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miskamm [114]
3 years ago
10

Which processes lead to the formation of the distorted, wavy patterns found in metamorphic rock?

Chemistry
1 answer:
vova2212 [387]3 years ago
5 0
<span>The process that lead to the formation of the distorted, wavy patterns found in metamorphic rock is regional metamorphism as heat and pressure increases, minerals become segregated into bands.

Hope this helps.
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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
xenn [34]
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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3 years ago
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How many moles are there in 987 grams of Ra(OH)2? id also love to know the simplest equation for converting moles --&gt; so i co
OlgaM077 [116]
Mole = \frac{mass}{molar mass}

Mass of Radium Hydroxide = 987g
Molar Mass of Ra(OH) _{2} = ((226) + (2 * 16) + (2 * 1))
                                            =  260 g / mol

∴  Moles of <span>Ra(OH) _{2} = </span>\frac{987 g }{ 260 g / mol }
                                            = 3.796 mols
6 0
3 years ago
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Advocard [28]

Answer:

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4 0
3 years ago
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Arturiano [62]
In the same period number 2
4 0
3 years ago
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Suppose that 25.0 mL of 0.10 M CH3COOH (aq) is titrated with 0.10 M NaOH (aq). What is the pH after the addition of 10.0 mL of 0
olya-2409 [2.1K]

Answer:

pH=-1.37

Explanation:

We are given that 25 mL of 0.10 M CH_3COOH is titrated with 0.10 M NaOH(aq).

We have to find the pH of solution

Volume of CH_3COOH=25mL=0.025 L

Volume of NaoH=0.01 L

Volume of solution =25 +10=35 mL=\frac{35}{1000}=0.035 L

Because 1 L=1000 mL

Molarity of NaOH=Concentration OH-=0.10M

Concentration of H+= Molarity of CH_3COOH=0.10 M

Number of moles of H+=Molarity multiply by volume of given acid

Number of moles of H+=0.10\times 0.025=0.0025 moles

Number of moles of OH^-=0.10\times 0.01=0.001mole

Number of moles of H+ remaining after adding 10 mL base = 0.0025-0.001=0.0015 moles

Concentration of H+=\frac{0.0015}{0.035}=4.28\times 10^{-2} m/L

pH=-log [H+]=-log [4.28\times 10^{-2}]=-log4.28+2 log 10=-0.631+2

pH=-1.37

6 0
3 years ago
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