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Dominik [7]
3 years ago
13

________ is a strong, parallel alignment of coarse mica flakes and/or of different mineral bands in a metamorphic rock.

Chemistry
1 answer:
kari74 [83]3 years ago
3 0

Answer:

Foliation is a strong, parallel alignment of coarse mica flakes and/or of different mineral bands in a metamorphic rock.

Explanation:

Foliation is most commonly prevalent in metamorphic rocks. Foliation is the parallel alignment of textural and structural features of a rock.

Differential stress plays a major role on the texture of metamorphic rocks because it forces the mineral constituent of the rocks to align parallel to each other.  Foliation can occur in different ways for example a mineral like mica which is usually platy can crystallize in rocks , due to differential stress the mineral grows in such a way it remains parallel to the movement in which the part of the rock slide relative to one another and parallel to the forces applied or the mineral might grow perpendicular to the direction of the compressed stress.  

Notice the foliation in the texture of gneiss .You can see the light and dark mineral found in separate or parallel layer.

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In an experiment, the molar mass of the compound was determined to
sasho [114]
Determine the mass in grams of each element in the sample. If you are given percent composition, you can directly convert the percentage of each element to grams.
For example, a molecule has a molecular weight of 180.18 g/mol. It is found to contain 40.00% carbon, 6.72% hydrogen and 53.28% oxygen.
Convert the percentages to grams.
40.00 grams of carbon
6.72 grams of hydrogen
53.28 grams of oxygen
3 0
3 years ago
A space air is at a temperature of 75 oF, and the relative humidity (RH) is 45%. Using calculations, find: (a) the partial press
earnstyle [38]

Answer:

A) Partial Pressure of dry air = 13.32 KPa

Partial Pressure of water vapour = 1.332 KPa

B) Humidity ratio; X = 0.0691

C) V_p = 0.8384 m³/Kg

Explanation:

A) We are given;

Temperature = 75°F

Relative Humidity = 45%

Now,to calculate the partial pressure, we will use the relationship;

Relative Humidity = (Partial Pressure/Vapour Pressure) × 100%

Making partial pressure the subject;

Partial Pressure = Relative Humidity × Vapour Pressure/100%

From the first table attached, at temperature of 75°F, the vapor pressure is 29.6 × 10^(-3) bar = 29.6 KPa

Thus;

Partial Pressure of dry air = (45 × 29.6)/100

Partial Pressure of dry air = 13.32 KPa

From online values, vapour pressure of water vapour at 75°F = 2.96 KPa

Thus;

Partial Pressure of water vapour = (45 × 2.96)/100 = 1.332 KPa

B) humidity ratio of moist air is given as;

X = 0.62198 pw / (pa - pw)  

where;

pw = partial pressure of the water vapor in moist air

pa = atmospheric pressure of the moist air

Thus;

X = (0.62198 × 1.332)/(13.32 - 1.332)  

X = 0.0691

C) Formula for moist air specific volume is;

V_p = (1 + (xRw/Ra) × RaT/p

Where;

V_p is specific volume

T is temperature = 75°F = 297.039 K

p is barometric pressure which in this case is standard sea level pressure = 101.325 KPa

pw is partial pressure of the water vapor in moist air = 1.332 KPa

Rw is individual gas constant for water = 0.4614 KJ/Kg.K

Ra is individual gas constant for air = 0.2869 KJ/Kg.K

V_p = (1 + (0.0691 * 0.4614/0.2869)) × 0.286.9 * 297.039/101.325

V_p = 0.8384 m³/Kg

6 0
3 years ago
A buret is filled with 0.1517 M naoh A 25.0 mL portion of an unknown acid and two drops of indicator are added to an Erlenmeyer
antiseptic1488 [7]

Answer:

molarity of acid =0.0132 M

Explanation:

We are considering that the unknown acid is monoprotic. Let the acid is HA.

The reaction between NaOH and acid will be:

NaOH+HA--->NaA+H_{2}O

Thus one mole of acid will react with one mole of base.

The moles of base reacted = molarity of NaOH X volume of NaOH

The volume of NaOH used = Final burette reading - Initial reading

Volume of NaOH used = 22.50-0.55= 21.95 mL

Moles of NaOH = 0.1517X21.95=3.33 mmole

The moles of acid reacted = 3.33 mmole

The molarity of acid will be = \frac{mmole}{volumne(mL)}=\frac{0.33}{25}=0.0132M

4 0
3 years ago
What is the difference between pure and applied chemistry?
bekas [8.4K]

Answer:

Explanation:

The major difference between pure and applied chemistry is the purpose and intent of the study.

Pure chemistry deals with the study of matter, matter transformations, and interactions between the different materials of the world, for only the sake of gaining empirical knowledge about the various substances that exist in the world. It does not really seek to apply this knowledge to do anything industrial.

Applied chemistry is the study of chemistry with the aim of utilizing this knowledge to solve the various problems that man faces. This approach of study is not for knowledge sake alone, rather it is for industrial application

3 0
3 years ago
A wooden tool is found at an archaeological site. Estimate the age of the tool using the following information: A 100 gram sampl
lord [1]

Answer:

2578.99 years

Explanation:

Given that:

100 g of the wood is emitting 1120 β-particles per minute

Also,

1 g of the wood is emitting 11.20 β-particles per minute

Given, Decay rate = 15.3 % per minute per gram

So,

Concentration left can be calculated as:-

C left = [A_t]=\frac{11.20\ per\ minute}{15.3\ per\ minute\ per\ gram}\times [A_0]= 0.7320[A_0]

Where,  

[A_t] is the concentration at time t

[A_0] is the initial concentration

Also, Half life of carbon-14 = 5730 years

t_{1/2}=\frac {ln\ 2}{k}

Where, k is rate constant

So,  

k=\frac {ln\ 2}{t_{1/2}}

k=\frac {ln\ 2}{5730}\ years^{-1}

The rate constant, k = 0.000120968 year⁻¹

Time =?

Using integrated rate law for first order kinetics as:

[A_t]=[A_0]e^{-kt}

So,  

\frac {[A_t]}{[A_0]}=e^{-0.000120968\times t}

\frac {0.7320[A_0]}{[A_0]}=e^{-0.000120968\times t}

0.7320=e^{-0.000120968\times t}

ln\ 0.7320=-0.000120968\times t

<u>t = 2578.99 years</u>

8 0
3 years ago
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