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sattari [20]
3 years ago
10

Ce procent de impuritati contine un minereu de siderit, daca din 1500kg minereu s-au obtinut 700kg fier 90% ?

Chemistry
1 answer:
Pachacha [2.7K]3 years ago
8 0

FeCO3  ==> Fe + produsi minoritari.


m Fe impur= 700 kg


puritatea (p) = masa pura (mp)/ masa impura (mi) x 100

mp= p x mi / 100 sau mp = p/100 x mi => mp Fe = 90/100 x 700 = 630 kg Fe pur.


M FeCO3= 115.85 kg/kmol

115.85 kg FeCO3 .... 55.85 kg Fe

     x kg FeCO3  ........630 kg Fe

x= 630 * 115.85 /55.85 = 1306.81 kg FeCO3 (mp in formula puritatii)

p=mp/mi x 100

mi FeCO3 = 1500 kg

mp FeCO3=1306.81 kg

p=1306.81 / 1500 x 100 = 87.12% puritate Siderit




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Answer:

Potassium permanganate has a molar mass of 158.04 g/mol. This figure is obtained by adding the individual molar masses of <em><u>four oxygen atoms</u></em>, <em><u>one manganese atom</u></em> and <em><u>one potassium atom</u></em>

Explanation:

6 0
2 years ago
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In the following equation:
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Answer:

FeCl₃

Explanation:

                 4FeCl₃  +   3O₂     => 2Fe₂O₃+ 6Cl₂

Given =>  7moles     9moles

A simple way to determine which reagent is the limiting reactant is to convert all given data to moles then divide by the respective coefficients of the balanced equation. The smaller value will be the limiting reactant.

                 4FeCl₃     +   3O₂     => 2Fe₂O₃+ 6Cl₂

Given =>  7/4 = 1.75*     9/3 = 3

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NOTE: However, when working problems, one must use original mole values given.

   

7 0
3 years ago
1‑Propanol ( P⁰ 1 = 20.9 Torr at 25 ⁰C ) and 2‑propanol ( P⁰ 2 = 45.2 Torr at 25 ⁰C ) form ideal solutions in all proportions. L
anastassius [24]

Answer : The mole fraction of vapor phase 1‑Propanol and 2‑Propanol is, 0.352 and 0.648 respectively.

Explanation : Given,

Vapor presume of 1‑Propanol (P^o_1) = 20.9 torr

Vapor presume of 2‑Propanol (P^o_2) = 45.2 torr

Mole fraction of 1‑Propanol (x_1) = 0.540

Mole fraction of 2‑Propanol (x_2) = 1-0.540 = 0.46

First we have to calculate the partial pressure of 1‑Propanol and 2‑Propanol.

p_1=x_1\times p^o_1

where,

p_1 = partial vapor pressure of 1‑Propanol

p^o_1 = vapor pressure of pure substance 1‑Propanol

x_1 = mole fraction of 1‑Propanol

p_1=(0.540)\times (20.9torr)=11.3torr

and,

p_2=x_2\times p^o_2

where,

p_2 = partial vapor pressure of 2‑Propanol

p^o_2 = vapor pressure of pure substance 2‑Propanol

x_2 = mole fraction of 2‑Propanol

p_2=(0.46)\times (45.2torr)=20.8torr

Thus, total pressure = 11.3 + 20.8 = 32.1 torr

Now we have to calculate the mole fraction of vapor phase 1‑Propanol and 2‑Propanol.

\text{Mole fraction of 1-Propanol}=\frac{\text{Partial pressure of 1-Propanol}}{\text{Total pressure}}=\frac{11.3}{32.1}=0.352

and,

\text{Mole fraction of 2-Propanol}=\frac{\text{Partial pressure of 2-Propanol}}{\text{Total pressure}}=\frac{20.8}{32.1}=0.648

Thus, the mole fraction of vapor phase 1‑Propanol and 2‑Propanol is, 0.352 and 0.648 respectively.

3 0
3 years ago
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Answer:

hope it helped you

Explanation:

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3 0
3 years ago
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ad-work [718]

Answer:

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Explanation:

this is because water includes 0H, which would mean that it is more than 6

4 0
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