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Goshia [24]
2 years ago
13

A) How much CO is required for the production of Fe from 1000 tonnes of magnetite (Fe2O4). Assume that the iron ore is 100% pure

.
B) How many kilograms of iron should be produced?

C) If the reaction yielded 555 kg of Fe, what is the percentage yield for the process
Chemistry
1 answer:
Kryger [21]2 years ago
5 0
Its b im pretty sure
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15
Vinil7 [7]

Answer:

barium and silicon has same valence electrons

Explanation:

barium-2,8,18,18,8,2

neon-2,8

silicon-2,8,2,2

carbon-2,4

5 0
2 years ago
How many atoms are 0.750 moles of zinc?
Darya [45]
<span>.750 moles X (6.02 x10^23 atoms/1 mol)= 4.52 X 10^23 atoms is the answer </span>
7 0
3 years ago
Interpret the block diagram seen here. Arranging the letters from oldest to youngest, the order should be
Ganezh [65]

Answer: option B. - A, B, D, E, C, H, F, G is correct using the principle of cross-cutting relationships.

The principle of cross-cutting relationships states that a fault or intrusion is younger than the rocks that it cuts through.

Explanation:

The full sequence of events is:

1. Layer A formed.

2. Layer B formed

3. Layer D formed.

4. Layer E formed

5. After layers A-B-D-E were present, intrusion C cut across all three.

6. Fault H formed, shifting rocks E through A and intrusion C.

7. Weathering and erosion created a layer of soil on top of layer F then G.

5 0
2 years ago
Read 2 more answers
I need help ASAP!!!
fiasKO [112]

Answer:

cells; energy

Explanation:

3 0
3 years ago
Read 2 more answers
***BRAINLIEST ASNWERRR***<br>How many grams are in 34.2 moles of Lithium (Li)?​
Step2247 [10]

Mass of Li = 237.38 g

<h3>Further explanation</h3>

The mole itself is the number of particles contained in a substance amounting to 6.02.10²³  

\large {\boxed {\boxed {\bold {mol = \frac {mass} {molar \: mass}}}}

<h3>Known</h3>

Moles of Li = 34.2

Molar mass(MW) of Li = 6.941 g/mol

then mass of Lithium (Li) :

\tt mol=\dfrac{mass}{MW}\\\\mass=mol\times MW\\\\mass=34.2\times 6,941~g/mol\\\\mass=\boxed{\bold{237.38~g}}

3 0
3 years ago
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