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2 years ago

What is the difference between a sea and an ocean

Chemistry

1 answer:

IceJOKER [234]2 years ago

8 0

Seas are smaller than oceans. Oceans are where land and water meet, and are usually bigger.

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A jet flies over the ocean. A sound wave travels from the jet to the surface of the ocean. Almost none of the energy from the so

lutik1710 [3]

We need to see the diagram

6 0

3 years ago

NItrogen in air reacts at high temperature to form NO2 according to the reaction:

Rina8888 [55]

Answer:

The correct answer is option E.

Explanation:

Structures for the reactants and products are given in an aimage ;

Number of double bonds in oxygen gas molecule = 1

Number of double bonds in nitro dioxide gas molecule = 1

Number of single bond in in nitro dioxide gas molecule = 1

Number of triple bonds in nitrogen gas molecule = 1

$N_2+2O_2\rightarrow 2NO_2,\Delta H=?$

$\Delta H=[2 mol\times \Delta H_{f,NO_2}]-[1 mol\times \Delta H_{f,N_2}-2 mol\times \Delta H_{f,O_2}]$

$\Delta H_{f,NO_2}=33.18 kJ/mol$

$\Delta H_{f,N_2}=0$ (pure element)

$\Delta H_{f,O_2}=0$ (pure element )

$\Delta H=2 mol\times 33.18 kJ/mol=66.36kJ=15.86 kcal$

The enthalpy of the given reaction is 15.86 kcal.

6 0

3 years ago

Calculate the work done when an ideal gas expands isothermally and reversibly in a piston and cylinder assembly for expansion of

pantera1 [17]

Answer:

$W=5743.1077\ J$

Explanation:

The expression for the work done is:

$W=RT \ln \left( \dfrac{P_1}{P_2} \right)$

Where,

W is the amount of work done by the gas

R is Gas constant having value = 8.314 J / K mol

T is the temperature

P₁ is the initial pressure

P₂ is the final pressure

Given that:

T = 300 K

P₁ = 10 bar

P₂ = 1 bar

Applying in the equation as:

$W=8.314\times 300 \ln \left( \dfrac{10}{1} \right)$

$W=300\times \:8.314\ln \left(10\right)$

$W=2.30258\times \:2494.2$

$W=5743.1077\ J$

5 0

2 years ago

1. The most useful ore of aluminum is bauxite, in which Al is present as hydrated oxides, Al2O3⋅xH2O The number of kilowatt-hour

lianna [129]

*Answer:

Option A: 59.6

Explanation:

Step 1: Data given

Mass of aluminium = 4.00 kg

The applied emf = 5.00 V

watts = volts * amperes

Step 2: Calculate amperes

equivalent mass of aluminum = 27 / 3 = 9

mass of deposit = (equivalent mass x amperes x seconds) / 96500

4000 grams = (9* amperes * seconds) / 96500

amperes * seconds = 42888888.9

1 hour = 3600 seconds

amperes * hours = 42888888.9 / 3600 = 11913.6

amperes = 11913.6 / hours

Step 3: Calculate kilowatts

watts = 5 * 11913.6 / hours

watts = 59568 (per hour)

kilowatts = 59.6 (per hour)

The number of kilowatt-hours of electricity required to produce 4.00kg of aluminum from electrolysis of compounds from bauxite is 59.6 kWh when the applied emf is 5.00V

5 0

2 years ago

When liquid silver nitrate and liquid sodium chloride or combine, solid silver chloride forms along with a new liquid, sodium ni

schepotkina [342]

Liquid silver nitrate and liquid sodium chloride combined forms the substance silver chloride.

3 0

2 years ago