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2 years ago

the various geologic settings in which rocks can melt, metamorphose, or become sediment are ultimately generated by

1 answer:

3 0

Geological settings whereby there is melting of rock as well as formation of sediments can be attributed to Plate tectonics .

Plate tectonics can be regarded as theory which explained the Earth's outer shell and how its been divided into solid rocks as well as slabs and its often referred to as Plate.

Plate tectonics explained that there is a large-scale motion plates which Earth's lithosphere is made up.

Tectonic processes started about 3.3 and 3.5 billion years ago.

Through the generation of Tectonic plate, there are melting of rocks as well as generation of sediments

The types of boundaries that exist between those tectonic plates can be categorized into 4, base on the the movement of the plates, and these are;

divergent boundaries
convergent boundaries
transform fault boundaries
plate boundary zones

Therefore, plate tectonic gives explanation about Earth's outer shell.

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Which instrument is used to observe space system from Earth?

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Answer:

The telescope is used.

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2 years ago

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Which metal can be used as a sacrificial electrode to prevent the rusting of an iron pipe?

Pepsi [2]

manganese The metal that is used as a sacrificial electrode to prevent the rusting of iron is manganese.

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3 years ago

A rigid tank contains 0.66 mol of oxygen (O2). Find the mass of oxygen that must be withdrawn from the tank to lower the pressur

dsp73

Answer:

12.8 g of $O_{2}$ must be withdrawn from tank

Explanation:

Let's assume $O_{2}$ gas inside tank behaves ideally.

According to ideal gas equation- $PV=nRT$

where P is pressure of $O_{2}$ , V is volume of $O_{2}$ , n is number of moles of $O_{2}$ , R is gas constant and T is temperature in kelvin scale.

We can also write, $\frac{V}{RT}=\frac{n}{P}$

Here V, T and R are constants.

So, $\frac{n}{P}$ ratio will also be constant before and after removal of $O_{2}$ from tank

Hence, $\frac{n_{before}}{P_{before}}=\frac{n_{after}}{P_{after}}$

Here, $\frac{n_{before}}{P_{before}}=\frac{0.66mol}{43atm}$ and $P_{after}=17atm$

So, $n_{after}=\frac{n_{before}}{P_{before}}\times P_{after}=\frac{0.66mol}{43atm}\times 17atm=0.26mol$

So, moles of $O_{2}$ must be withdrawn = (0.66 - 0.26) mol = 0.40 mol

Molar mass of $O_{2}$ = 32 g/mol

So, mass of $O_{2}$ must be withdrawn = $(32\times 0.40)g=12.8g$

7 0

2 years ago

students used a balance and a graduated cylinder to collect the data shown in table 7. calculate the density of the sample. if t

Licemer1 [7]

Answer:

Percentage error = 1.88 %

Solution:

Data Given:

Mass of Sample = 20.46 g

Volume of Sample = 43.0 mL - 40.0 mL = 3.0 mL

Formula Used:

Density = Mass / Volume

Putting values,

Density = 20.46 g / 3.0 mL

Density = 6.82 g.mL⁻¹

Percentage Error:

Experimental Value = 6.82 g.mL⁻¹

Accepted Value = 6.95 g.mL⁻¹

= 6.82 g.mL⁻¹ / 6.95 g.mL⁻¹ × 100 = 98.12 %

Percentage Error = 100 % - 98.12 %

Percentage error = 1.88 %

3 0

2 years ago

A cylinder contains 250 g of Helium at 200 K. The external pressure is constant at 1 atm. The temperature of the system is raise

Black_prince [1.1K]

Answer:

There is 96200 J or 96 kJ of heat released.

Explanation:

Step 1: Data given

Mass of helium = 250 grams

Temperature = 200 K

Temperature is raised by 74 K

The heat capacity of helium = 20.8 J/mol*K

Molecular weight for helium = 4 g/mol

Step 2: Calculate moles of Helium

Moles of Helium = mass of Helium / molar mass of helium

Moles of helium = 250 grams / 4g/mol

Moles of helium = 62.5 moles

Step 3: Calculate heat

Q = n*c*ΔT or

⇒ with Q = the heat released in Joule

⇒ with n = moles of helium = 62.5 moles

⇒ with c = the heat capacity of helium = 20.8 J/mol*K

⇒ with ΔT = the change in temperature = T2 - T1 = 74 K

Q = 62.5 mol * 20.8 J/mol*K * 74 K

Q = 96200 J = 96 kJ

Since the temperature is raised, this is an exothermic reaction. The heat is released.

There is 96200 J or 96 kJ of heat released.

8 0

3 years ago