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ad-work [718]
3 years ago
5

The crystals that form in a magma that cools more slowly than those from another magma will be _________.

Chemistry
2 answers:
maksim [4K]3 years ago
8 0

Answer:

The correct answer is A.

Explanation:

Magmatic rocks are formed by the cooling effect that affects a magma. If the cooling that occurs is slow, in the interior of the Earth, the minerals have a long time to crystallize and form compact rocks of large and long crystals. If this cooling occurs faster when magma comes to the surface, the rocks have small crystals or no crystals at all.

Have a nice day!

xenn [34]3 years ago
5 0

The answer to this is A, Larger.

If the Magma cools quickly, the crystals do not have much time to form but if it cools slowly it will then form into a large crystal.

Hope this helped :) have a great day

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Suggest why sodium wasn’t produced until the early 19th century.
AlexFokin [52]

Answer:

The reason is that sodium attaches itself very strongly to other elements. Its compounds are very difficult to break apartand also because it is so reactive.

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3 years ago
One degree Celsius indicates the same temperature change as
Harlamova29_29 [7]
B 5/9 degree Fahrenheit
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3 years ago
Estimate the number of moles of water in all the Earth's oceans. Assume water covers 75% of the Earth to an average depth of 3 k
o-na [289]

Answer:

there are approximately n ≈ 10²² moles

Explanation:

Since the radius of the earth is approximately R=6378 km= 6.378*10⁶ m , then the surface S of the earth would be

S= 4*π*R²

since the water covers 75% of the Earth's surface , the surface covered by water Sw is

Sw=0.75*S

the volume for a surface Sw and a depth D= 3 km = 3000 m ( approximating the volume through a rectangular shape) is

V=Sw*D

the mass of water under a volume V , assuming a density ρ= 1000 kg/m³ is

m=ρ*V

the number of moles n of water ( molecular weight M= 18 g/mole = 1.8*10⁻² kg/mole ) for a mass m is

n = m/M

then

n = m/M = ρ*V/M = ρ*Sw*D/M = 0.75*ρ*S*D/M = 3/4*ρ*4*π*R² *D/M = 3*π*ρ*R² *D/M

n=3*π*ρ*R² *D/M

replacing values

n=3*π*ρ*R² *D/M = 3*π*1000 kg/m³*(6.378*10⁶ m)² *3000 m /(1.8*10⁻² kg/mole) = 3*π*6.378*3/1.8 * 10²⁰ = 100.18 * 10²⁰ ≈ 10²² moles

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8 0
3 years ago
calculate the difference in slope of the chemical potential against temperature on either side of the normal freezing point of w
kipiarov [429]

Answer:

(a) The normal freezing point of water (J·K−1·mol−1) is -22Jmole^-^1k^-^1

(b) The normal boiling point of water (J·K−1·mol−1) is -109Jmole^-^1K^-^1

(c) the chemical potential of water supercooled to −5.0°C exceed that of ice at that temperature is  109J/mole

Explanation:

Lets calculate

(a) - General equation -

      (\frac{d\mu(\beta )}{dt})p-(\frac{d\mu(\alpha) }{dt})_p = -5_m(\beta )+5_m(\alpha ) =  -\frac{\Delta H}{T}

 \alpha ,\beta → phases

ΔH → enthalpy of transition

T → temperature transition

 (\frac{d\mu(l)}{dT})_p -(\frac{d\mu(s)}{dT})_p == -\frac{\Delta_fH}{T_f}

            = \frac{-6.008kJ/mole}{273.15K} ( \Delta_fH is the enthalpy of fusion of water)

           = -22Jmole^-^1k^-^1

(b) (\frac{d\mu(g)}{dT})_p-(\frac{d\mu(l)}{dT})_p= -\frac{\Delta_v_a_p_o_u_rH}{T_v_a_p_o_u_r}

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(c) \Delta\mu =\Delta\mu(l)-\Delta\mu(s) =-S_m\DeltaT

[\mu(l-5°C)-\mu(l,0°C)] =  [\mu(s-5°C)-\mu(s,0°C)]=-S_mΔT

\mu(l,-5°C)-\mu(s,-5°C)=-Sm\DeltaT [\mu(l,0

\Delta\mu=(21.995Jmole^-^1K^-^1)\times (-5K)

     = 109J/mole

6 0
2 years ago
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Natali5045456 [20]

Answer:

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6. What are laws? the protection of all people within a society EVERYBODY SHOULD BE EQUALLY PROTECTED UNDER THE LAW

6 0
3 years ago
Read 2 more answers
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