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4 years ago

Give me an example of oxide ore

Chemistry

2 answers:

user100 [1]4 years ago

7 0

example of oxide ore are hematite ,magnetite,chromite, zincite

Natalka [10]4 years ago

6 0

Explanation:

Several species of oxide minerals are also quite useful for industrial purposes, and hold significant economic value. These include chief ores such as Hematite and Magnetite (Iron), Chromite (Chromium), Manganite (Manganese), Cassiterite (Tin), Ilmenite and Rutile (Titanium), Franklinite and Zincite (Zinc).

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The empirical formula of a compound of iron (Mg) and sulfur (S) was determined using the following data. A sample of Mg was weig

Fudgin [204]

Answer:

The empirical formula is MgS

The number of moles Mg is 0.25 moles

There will react 0.25 moles

Option A is correct

Explanation:

Step 1: data given

The molecular weight of Mg = 24.305 g/mol

The molecular weight of S is 32.065 g/mol

mass of crucible and cover = 27.631 grams

mass of crucible, cover and Mg = 33.709 grams

mass of crucible, cover, and the compound formed 41.725 grams

Step 2: Calculate mass of Mg

Mass of Mg = mass of crucible, cover and Mg - mass of crucible and cover

Mass of Mg = 33.709 grams - 27.631 grams

Mass of Mg = 6.078 grams

Step 3: Calculate moles Mg

Moles Mg = mass Mg / atomic mass Mg

Moles Mg = 6.078 grams 24.305 g/mol

Moles Mg = 0.250 moles

Step 4: Calculate mass compound

Mass compound = mass of crucible, cover, and the compound formed - mass of crucible and cover

Mass compound = 41.725 grams - 27.631 grams

Mass compound = 14.094 grams

Step 5: Calculate mass sulfur

Mass sulfur = 14.094 grams - 6.078 grams

Mass sulfur = 8.016 grams

Step 6: Calculate moles Sulfur

Moles sulfur = 8.016 grams / 32.065 g/mol

Moles sulfur = 0.250 moles

Step 7: Calculate the mol ratio

We diviide by the smallest amount of moles

Mg: 0.250 moles / 0.250 moles = 1

S: 0.250 moles / 0.250 moles = 1

The empirical formula is MgS

The number of moles Mg is 0.25 moles

There will react 0.25 moles

Option A is correct

4 0

3 years ago

Read 2 more answers

How many moles are in 1.9 x 1046 atoms of chromium?

faust18 [17]

Answer:

<h2>3.156 × 10²² atoms</h2>

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

$n = \frac{N}{L} \\$

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

$n = \frac{1.9 \times {10}^{46} }{6.02 \times {10}^{23} } \\ = 3.156 \times {10}^{22}$

We have the final answer as

<h3>3.156 × 10²² atoms</h3>

Hope this helps you

8 0

3 years ago

Someone please help me??

Scrat [10]

The answer is 0.18

Because the equation for finding the molarity of a solution is:

Moles divided by Liter of solution,

so, you dissolve 0.32 mol of KCL in 1759 ml of Water 1759 ml is 1.759 liters when you plug the numbers in you get 0.18

Hope this helps,

kwrob

6 0

3 years ago

Help NEED THE ANSWER OR HELP PLEASE !!!!<br><br> 3. What is the mass of 8.0 cm3 of metal A?

Anna35 [415]

Answer:

now this area isn't my expertise so take this with a grain of salt but I believe its about 91125 g

Explanation:

8 cm is about 45g cube that and its 91125 g

6 0

3 years ago

Learn how changes in binding free energy affect binding and the ratio of unbound and bound molecules.

Delvig [45]

C)[D]/[ED] = 5.20

D)[D]/[ED] = 5.20

E)[D']_T = 1.495* 10 ^-7 M

F)[D'] / [ED'] = 0.0579

Explanation:

E = 250 nM =2.5* 10 ^-7 mol/L , T=298.15 K

Dissociation constant of K_D = 1.3 μM (1.3 *10 ^-6 M)

E + D ⇄ ED → K_a = [ED] / [D][E] (association constant)

ED ⇄ E + D → K_D = [E][D] / [ED] (dissociation constant)

[E] =2.5*10^-7 mol/L

K_D = 1.3* 10^-6 M

K_D = [E][D] / [ED] → [D]/[ED] = K_D / [E]

= [D]/[ED] = 1.3* 10 ^-6 / 2.5 *10^-7

= 13/25 * 10

=130/25 = 5.20

[D]/[ED] = 5.20

ΔG =RTln Kd

ΔG_2 for E and D = 1.987 * 298.15 * ln 1.3*10^-6

ΔG_2 592.454 * [ln 1.3 +ln 10^-6]

ΔG_1 = 592.424 [0.2623 - 13.8155]

ΔG_2 = -592.424 * 13.553

ΔG_1 = -8184.633 cal/ mol

ΔG_1 = -8184.633 * 4.18 J/mol = -34244.508 J?mol

ΔG_1 = -34.245 KL/mol

so, ΔG_2 = ΔG_1 - 10.5 KJ/mol

ΔG_2 = -34.245 - 10.5

ΔG_2 = -44.745KJ / mol

ΔG_2 =RT ln K_D

-44.745 *10^3

=8.314 *298.15 lnK_D

lnK_D' = - 44745 / 2478.81 g

ln K_D' = -18.051

K_D' = -18.051

K_D' = e^-18.051

[D]/[ED] = 5.20

[E] = 2.5* 10 ^-7 mol/ L = a

K_D' = [E][D] / [ED'] E +D' → ED'

K_D' = a/2(x-(a/2) / (a/2)

KD' = x - a/2

=2.447 *10^-8 = (2.5/2) * 10^-7

x=2.447 * 10^-8 + 1.25 * 10^-7

x = 2.447 *10^-8 + 1.25 * 10 ^-7

x= 10^-7 [1.25 + 0.2447]

x = 1.4947 * 10^-7

[D']_T = 1.495* 10 ^-7 M

K_D' = [E][D'] / [ED']

[D'] / [ED'] = KD' / [E]

[D'] / [ED'] = 1.447 *10^-8 / 2.5* 10^-7

[D'] / [ED'] = 0.5788 * 10^-1

[D'] / [ED'] = 0.0579

5 0

3 years ago

Give me an example of oxide ore​

Give me an example of oxide ore