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charle [14.2K]
3 years ago
10

Which of the following characteristics must be present for a substance to be called a mineral?

Chemistry
2 answers:
Komok [63]3 years ago
6 0

Question options: <em> a. have a crystal structure b. has an unpredictable chemical structure c. can be liquid or solid d. can be organic or inorganic .</em>

Answer:

options (a) and (c) applies

Explanation:

  • Minerals usually have a definite crystalline structure, however this is not always the case, as some naturally occurring minerals are not crystal like.
  • the chemical structure of minerals are predictable.
  • Minerals can only exist as solids. if any substance is in its liquid form then its not a mineral.
  • finally minerals can either be inorganic or organic.Inorganic minerals do not contain carbon, while organic minerals are made up of carbon

<em />

<em />

MArishka [77]3 years ago
4 0
Minerals are solid, naturally occurring, inorganic compounds that possess an orderly internal structure and a regular chemical composition. Minerals should occur naturally. Hope this answers the question. Have a nice day. Feel free to ask more questions.
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calculate the ph of a 0.020 m carbonic acid solution, h2co3(aq), that has the stepwise dissociation constants ka1
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The calculated pH is 3.79. therefore, the solution is acidic.

No, carbonic acid is not a strong acid. H2CO3 is a weak acid that dissociates into a proton (H+ cation) and a bicarbonate ion (HCO3- anion). This compound only partly dissociates in aqueous solutions.

H2 CO3    =    H (+) + HCO3(-)          Ka1 = 4.3 * 10^ -7

   0.06 - x              x          x

Ka1 = x^2 / (0.06 - x) = 4.3 * 10^ - 7

A low Ka => x << 0.06 => 0.06 -x ≈ 0.06

=> Ka1 ≈ x^2 / 0.06 => x^2 ≈ 0.06 * Ka1 = 0.06 * 4.3 * 10^-7

=> x ≈ √ [ 2.58 * 10 ^ -8] = 1.606 * 10^ - 4 = 0.0001606

Second dissociation

HCO3(-)    =   H (+) + CO3(2-)         Ka2 = 5.6 * 10^ - 11

0.0001606 - y            y             y

Ka2 ≈ y^2 / 0.0001606 => y = √ [0.0001606 * 5.6* 10^ -11]

y = 9.48 * 10^ -8

An acidic solution has a high concentration of hydrogen ions (H +start superscript, plus, end superscript), greater than that of pure water.

[H+] = x + y = 1.607 * 10^ -4

pH = - log [H+] = 3.79

Learn more about concentration here-

brainly.com/question/10725862

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5 0
2 years ago
Used oil needs to be kept in what kind of containers
Paul [167]

Answer:

Never store used oil in anything other than tanks and storage containers. Used oil may also be stored in units that are permitted to store regulated hazardous waste. Tanks and containers storing used oil do not need to be RCRA permitted, however, as long as they are labeled and in good condition.

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3 years ago
When 3-methyl-3-pentanol is treated with chromic acid, Group of answer choices
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Answer: The orange color remains unchanged. (B)

Explanation:

7 0
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iron combines with oxygen to form rust. given the chemical reaction, how many grams of rust would be produced if 3 grams of reac
tia_tia [17]

Answer:

I believe its 1 gram

Explanation:

8 0
3 years ago
Carbon tetrachloride can be produced by the following reaction: Suppose 1.20 mol of and 3.60 mol of were placed in a 1.00-L flas
hjlf

The given question is incomplete. The complete question is :

Carbon tetrachloride can be produced by the following reaction:

CS_2(g)+3Cl_2(g)\rightleftharpoons S_2Cl_2(g)+CCl_4(g)

Suppose 1.20 mol CS_2(g) of and 3.60 mol of Cl_2(g)  were placed in a 1.00-L flask at an unknown temperature. After equilibrium has been achieved, the mixture contains 0.72 mol  of CCl_4. Calculate equilibrium constant at the unknown temperature.

Answer: The equilibrium constant at unknown temperature is 0.36

Explanation:

Moles of  CS_2 = 1.20 mole

Moles of  Cl_2 = 3.60 mole

Volume of solution = 1.00  L

Initial concentration of CS_2 = \frac{moles}{volume}=\frac{1.20mol}{1L}=1.20M

Initial concentration of Cl_2 = \frac{moles}{volume}=\frac{3.60mol}{1L}=3.60M

The given balanced equilibrium reaction is,

                 CS_2(g)+3Cl_2(g)\rightleftharpoons S_2Cl_2(g)+CCl_4(g)

Initial conc.         1.20 M        3.60 M                  0                  0

At eqm. conc.     (1.20-x) M   (3.60-3x) M   (x) M        (x) M

The expression for equilibrium constant for this reaction will be,

K_c=\frac{[S_2Cl_2]\times [CCl_4]}{[Cl_2]^3[CS_2]}

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

K_c=\frac{(x)\times (x)}{(3.60-3x)^3\times (1.20-x)}

Given :Equilibrium concentration of CCl_4 , x = \frac{moles}{volume}=\frac{0.72mol}{1L}=0.72M

K_c=\frac{(0.72)\times (0.72)}{(3.60-3\times 0.72)^3\times (1.20-0.72)}

K_c=0.36

Thus equilibrium constant at unknown temperature is 0.36

4 0
3 years ago
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