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emmainna [20.7K]
3 years ago
15

Calculate the mass (in g) of 2.1×1024 atoms of w. Calculate the mass (in ) of atoms of . 1.5 ×102g 3.9×102g 2.4 ×102g 6.4 ×102g

3.2 ×102g
Chemistry
1 answer:
storchak [24]3 years ago
6 0

W is the symbol of tungsten.

Molar mass of W  = 183.84 g/mol

Now, number of atoms are converted to number of moles by Avogadro number i.e. 6.022\times 10^{23}

Number of moles of tungsten  =\frac{number of atoms}{Avogadro number}

number of moles  = \frac{2.1\times 10^{24}atoms}{6.022\times 10^{23} atoms/mol}

= 0.3487\times 10 mole\simeq 3.49 mole

Therefore,  1 mole of tungsten consist of  6.022\times 10^{23} atoms

So,  3.49 mole of tungsten consist of 2.1\times 10^{24} atoms.

Number of moles is also equal to \frac{mass in g}{molar mass}

3.49 mole =\frac{mass in g}{183.84 g/mol}

mass in g  =\frac{3.49 mole\times 183.84 g/mol}

= 641.6016 g or 6.4\times 10^{2} g

Thus, mass in g of 2.1\times 10^{24}atoms = 6.4\times 10^{2} g.


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lianna [129]

Answer:

The mass of hydrogen gas formed is 0.205 grams

Explanation:

<u>Step 1:</u> Data given

Mass of 1.83 grams of Al

Mass of NaOH = 4.30 grams

Molar mass of Al = 26.98 g/mol

Molar mass of NaOH = 40 g/mol

<u>Step 2:</u> The balanced equation:

2 Al(s) + 2 NaOH(aq) + 6 H2O(l) → 2 NaAl(OH)4(aq) + 3 H2(g)

<u>Step 3:</u> Calculate moles of Al

Moles Al = mass Al / Molar mass Al

Moles Al = 1.83 grams / 26.98 g/mol

Moles Al = 0.0678 moles

<u>Step 4:</u> Calculate moles of NaOH

Moles NaOH = 4.30 grams / 40 g/mol

Moles NaOH = 0.1075 moles

<u>Step 5</u>: Calculate limiting reactant

For 2 moles of Al, we need 2 moles of NaOH

Aluminium is the limiting reactant. It will completely be consumed ( 0.0678 moles)

NaOH is in excess. There will react 0.0678 moles

There will remain 0.1075 - 0.0678 = 0.0397 moles

<u>Step 6</u>: Calculate moles of hydrogen

For 2 moles of Al, we need 2 moles of NaOH, to produce 3 moles of hydrogen

For 0.0678 moles of Al, there is produced 0.0678 *3/2 = 0.1017 moles of H2

<u>Step 7</u>: Calculate mass of H2

Mass of H2 = Moles H2 * Molar mass of H2

Mass of H2 = 0.1017 moles * 2.02 g/mol

Mass of H2 = 0.205 grams

The mass of hydrogen gas formed is 0.205 grams

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How much energy (in Joules) is required to convert 129 grams of ice at −23.0 °C to liquid water at 18.0 °C?
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Answer:

The energy that is required for the process is:

6230.7 J + 42957 J + 9715.2 J = <u>58902.9 joules</u>

Explanation:

This is a calorimetry problem:

Q = m . C . ΔT

Q = heat; m = mas; C is the specific heat and

ΔT = Final T° - Initial T°

Q = C lat . m

Q = Heat

m = mass

C lar = Latent heat of fusion

First of all we calculate the heat for ice, before it takes the melting point. (from -23°C  to 0°C)

Q = 129 g . 2.10 J/g°C . (0°C - (-23°C)

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Q = 333 J/°C . 129 g → 42957 joules

And in the end, we have water that changed its T° from O°C to 18°C

Q = 129 g . 4.184 J/g °C . (18°C - 0°C)

Q = 9715.2 Joules

The energy that is required for the process is:

6230.7 J + 42957 J + 9715.2 J = 58902.9 joules

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