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2 years ago

The surface of the earth is divided into more than 50 large plates

Chemistry

1 answer:

Romashka-Z-Leto [24]2 years ago

5 0

Major Plates

Africa Plate
Antarctic Plate
Indo-Australian Plate
Australian Plate
Eurasian Plate
North American Plate
South American Plate
Pacific Plate
Minor Plates
There are dozens of smaller plates, the seven largest of which are:
Arabian Plate
Caribbean Plate
Juan de Fuca Plate
Cocos Plate
Nazca Plate
Philippine Sea Plate
Scotia Plate

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The planet Mercury takes 176 days to do one spin around its pole. Compared to the Earth,

Tanya [424]

Answer:

The sun would appear to move more slowly across Mercury's sky.

Explanation:

This is because, the time it takes to do one spin or revolution on Mercury is 176 days (which is its period), whereas, the time it takes to do one spin or revolution on the Earth is 1 day.

Since the angular speed ω = 2π/T where T = period

So on Mercury, T' = 176days = 176 days × 24 hr/day × 60 min/hr × 60 s/min = ‭15,206,400‬ s

So, ω' = 2π/T'

= 2π/15,206,400‬ s

= 4.132 × 10⁻⁷ rad/s

So on Earth, T" = 1 day = 1 day × 24 hr/day × 60 min/hr × 60 s/min = ‭‭86,400‬ s

So, ω" = 2π/T"

= 2π/86,400‬ s

= 7.272 × 10⁻⁵ rad/s

Since ω' = 4.132 × 10⁻⁷ rad/s << ω" = 7.272 × 10⁻⁵ rad/s, the sun would appear to move more slowly across Mercury's sky.

4 0

2 years ago

HELP WITH CHEMISTRY PLEASE!

maria [59]

Answer:

1) 1.52 atm.

2) 647.85 K.

3) 20.56 L.

4) 1.513 mole.

5) 254.22 K = -18.77 °C.

Explanation:

In all this points, we should use the law of ideal gas to solve this problem: PV = nRT.
Where, P is the pressure (atm), V is the volume (L), n is the number of moles, R is the general gas constant (0.082 L.atm/mol.K), and T is the temperature (K).

1) In this point; n, R, and T are constants and the variables are P and V.

P and V are inversely proportional to each other that if we have two cases we get: P1V1 = P2V2.

In our problem:

P1 = ??? (is needed to be calculated) and V1 = 45.0 L.

P2 = 5.7 atm and V2 = 12.0 L.

Then, the original pressure (P1) = P2V2 / V1 = (5.7 atm x 12.0 L) / (45.0 L) = 1.52 atm.

2) In this case, n and R are the constants and the variables are P, V, and T.

P and V are inversely proportional to each other and both of them are directly proportional to the temperature of the gas that if we have two cases we get: P1V1T2 = P2V2T1.

In our problem:

P1 = 212.0 kPa, V1 = 32.0 L, and T1 = 20.0 °C = (20 °C + 273) = 293 K.

P2 = 300.0 kPa, V2= 50.0 L, and T2 = ??? (is needed to be calculated)

Then, the temperature in the second case (T2) = P2V2T1 / P1V1 = (300.0 kPa x 50.0 L x 293 K) / (212.0 kPa x 32.0 L) = 647.85 K.

3) In this case, P, n and R are the constants and the variables are V, and T.

V and T are directly proportional to each other that if we have two cases we get: V1T2 = V2T1.

In our problem:

V1 = 25.0 L and T1 = 65.0 °C + 273 = 338 K.

V2 = ??? (is needed to be calculated) and T2 = 5.0 °C + 273 = 278 K.

Herein, there is no necessary to convert T into K.

Then, the volume in the second case (V2) = V1T2 / T1 = (25.0 L x 278 °C) / (338 °C) = 20.56 L.

4) We can get the number of moles that will fill the container from: n = PV/RT.

P = 250.0 kPa, we must convert the unit from kPa to atm; 101.325 kPa = 1.0 atm, then P = (1.0 atm x 250.0 kPa) / (101.325 kPa) = 2.467 atm.

V = 16.0 L.

R = 0.082 L.atm/mol.K.

T = 45 °C + 273 = 318 K.

Now, n = PV/RT = (2.467 atm x 16.0 L) / (0.082 L.atm/mol.K x 318 K) = 1.513 mole.

5) In this case, V, n and R are the constants and the variables are P, and T.

P and T are directly proportional to each other that if we have two cases we get: P1T2 = P2T1.

In our problem:

P1 = 2200.0 mmHg and T1 = ??? (is needed to be calculated) .

P2 = 2700.0 mmHg and T2 = 39.0 °C + 273 = 312.0 K.

Herein, there is no necessary to convert P into atm.

Then, the temperature in the morning (T1) = P1T2 / P2 = (2200.0 mmHg x 312.0 K) / (2700.0 mmHg) = 254.22 K = -18.77 °C.

6 0

3 years ago

Read 2 more answers

15.1 L N2 at 25 °C and 125 kPa and 44.3 L O2 at 25 °C and 125 kPa were transferred to a tank with a volume of 6.25 L. What is th

Y_Kistochka [10]

Answer:

The total pressure of the mixture in the tank of volume 6.25 litres at 51°C is 1291.85 kPa.

Explanation:

For N2,

Pressure(P₁)=125 kPa

Volume(V₁)=15·1 L

Temperature (T₁)=25°C=25+273 K=298 K

Similarly, for Oxygen,

Pressure(P₂)= 125 kPa

Volume(V₂)= 44.3 L

Temperature(T₂)=25°C= 298 K

Then, for the mixture,

Volumeof the mixture( V)= 6.25 L

Pressure(P)=?

Temperature (T)= 51°C = 51+273 K=324 K

Then, By Combined gas laws,

$\frac{P_{1} V_{1} }{T_{1} } +\frac{P_{2} V_{2} }{T_{2} } =\frac{PV}{T}$

or, $\frac{15.1*125}{298} +\frac{44.3*125}{298} =\frac{P*6.25}{324}$

or, $6.34+18.58=\frac{P*6.25}{324}$

or, $P=\frac{24.92*324}{6.25}$

∴P=1291.85 kPa

So the total pressure of the mixture in the tank of volume 6.25 litres at 51°C is 1291.85 kPa.

3 0

2 years ago

Be able to describe the relationship between Freedom of Movement and phase change.

lisabon 2012 [21]

Ion knoe Wdym by “be able to describ’ so ima put it in my own words idr lol:)

if you talm bout some kentic energy or sum ok but other Dan dat ion knoe tbh

I can explain how transferring kinetic energy in and out of a substance can cause a change

7 0

3 years ago

What mass of aluminum is produced by the decomposition of 5.0 kg al2o3?

Nina [5.8K]

The mass for of aluminum that is produced by the decomposition of 5.0 Kg Al2O3 is 2647 g or 2.647 Kg

calculation
Write the equation for decomposition of Al2O3

Al2O3 = 2Al + 3 O2

find the moles of Al2O3 = mass/molar mass

convert 5 Kg to g = 5 x1000 = 5000 grams
molar mass of Al2O3 = 27 x2 + 16 x3 = 102 g/mol

moles =5000 g/ 102 g/mol = 49.0196 moles

by use of mole ratio between Al2O3 to Al which is 1:2 the moles of Al = 49.0196 x2 =98.0392 moles

mass of Al = moles x molar mass

= 98.0392 moles x 27g/mol = 2647 grams or 2647/1000 = 2.647 Kg

7 0

3 years ago

Read 2 more answers