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3 years ago

What is the term for an irregular way a mineral breaks apart?

Chemistry

1 answer:

kari74 [83]3 years ago

6 0

Answer:

That would be fracture

Explanation:

A way that a mineral breaks apart curved at the edges or split into uneven pieces

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Consider the reaction

SOVA2 [1]

Answer :

(a) The average rate will be:

$\frac{d[Br_2]}{dt}=9.36\times 10^{-5}M/s$

(b) The average rate will be:

$\frac{d[H^+]}{dt}=1.87\times 10^{-4}M/s$

Explanation :

The general rate of reaction is,

$aA+bB\rightarrow cC+dD$

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

$\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}$

$\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}$

$\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}$

$\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}$

$Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}$

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

$5Br^-(aq)+BrO_3^-(aq)+6H^+(aq)\rightarrow 3Br_2(aq)+3H_2O(l)$

The expression for rate of reaction :

$\text{Rate of disappearance of }Br^-=-\frac{1}{5}\frac{d[Br^-]}{dt}$

$\text{Rate of disappearance of }BrO_3^-=-\frac{d[BrO_3^-]}{dt}$

$\text{Rate of disappearance of }H^+=-\frac{1}{6}\frac{d[H^+]}{dt}$

$\text{Rate of formation of }Br_2=+\frac{1}{3}\frac{d[Br_2]}{dt}$

$\text{Rate of formation of }H_2O=+\frac{1}{3}\frac{d[H_2O]}{dt}$

Thus, the rate of reaction will be:

$\text{Rate of reaction}=-\frac{1}{5}\frac{d[Br^-]}{dt}=-\frac{d[BrO_3^-]}{dt}=-\frac{1}{6}\frac{d[H^+]}{dt}=+\frac{1}{3}\frac{d[Br_2]}{dt}=+\frac{1}{3}\frac{d[H_2O]}{dt}$

Part (a) :

Given:

$\frac{1}{5}\frac{d[Br^-]}{dt}=1.56\times 10^{-4}M/s$

As,

$-\frac{1}{5}\frac{d[Br^-]}{dt}=+\frac{1}{3}\frac{d[Br_2]}{dt}$

and,

$\frac{d[Br_2]}{dt}=\frac{3}{5}\frac{d[Br^-]}{dt}$

$\frac{d[Br_2]}{dt}=\frac{3}{5}\times 1.56\times 10^{-4}M/s$

$\frac{d[Br_2]}{dt}=9.36\times 10^{-5}M/s$

Part (b) :

Given:

$\frac{1}{5}\frac{d[Br^-]}{dt}=1.56\times 10^{-4}M/s$

As,

$-\frac{1}{5}\frac{d[Br^-]}{dt}=-\frac{1}{6}\frac{d[H^+]}{dt}$

and,

$-\frac{1}{6}\frac{d[H^+]}{dt}=\frac{3}{5}\frac{d[Br^-]}{dt}$

$\frac{d[H^+]}{dt}=\frac{6}{5}\times 1.56\times 10^{-4}M/s$

$\frac{d[H^+]}{dt}=1.87\times 10^{-4}M/s$

5 0

3 years ago

Explain this method (Froth floatation method)..........

Svetach [21]

Answer:

froth flotation is a technique commonly used in the mining industry. In this technique, particles of interest are physically separated from a liquid phase as a result of differences in the ability of air bubbles to selectively adhere to the surface of the particles, based upon their hydrophobicity.

Explanation:

Froth floatation method is commonly used to concentrate sulphide ore such as galena (PbS), zinc blende (ZnS) etc. (ii) In this method, the metaalic ore particles which are perferentially wetted by oil can be separated from gangue. (iii) In this method, the crushed ore is suspended in water and mixed with frothing agent such as pine oil, eucalyptus oil etc. (iv) A small quantity of sodium ethyl xanthate which act as a collector is also added. (v) A froth is generated by blowing air through this mixture. (vi) The collector molecules attach to the ore particles and make them water repellent. (vii) As a result, ore parrticles, wetted by the oil, rise to the surface along with the froth. (viii) The froth is skimmed off and dried to recover the concentration ore. (ix) The gangue particles that are preferentially wetted by water settle at the bottom.

3 0

2 years ago

Need help plz help any one

dimulka [17.4K]

Hi!

Composition: the way in which a whole is made up

Hope I helped!
Have a nice day!

6 0

3 years ago

Calcular la presión final de un gas que inicialmente está a 21°C y 0,98 atm. Sabiendo que su temperatura aumenta a 37°C.

KatRina [158]

Answer:

1.03 atm

Explanation:

Primero convertimos 21 °C y 37 °C a K:

21 °C + 273.16 = 294.16 K

37 °C + 273.16 = 310.16 K

Una vez tenemos las temperaturas absolutas, podemos resolver este problema usando la ley de Gay-Lussac:

T₁P₂ = T₂P₁

En este caso:

T₁ = 294.16 K

P₂ = ?

T₂ = 310.16 K

P₁ = 0.98 atm

Colocando los datos:

294.16 K * P₂ = 310.16 K * 0.98 atm

Y despejando P₂:

P₂ = 1.03 atm

7 0

2 years ago

What are the hydronium and hydroxide ion concentrations in a solution whose ph is 6.52?

erica [24]

A solution with a pH of 6.52 has a hydronium ion concentration of 3.02x10-7 mol/L and a hydroxide ion concentration of 3.31x10-8 mol/L.

The hydronium ion concentration of a solution can be calculated from pH by using $10^{-pH}$ . For a pH of 6.52, hydronium ion concentration is 3.02x10-7 mol/L.

The concentration of hydroxide ions can be determined by identifying the value of pOH. The sum of pOH and pH is equal to 14, which is based on the negative logarithm of the ion-product constant of water. At a pH of 6.52, pOH is equal to 7.48.

The relationship between pOH and hydroxide ion concentration is the same as the relationship between pH and hydronium ion concentration. With this, the hydroxide ion concentration at pOH of 7.48 is $10^{-7.48}$ or 3.31x10-8 mol/L.

For more information regarding pH and pOH, please refer to the link brainly.com/question/13557815.

#SPJ4

3 0

1 year ago