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3 years ago

Wind in the atmosphere and currents in the ocean are caused by uneven heating that produces _____.

2 answers:

5 0

Hello,

Your answer to this question would be Convection.

Wind in the atmosphere and currents in the ocean are caused by uneven heating that produces convection.

Hope this helps

gulaghasi [49]3 years ago

4 0

The best answer would be Convection

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In an electron-dot structure of an element, the dots are used to represent ________.

Vlad [161]

Dots are used to represent electrons.

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4 years ago

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denis23 [38]

Answer: The answer is the second one

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3 years ago

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Catalyst

Hoochie [10]

a. volume of NO : 41.785 L

b. mass of H2O : 18 g

c. volume of O2 : 9.52 L

<h3>Further explanation</h3>

Given

Reaction

4 NH₃ (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (l)

Required

a. volume of NO

b. mass of H2O

c. volume of O2

Solution

Assume reactants at STP(0 C, 1 atm)

Products at 1000 C (1273 K)and 1 atm

a. mol ratio NO : O2 from equation : 4 : 5, so mo NO :

$\tt \dfrac{4}{5}\times 0.5=0.4$

volume NO at 1273 K and 1 atm

$\tt V=\dfrac{nRT}{P}=\dfrac{0.4\times 0.08206\times 1273}{1}=41.785~L$

b. 15 L NH3 at STP ( 1mol = 22.4 L)

$\tt \dfrac{15}{22.4}=0.67~mol$

mol ratio NH3 : H2O from equation : 4 : 6, so mol H2O :

$\tt \dfrac{6}{4}\times 0.67=1$

mass H2O(MW = 18 g/mol) :

$\tt mass=mol\times MW=1\times 18=18~g$

c. mol NO at 1273 K and 1 atm :

$\tt n=\dfrac{PV}{RT}=\dfrac{1\times 35.5}{0.08206\times 1273}=0.34$

mol ratio of NO : O2 = 4 : 5, so mol O2 :

$\tt \dfrac{5}{4}\times 0.34=0.425$

Volume O2 at STP :

$\tt 0.425\times 22.4=9.52~L$

5 0

3 years ago

Investigate Weathering and Erosion 1. Describe at least one part of the experimental procedure that you thought was essential to

zvonat [6]

Answer:

Controlling the environment is the most key procedures for getting good results.

Explanation:

The control environment for an experiment is the essential part for getting good results. In control environment, there is no or less chances of disruption

from the external environment which can cause the results of the data more acceptable. So the scientists prefers laboratory for performing experiment as compared to outer environment. So in my opinion for getting better results, the control environment is the most necessary experimental procedure.

4 0

3 years ago

Read 2 more answers

A 200.0mL closed flask contains 2.000mol of carbon monoxide gas and 2.000mol of oxygen gas at the temperature of 300.0K. How man

max2010maxim [7]

Answer:

There will react 0.400 moles of oxygen.

Explanation:

Step 1: Data given

Volume of the closed flask = 200.00 mL = 0.2 L

Number of moles of CO = 2.000 mol

Number of moles of O2 = 2.000 mol

Temperature = 300.0 K

Pressure decreases with 10%

Step 2: The balanced equation

2CO(g)+O2(g)⟶2CO2(g)

Step 3: Calculate the initial pressure of the flask before the reaction

P = nRT/V

⇒ with n = the number of moles (2.000 moles CO + 2.000 moles O2 = 4.000 moles)

⇒ R is gas constant (0.08206 atm*L/mol*K)

⇒T = the temperature = 300.0K

⇒ V = the volume = 200.0 mL = 0.2 L

P = (4 * 0.08206*300)/0.2

P = 492.36 atm

Step 4: When the pressure is 10 % decreased:

The final pressure = 492.36 - 49.236 = 443.124 atm

Step 5: Calculate the number of moles

n = PV/RT

⇒ with n = the number of moles

⇒ with P = the pressure = 443.124 atm

⇒ V = the volume = 200.0 mL = 0.2 L

⇒ R is gas constant (0.08206 atm*L/mol*K)

⇒T = the temperature = 300.0K

n =(443.124*0.2)/(0.08206*300)

n = 3.6 moles = total number of moles

Step 6: Calculate number of moles

For the reaction :2CO(g) + O₂(g) ⟶ 2CO₂(g)

For each mole of O2 we have 2 moles of CO, to produce 2 moles of CO2

Moles CO = (2 -2X) moles

Moles O2 = (2-X) moles

Moles CO2 = 2X

The total number of moles (4 -X)= 3.6 moles

Where X are moles that react

X = 0.400 moles

There will react 0.400 moles of oxygen.

6 0

3 years ago