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deff fn [24]
3 years ago
12

What is a zone of weak, variable winds located at 30 degrees north and 30 degrees south?

Chemistry
1 answer:
timurjin [86]3 years ago
4 0

Answer:

Horse latitude, trade winds

Explanation:

  • The area of the low pressure or the calm consists of the variable light winds that blow near the equator are known to the marines as the doldrums and they form a circuital pattern near the earth atmosphere.
  • Forms at a center of the near the higher pressure systems called as the horse latitudes where the  trade winds at the surface are weak and variable and this zone is found generally in latitudes of the 30° North and South of the equator and move in an east to west direction.

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1.42 mol sample of neon gas at a temperature of 13.0 °C is found to occupy a volume of 25.5 liters. The pressure of this gas sam
Dima020 [189]

Answer: 996 mmHg

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at NTP, weighs equal to the molecular mass and contains avogadro's number 6.023\times 10^{23} of particles.

According to the ideal gas equation:

PV=nRT

P = Pressure of the gas = ?

V= Volume of the gas = 25.5 L

T= Temperature of the gas = 13°C = (273+13) K  = 286K

R= Gas constant = 0.0821 atmL/K mol

n=  moles of gas= 1.42

P=\frac{nRT}{V}=\frac{1.42\times 0.0821\times 286}{25.5}=1.31atm=996mmHg      (760mmHg=1atm)

Thus pressure of this gas sample is 996 mm Hg.

3 0
3 years ago
What happened to the salt molecules when they were added to cold water
Nimfa-mama [501]
Cooler water molecules are denser than warm water and will not allow much of the salt to dissolve
8 0
2 years ago
Which of the following is best revealed by the bohr model?
N76 [4]
D. Electrons orbit in a fixed shell
4 0
3 years ago
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The reaction of nitrogen with methane in excess oxygen produces water, ________, and ___________. N2(g) + O2(g) + CH4(g) → CO2(g
IrinaVladis [17]

Answer:

C.

Explanation:

3 0
2 years ago
Nitrogen and hydrogen combine to form ammonia in the Haber process. Calculate (in kJ) the standard enthalpy change ∆H° for the r
Kazeer [188]

Answer:

∆H° rxn = - 93 kJ

Explanation:

Recall that a change in standard in enthalpy, ∆H°, can be calculated from the inventory of the energies, H, of the bonds  broken minus bonds formed (H according to Hess Law.

We need to find in an appropiate reference table the bond energies for all the species in the reactions and then compute the result.

              N₂ (g)   +            3H₂ (g)   ⇒                          2NH₃ (g)

1 N≡N = 1(945 kJ/mol)     3 H-H = 3 (432 kJ/mol)       6 N-H = 6 ( 389 kJ/mol)

∆H° rxn = ∑  H bonds broken  - ∑ H bonds formed

∆H° rxn = [ 1(945 kJ)   + 3 (432 kJ) ] - [ 6 (389 k J]

∆H° rxn = 2,241 kJ -2334 kJ = -93 kJ

be careful when reading values from the reference table since you will find listed N-N bond energy (single bond), but we have instead a triple bond,  N≡N,  we have to use this one .

8 0
3 years ago
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