the answer. Thank you!)
Dominik [7]

Sample cost = $15.796

<h3>Further explanation</h3>

Given

Mass = 2 g

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Cost =  $5.67 per mL

Required

Cost

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7 0
3 years ago
The compound 1,1-difluoroethane decomposes at elevated temperatures to give fluoroethylene and hydrogen fluoride: CH3CHF2(g) → C
Maru [420]

Answer : The final temperature would be, 791.1 K

Explanation :

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K_2 = rate constant at T_2 = 4\times K_1

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T_2 = final temperature = ?

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

\log (\frac{4\times K_1}{K_1})=\frac{265000J/mol}{2.303\times 8.314J/mole.K}[\frac{1}{733K}-\frac{1}{T_2}]

T_2=791.1K

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