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Lynna [10]
3 years ago
11

How would you expect the appearance of a rock high in iron and magnesium to differ from a rock with very little iron and magnesi

um?
Chemistry
2 answers:
kondor19780726 [428]3 years ago
8 0

Answer:

As explained below.

Explanation:

  • The rocks that have a higher content of magnesium and iron content are found in igneous rocks composition that is mafic and felsic in nature have a higher concentration of silica content also and are highly intrusive n nature are highly explosive and have a more acidic base.
  • These rocks like the magmas of 750 to 950 degrees temperature are dark in color and higher pressure and others that don't have this high content are characterized as sedimentary or metamorphic in nature.  
  • <u>These have less density and light color with layered structures present in them.</u>
xenn [34]3 years ago
5 0
High iron and magnesium rocks have a high percentage of dark-colored (mafic) minerals.

So the high iron and magnesium would be darker than the low iron and magnesium.
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Do your cells have access to everything they need over the first 4 or 5 hours that you are lost? Yes or No?
77julia77 [94]

Explanation: cell in the body is enclosed by a cell (Plasma) membrane. The cell membrane separates the material outside the cell, extracellular, from the material inside the cell, intracellular. ... All materials within a cell must have access to the cell membrane (the cell's boundary) for the needed exchange

**Answer**: The answer would be Yes I believe

4 0
2 years ago
11C. Potassium hydrogen phthalate is a solid, monoprotic acid frequently used in the laboratory to standardize strong base solut
Llana [10]

For the reactants,

  • The oxidation number of hydrogen = +1
  • The oxidation number of oxygen = -2
  • The oxidation number of arsenic = +5
  • The oxidation number of carbon = +3

For the products,

  • The oxidation number of hydrogen = +1
  • The oxidation number of oxygen = -2
  • The oxidation number of arsenic = +3
  • The oxidation number of carbon = +4

Here, arsenic (+5 to +3) and carbon (+3 to +4) are the only oxidation numbers changing.

Note that an increase in oxidation number means electrons are lost. Thus oxidation is occurring, and a decrease in oxidation number means electrons are being gained, and thus reduction is occurring.

Also, the compound that contains the element being oxidized is the reducing agent, and the compound that contains the element being reduced is the oxidizing agent.

So, the answers are:

name of the element oxidized: Carbon

name of the element reduced: Arsenic

formula of the oxidizing agent: \text{H}_{3}\text{AsO}_{4}

formula of the reducing agent: \text{H}_{2}\text{C}_{2}\text{O}_{4}

6 0
2 years ago
The average adult male has a total blood volume of 5.0 L. After drinking a few beers, he has a BAC of 0.10. What mass of alcohol
stellarik [79]
  B A C ( Blood Alcohol Content ) of 0.10 means that there are 0.10 g of alcohol for every  dl  of blood.
  5 L = 50 dl
  50 * 0.10 g = 5 g
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7 0
3 years ago
Read 2 more answers
Can someone help with this please?
PtichkaEL [24]
A and b are related because if u look carefully at what its showing u that there both the same but what its telling u is not the same.

a and c are the same but the picture is different and also the way they describe it is different and what they want u to look at is that if you look at it closely then youll know the difference and how to find it as well
8 0
3 years ago
Iron(II) sulfide has a primitive cubic unit cell with sulfide ions at the lattice points.
masya89 [10]

We have that the  the density of FeS  is mathematically given as

  •  \phi=2.56h/cm^3

From the question we are told

Iron(II) sulfide has a primitive <em>cubic</em> unit cell with <em>sulfide</em> ions at the <em>lattice points.</em>

The ionic radii of iron(II) ions and sulfide ions are 88 pm and 184 pm, respectively.

What is the density of FeS (in g/cm3)?

<h3>Density</h3>

Generally the equation for the Velocity  is mathematically given as

V_c=a^3=(2\pi Fe^{2+}+2\pi S^{2-})^3\\\\Therefore\\\\V=a^3(2\pi*0.088+2\pi 0.184)^3\\\\V=16.98*10^{-23}\\\\Therefore\\\\\phi=n\frac{PFeion+PSion}{VNa}\\\\\phi=3*\frac{55.85+32}{16.9*10^{-23}*6.023*10^{23}}

  • \phi=2.56h/cm^3
  • \phi=2.56h/cm^3

For more information on ionic radii visit

brainly.com/question/13981855

4 0
2 years ago
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