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4 years ago

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Chemistry

1 answer:

iogann1982 [59]4 years ago

4 0

The molecules of hydrogen gas that are formed is when 48.7 g of sodium are added to water is 6.375 x 10²³ molecules

calculation

2 Na +2H₂O → 2 NaOH +H₂

Step 1: find the moles of sodium (Na)

moles =mass÷ molar mass

from periodic table the molar mass of Na = 23 g/mol

moles= 48.7 g÷ 23 g/mol =2.117 moles

Step 2:use the mole ratio to determine the moles of H₂

from given equation Na:H₂ is 2:1

therefore the moles of H₂ = 2.117 moles x 1/2=1.059 moles

Step 3: find the molecules of H₂ using the Avogadro's law

According to Avogadro's law 1 mole = 6.02 x 10²³ molecules

1.059 moles = ? molecules

by cross multiplication

= [(1.059 moles x 6.02 x10²³ molecules) / 1 mole] =6.375 x 10²³ molecules

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Rank the following transitions in a hydrogen atom in order of increasing wavelength of electromagnetic radiation that could prod

Dmitrij [34]

Answer:

n=2 to n=4 < n=6 to n=8 < n=10 to n=12 < n=14 to n=16

Explanation:

According to Neils Bohr, electrons in an atom are found in specified energy levels. Transitions are possible from one energy level to another when the electron receives sufficient energy usually in the form of a photon of electromagnetic radiation of appropriate frequency and wavelength. The energy of this photon corresponds to the energy difference between the two energy levels. Thus the higher the energy difference between energy levels, the greater the energy of the photon required to cause the transition and the shorter the wavelength of the photon.

High energy photons have a very short wavelength. It should be noted that as n increases, the energy of successive energy levels decreases and transitions between them now occurs at longer wavelengths. Hence, the highest energy and shortest wavelength of photons are required for transition involving lower values of n because such electrons are closer to the nucleus and are more tightly bound to it than electrons found at a greater distance from the nucleus.

Hence transition involving electrons at higher energy levels occur at a longer wavelength compared to transition involving electrons closer to the nucleus. This is the basis for the arrangement of wavelengths required to effect the various electronic transitions shown in the answer.

3 0

3 years ago

. The heat capacity of a bomb calorimeter was determined by burning 6.79 g methane (energy of combustion 802 kJ/mol CH4) in the

kolbaska11 [484]

Answer:

a) The heat capacity of the calorimeter is 31.4 kJ/ºC.

b) The energy of combustion of acetylene in kJ/mol is 1097 kJ/mol.

Explanation:

The heat capacity ( C ) of a substance is the amount of heat required to raise the temperature of a given quantity of the substance by one degree Celsius. Its units are J/°C. or kJ/ºC.

If we know the heat capacity and the amount of a substance, then the change in the sample’s temperature (Δt ) will tell us the amount of heat (q) that has been absorbed or released in a particular process. One of the equations for calculating the heat change is given by:

$q=C.ΔT$

Where ΔT is the temperature change: ΔT= tfinal - tinitial, and C the heat capacity.

In the calorimeter, the heat given off by the sample is absorbed by the water and the bomb. The special design of the calorimeter enables us to assume that no heat (or mass) is lost to the surroundings during the time it takes to make measurements.

Therefore, we can call the bomb and the water in which it is submerged an isolated system. Because no heat enters or leaves the system throughout the process, the heat change of the system ( q system ) must be zero and we can write:

$qsystem = qrxn + qcal$

$qsystem = 0$

where q cal and q rxn are the heat changes for the calorimeter and the reaction, respectively. Thus, $qrxn = -qcal$

To calculate qcal , we need to know the heat capacity of the calorimeter ( Ccal ) and the temperature rise, that is, $qcal = Ccal. ΔT$

a. The quantity Ccal is calibrated by burning a substance with an accurately known heat of combustion. In order to do this, we need to convert the molar heat of combustion (expressed in kJ/mol) into heat of combustion (expressed in kJ). For that matter, we transform the 6.79 grams of methane into moles:

$1 mol CH₄÷16.04 g CH₄ × 6.79 g CH₄ = 0.423 mol CH₄$

And then multiply it by the molar heat of combustion:

$802 kJ/mol × 0.423 mol = 339 kJ$

Now we know that that the combustion of 6.79 g of methane releases 339 kJ of heat. If the temperature rise is 10.8ºC, then the heat capacity of the calorimeter is given by

$Ccal= qcal/ΔT = 339 kJ/10.8ºC = 31.4 kJ/ºC$

Once C cal has been determined, the calorimeter can be used to measure the heat of combustion of other substances. Note that although the combustion reaction is exothermic, q cal is a positive quantity because it represents the heat absorbed by the calorimeter.

b. The heat absorbed by the bomb and water is equal to the product of the heat capacity and the temperature change. Working with the same equation, and assuming no heat is lost to the surroundings, we write

$qcal=Ccal.ΔT= 31.4 kJ/°C × 16.9 °C = 531kJ$

Now that we have the heat of combustion, we need to calculate the molar heat. Because qsystem = qrxn + qcal and qrxn = -qcal, the heat change of the reaction is -531 kJ.

This is the heat released by the combustion of 12.6 g of acetylene ; therefore, we can write the conversion factor as 531 kJ÷12.6 g

The molar mass of acetylene is 26.04 g, so the heat of combustion of 1 mole of acetylene is

$molar heat of combustion= -531 kJ÷12.6 g × 26.04 g÷ 1 mol= 1097 kJ/mol$

Therefore, the energy of combustion of acetylene in kJ/mol is 1097 kJ/mol.

7 0

4 years ago

What is the density of a cube of water measuring 2cmX4cmX1cm, with a mass of 8g?

Vinil7 [7]

The actual formula for volume for a cube is the length multiplied by the width and then multiplied by the height. Since all three measurements are the same, the formula results in the measurement of one side cubed. For the example, 5^3 is 125 cm^3. Multiply the volume by the known density, which is the mass per volume.

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3 years ago

A sample of He gas (3.0 L) at 5.6 atm and 25°C was combined with 4.5 L of Ne gas at 3.6 atm and 25°C at constant temperature in

In-s [12.5K]

Answer:

3.6667

Explanation:

For helium gas:

Using Boyle's law

${P_1}\times {V_1}={P_2}\times {V_2}$