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shepuryov [24]
4 years ago
9

Metamorphic rocks that melt in a hot fluid can re-crystallize into a different form of rock. true or false

Chemistry
1 answer:
Scilla [17]4 years ago
5 0
Yes metamorphic rocks melt in hot fluid
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Listed in the Item Bank are some important labels for sections of the image below. To find out more information about labels, so
daser333 [38]

Answer:

Explanation:

Electrons have negative charge and they orbit around the nucleus.

Protons have positive charge and are inside the nucleus

Neutrons have neutral charge they are also inside the nucleus

the nucleus is the middle of the atom

the electron cloud is a cloud which the electrons travel around

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3 years ago
Which of the following is the correct definition of conduction?
KIM [24]

Answer:

B. The transition of heat across matter

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How many grams of fosforic acid (H3PO4) are required to prepare 500 mL of a 0.2 M solution?
navik [9.2K]

Answer:

Mass = 9.8 g

Explanation:

Given data:

Molarity of solution = 0.2 M

Volume of solution = 500 mL

Number of grams of phosphoric acid = ?

Solution:

First of all we will convert the volume milliliter to litter.

500 mL × 1 L/1000 mL

0.5 L

Molarity = moles of solute / volume in litter

0.2 M =  number of moles / volume in litter

Number of moles = 0.2 mol/L × 0.5 L

Number of moles = 0.1 mol

Number of grams:

Number of moles = mass/molar mass

Mass = number of moles × molar mass

Mass = 0.1 mol × 98 g/mol

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7 0
3 years ago
Assuming dopant atoms are uniformly distributed in a silicon crystal, how far apart are these atoms when the doping concentratio
enyata [817]

Answer:

d =~ 5.8μm

d =~ 0.13 μm

Explanation:

when the doping concentrations are 5 × 10^15 cm^-3

d = v^-1/3  ; where d represent the distance between the atoms , and v  represent the volume

d =1/ ∛v

d = 1/ ∛5 × 10^15

d = 1/ 170997.5

d = 5.85 × 10 ^ -6

d =~ 5.8μm

when the doping concentrations are 5 × 10^20 cm^-3

d = v^-1/3  ; where d represent the distance between the atoms , and v  represent the volume

d =1/ ∛v

d = 1/ ∛5 × 10^20

using the principle of surds and standard forms, we have

d = 1/ ∛0.5 × 10^21  

d = 1/7937005.26

d = 1.26 × 10 ^ -7

d = 0.126 × 10 ^ -6

d =~ 0.13 μm

8 0
3 years ago
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