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expeople1 [14]
3 years ago
9

Which of the following processes is involved in the formation of sedimentary rocks?

Chemistry
2 answers:
AlladinOne [14]3 years ago
6 0
D. deposition

hope i can help
riadik2000 [5.3K]3 years ago
4 0

Answer:

deposition

Explanation:

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Under certain conditions, the solid and liquid states of water can exist in equilibrium. How are these conditions indicated on t
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In the phase diagram the line between solid and liquid is where both phases available at the same time
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If 13.0 g of MgSO4⋅7H2O is thoroughly heated, what mass of anhydrous magnesium sulfate will remain?
Paul [167]

6.349 g mass of anhydrous magnesium sulfate will remain.

<h3>What are moles?</h3>

A mole is defined as 6.02214076 × 1023 of some chemical unit, be it atoms, molecules, ions, or others. The mole is a convenient unit to use because of the great number of atoms, molecules, or others in any substance.

Molar mass MgSO₄.7 H₂O = 246.52 g/mol

Moles =\frac{mass}{molar \;mass}

Moles =\frac{13.0 g}{246.52}

0.0527 moles

Molar mass MgSO₄ = 120.4 g/mol

Mass of anhydrous magnesium sulfate :

( 0.0527 x 120.4 ) => 6.349 g

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brainly.com/question/8455949

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2 years ago
Over-encouraging lesser skilled players can lead to embarrassment.<br> a. True<br> b. False
Margarita [4]
My answer would be a. True
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4 years ago
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Convert 11.03 moles of calcium nitrate to grams
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Im pretty sure the answer is <span> 0.01218859659g

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6 0
3 years ago
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A weather balloon is filled with helium that occupies a volume of 5.37 104 L at 0.995 atm and 32.0°C. After it is released, it r
svp [43]

Answer:

The new volume is 63583 L

Explanation:

Step 1: Data given

The initial volume of the balloon = 5.37 * 10^4 L

The initial pressure = 0.995 atm

The initial temperature = 32.0 °C = 305.15 K

The pressure decreased to 0.720 atm

The temperature decreased to -11.7 °C = 261.45 K

Step 2: Calculate the new volume

P1*V1 / T1 = P2*V2/T2

⇒with P1 = the initial pressure = 0.995 atm

⇒with V1 = the initial volume = 5.37 *10^4 L

⇒with T1 = the initial temperature = 305.15 K

⇒with P2 = the decreased pressure = 0.720 atm

⇒with V2 = the new volume = TO BE DETERMINED

⇒with T2 = the decreased temperature : 261.45 K

(0.995 * 5.37*10^4)/305.15 = (0.720 * V2) / 261.45

V2 = 63583 L

The new volume is 63583 L

8 0
3 years ago
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