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2 years ago

Which statement accurately describes the movement of Earth's plates due to convection currents?

Chemistry

1 answer:

ValentinkaMS [17]2 years ago

3 0

Answer:

The correct answer is the third statement, that is, when the heated substance gets near to the surface, it starts to cool.

Explanation:

A current that arises within a fluid due to convection is termed as the convection current. In comparison to the surface, the core is extremely hot, thus, both core and the surface exhibit very distinct temperatures. The mantle rises towards the surface when it gets heated as it moves close to the core. However, when it moves towards the surface, the mantle starts to turn cool and denser.

Afterward, the magma, which is cool, again then begins to move close to the core, and gets heated and rises again, this phenomenon results in the emergence of convection currents.

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Ammonia gas is compressed from 21°C and 200 kPa to 1000 kPa in an adiabatic compressor with an efficiency of 0.82. Estimate the

Evgen [1.6K]

Explanation:

It is known that efficiency is denoted by $\eta$ .

The given data is as follows.

$\eta$ = 0.82, $T_{1}$ = (21 + 273) K = 294 K

$P_{1}$ = 200 kPa, $P_{2}$ = 1000 kPa

Therefore, calculate the final temperature as follows.

$\eta = \frac{T_{2} - T_{1}}{T_{2}}$

0.82 = $\frac{T_{2} - 294 K}{T_{2}}$

$T_{2}$ = 1633 K

Final temperature in degree celsius = $(1633 - 273)^{o}C$

= $1360^{o}C$

Now, we will calculate the entropy as follows.

$\Delta S = nC_{v} ln \frac{T_{2}}{T_{1}} + nR ln \frac{P_{1}}{P_{2}}$

For 1 mole, $\Delta S = C_{v} ln \frac{T_{2}}{T_{1}} + R ln \frac{P_{1}}{P_{2}}$

It is known that for $NH_{3}$ the value of $C_{v}$ = 0.028 kJ/mol.

Therefore, putting the given values into the above formula as follows.

$\Delta S = C_{v} ln \frac{T_{2}}{T_{1}} + R ln \frac{P_{1}}{P_{2}}$

= $0.028 kJ/mol \times ln \frac{1633}{294} + 8.314 \times 10^{-3} kJ \times ln \frac{200}{1000}$

= 0.0346 kJ/mol

or, = 34.6 J/mol (as 1 kJ = 1000 J)

Therefore, entropy change of ammonia is 34.6 J/mol.

3 0

2 years ago

Consider the reaction: 2 Al + 3Br2 → 2 AlBr3 Suppose a reaction vessel initially contains 5.0 mole Al and 6.0 mole Br2. What is

timama [110]

Answer:

4 moles of $AlBr_3$ and 1 mole of $Al$ will be present in the reaction vessel

Explanation:

The reaction given in the question is

$2Al + 3 Br_2$ ⇒ $2AlBr_3$

According to the stoichiometric coefficients of the reaction, 2 moles of $Al$ requires 3 moles of $Br_2$ so in this reaction, $Br_2$ is a limiting reagent. So we will consider that $Al$ is in excess.

Now,

Since 3 moles of $Br_2$ requires 2 moles of $Al$

So, for 6 moles of $Br_2$ the moles of $Al$ required = $\frac{2}{3} \times 6$ = 4 moles.

Moles of $Al$ remaining after the completion of reaction = 5 - 4 = 1 mole.

Again,

Since 3 moles of $Br_2$ produces 2 moles of $AlBr_3$

So, moles of $AlBr_3$ produced by 6 moles of $Br_2$ = $\frac{2}{3} \times 6$ = 4 moles.

Therefore, after the completion of reaction, 4 moles of $AlBr_3$ and 1 mole of $Al$ will be present in the reaction vessel.

5 0

2 years ago

You can purchase nitric acid in a concentrated form that is 70.3% HNO3 by mass and has a density of 1.41 g/mL. Describe exactly

Andrej [43]

The way to working out the numbers is to increase the measure of HNO3 required by the molarity to discover what number of moles you require: 0.115. You ought to have the capacity to make sense of the recipe weight H is 1, N is 14, O is 16. The result of the quantity of moles duplicated by the recipe weight ought to give an esteem in grams. You can utilize the thickness to change over to a volume of HNO3 to add to the right volume of water.

6 0

2 years ago

Read 2 more answers

When 100 ml of 1.0 M Na3PO4 is mixed with 100 ml of 1.0 M AgNO3,a yellow precipitate forms and Ag+ becomes negligibly small. Whi

jok3333 [9.3K]

Answer:

Option A

Explanation:

Number of millimoles of Na3PO4 = 1 × 100 = 100

Number of millimoles of AgNO3 = 1 × 100 = 100

When 1 mole of Na3PO4 is dissociated we get 3 moles of sodium ions and 1 mole of phosphate ion

When 1 mole of AgNO3 is dissociated, we get 1 mole of Ag+ and 1 mole of NO3-

As Ag+ concentration is negligible, the dissociated Ag+ ion must have form the precipitate with phosphate ion and as number of moles of Ag+ and phosphate ion are same, therefore the concentration of phosphate ion must be negligible

Here as 100 millimoles of Na3PO4 is there, we get 300 millimoles of Na+ and 100 millimoles of PO43-

And as 100 millimoles of AgNO3 is there, we get 100 millimoles of Ag+ and 100 millimoles of NO3-

∴ Increasing order of concentration will be PO43- < NO3- < Na+

4 0

2 years ago

Read 2 more answers

Ron and Hermione begin with 1.50 g of the hydrate copper(II)sulfate ∙ x-hydrate (CuSO4 ∙ xH2O), where x is an integer. Part of t

Gwar [14]

Answer

Explanation:

We can go about this using the percentage compositions.

First, we calculate the percentage composition of the copper sulphate. This is obtainable by using the mass.

0.96/1.5 * 100 = 64%

Hence the percentage by mass of the water present is 36%

The molar mass of the anhydrous sulphate is 64 + 32 +4(16) = 160g/mol

The molar mass of the water is 2(1) + 16 = 18g/mol

Not forgetting that it is in multiples of x, the total molar mass of the water is 18x moles

The total mass of the copper sulphate hydrate is 160+ 18x

Now how do we get x? Like it is said earlier, the percentage composition is constant.

Hence, 64/100 * (160 + 18x) = 160

16000 = 64(160 + 18x)

16000 = 10,240 + 1152x

16,000 - 10,240 = 1152x

1152x = 5760

x = 5760/1152

x = 5

7 0

2 years ago