In what two places minerals are form?

abruzzese [7]

I think it is D the last choice

5 0

3 years ago

Recovery standards are a necessary tool for determining exactly how much of a particular analyte you are able to extract from a

Dafna11 [192]

Answer:

2.05mg Fe/ g sample

Explanation:

In all chemical extractions you lose analyte. Recovery standards are a way to know how many analyte you lose.

In the problem you recover 3.5mg Fe / 1.0101g sample: 3.465mg Fe / g sample. As real concentration of the standard is 4.0 mg / g of sample the percent of recovery extraction is:

3.465 / 4×100 = 86,6%

As the recovery of your sample was 1.7mg Fe / 0.9582g, the Iron present in your sample is:

1.7mg Fe / 0.9582g sample× (100/86.6) = 2.05mg Fe / g sample

I hope it helps!

5 0

3 years ago

Consider the following reaction at a high temperature. Br2(g) ⇆ 2Br(g) When 1.35 moles of Br2 are put in a 0.780−L flask, 3.60 p

UNO [17]

Answer : The equilibrium constant $K_c$ for the reaction is, 0.1133

Explanation :

First we have to calculate the concentration of $Br_2$ .

$\text{Concentration of }Br_2=\frac{\text{Moles of }Br_2}{\text{Volume of solution}}$

$\text{Concentration of }Br_2=\frac{1.35moles}{0.780L}=1.731M$

Now we have to calculate the dissociated concentration of $Br_2$ .

The balanced equilibrium reaction is,

$Br_2(g)\rightleftharpoons 2Br(aq)$

Initial conc. 1.731 M 0

At eqm. conc. (1.731-x) (2x) M

As we are given,

The percent of dissociation of $Br_2$ = $\alpha$ = 1.2 %

So, the dissociate concentration of $Br_2$ = $C\alpha=1.731M\times \frac{1.2}{100}=0.2077M$

The value of x = 0.2077 M

Now we have to calculate the concentration of $Br_2\text{ and }Br$ at equilibrium.

Concentration of $Br_2$ = 1.731 - x = 1.731 - 0.2077 = 1.5233 M

Concentration of $Br$ = 2x = 2 × 0.2077 = 0.4154 M

Now we have to calculate the equilibrium constant for the reaction.

The expression of equilibrium constant for the reaction will be :

$K_c=\frac{[Br]^2}{[Br_2]}$

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

$K_c=\frac{(0.4154)^2}{1.5233}=0.1133$

Therefore, the equilibrium constant $K_c$ for the reaction is, 0.1133

7 0

3 years ago

PURRRRRRRRRRRRRRRRPURRRRRRRRRRRRRRRRRRPURRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR

natulia [17]

Answer:

purrrrr cat

Explanation:

7 0

3 years ago

The total pressure of gases a, b, and c in a closed container is 4.1 atm. if the mixture is 36% a, 42% b, and 22% c by volume, w

Natali [406]

The partial pressure of gas C is 0.902 atm

calculation

partial pressure of gas c =[( percent by volume of gas C / total percent) x total pressure]

percent by volume of gas C= 22%

Total percent = 36% +42% + 22% = 100 %

Total pressure = 4.1 atm

partial pressure of gas C is therefore = 22/100 x 4.1 atm = 0.902 atm

3 0

3 years ago

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