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Free_Kalibri [48]
3 years ago
6

During the processes of erosion and deposition, sediments that are the ________ in size will be carried the greatest distances b

efore being deposited
(There's no choices it's just fill in the blank)
Also
Most _______ rocks form under conditions found a few kilometers under earths surface
(No choices,fill in the blank)
Chemistry
2 answers:
Mila [183]3 years ago
4 0
<span>During the processes of erosion and deposition, sediments that are the smaller </span>in size will be carried the greatest distances before being deposited. Most metamorphic rocks form under conditions found a few kilometers under earths surface
Ahat [919]3 years ago
3 0

The full sentences are given below:

1. During the process of erosion and deposition, sediments that are the SMALLEST in size will be carried the greatest distance before being deposited.

Erosion and deposition are the methods by which sand and rock particles are moved from one place to another. The erosion can be caused by water or wind. Water and wind have the capacity to transport particle from one location and deposit them in another location. How far the erosion is able to move the particles depend on the weight of the particles. It is easier for erosion to carry small particles over a long distance than for it to carry large particles over the same distance.

2. Most METAMORPHIC rocks form under conditions found a few kilometer under the earth surface.

Metamorphic rocks generally are formed from existing rocks. The existing rocks are usually subjected to heat and pressure, which cause radical changes in the chemical and physical properties of the rock. Metamorphic rocks can be formed underneath the earth surface if they are subjected to high temperature and pressure by the rock layers above them.

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Mass = 200.0 g volume = 100.0cm3 What is the density?
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The density of an object is defined as its mass divided by its volume. Mathematically, density = Mass / Volume. The unit of density is kilogram per cubic meter, kg / m^3 or g /cm^3.
For the question given above: the 
Mass = 200.0 g
Volume = 100.0 cm^3
Therefore, Density = Mass / Volume = 200 / 100  = 2
Thus,  the density of the object is 2 g /cm^3.
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Use the standard reaction enthalpies given below to determine ΔH°rxn for the following reactionP4(g) + 10 Cl2(g) → 4PCl5(s) ΔH°r
weqwewe [10]

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Therefore  \bigtriangledown H^\circ_{rxn}= -1835 KJ

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Enthalpy is denoted by H.

Enthalpy: Total heat change in a chemical reaction is called enthalpy.

The change of entalpy of a reaction is denoted by \bigtriangledown H^\circ_{rxn}

Hass's Law:The change in enthalpy of any process can be determined by calculating the sum of change in enthalpy of each of the steps involved in the process.

g= gas

S= solid

P₄(g)+10Cl₂(g)→ 4Cl₅(s)       \bigtriangledown H^\circ_{rxn}=?

PCl₅(s)→ PCl₃(g)+Cl₂(g) .......(1)       \bigtriangledown H^\circ_{rxn}= +157KJ

P₄(g)+6Cl₂(g)→  4PCl₃(g).............(2)     \bigtriangledown H^\circ_{rxn}= -1207 KJ

If we flip a reaction the value of enthalpy will be change positive to negative or nagative to positive but the numerical value will be remain same.

We need rearrange the equation (1) because in the required equation Cl₂ is on the left side. So we flip the first equation.

PCl₃(g)+Cl₂(g)→PCl₅(s)......(3)          \bigtriangledown H^\circ_{rxn}= -157KJ

Multiplying 4 with equation (3)

4 PCl₃(g)+4Cl₂(g)→4PCl₅(s)......(4)          \bigtriangledown H^\circ_{rxn}=4×( -157)KJ= -628 KJ

Adding equation (2) and (4) we get

P₄(g)+6Cl₂(g)+4 PCl₃(g)+4Cl₂(g)→4PCl₃(g)+4PCl₅(s)    \bigtriangledown H^\circ_{rxn}=( -1207-628)KJ

⇒P₄(g)+10Cl₂(g)→4PCl₃(g)-4PCl₃(g)+4PCl₅(s)      \bigtriangledown H^\circ_{rxn}= - 1835KJ

⇒P₄(g)+10Cl₂(g)→ 4Cl₅(s)       \bigtriangledown H^\circ_{rxn}= -1835 KJ

Therefore  \bigtriangledown H^\circ_{rxn}= -1835 KJ

5 0
3 years ago
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