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2 years ago

Silicon has 2 isotopes, Silicon-28 with an abundance of 90% and Silicon-30 with an abundance of 10%. Find the AAM for Silicon.

Chemistry

1 answer:

JulsSmile [24]2 years ago

5 0

Answer:

RAM

$RAM = (mass \: of \: ^{28} Si \: \times \%abundance) + (mass \: of \: ^{30} Si \: \times \%abundance) \\ RAM = (28 \times 90\%) + (30 \times 10\%) \\ RAM = 25.2 + 3 \\ RAM = 28.2$

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What is the crystalline arrangement of atoms shown here?

mixer [17]

Answer: Answer:

"The arrangement of atoms or ions in a crystal " is described by the terms body-centered cubic and face-centered cubic.

Explanation:

Face centred cubic system explains the crystal structure where an atom is present at each cubic corner of the crystal and the centre of each cube face. Meaningfully, a closed packed plane where at each "face of the cube" atoms touch the alongside face diagonals.

Whereas in body centric cube system has the lattice point present at the 8 corners of cell and an additional one at the center of the cell. Thus, both explains how the atom or ions are placed or arranged in a crystal.

Explanation: Hope this helps :)

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2 years ago

Which of the following locations would experience the highest rate of sublimation?

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I think its the arctic ice caps

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3 years ago

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Identify the ion ratios needed to form neutral compounds from the ions below. The metal ion (cation) is listed first, and the no

Natasha2012 [34]

To figure out the ratios of these compounds, it is important to remember that the charge of these compounds must be neutral.

So in order to make them neutral, you must have specific ratios:

$Na^{+}: Br^{-} =1:1$ ; This is true because they both have a charge of magnitude of 1.

$Al^{3+}: Cl^{-}=1:3$ ; We need 3 chlorine atoms because we need to balance out the charge from the 3+ charge of aluminum - therefore since chlorine has a 1- charge, we need 3 atoms.

$Mg^{2+}: O^{2-}=1:1$ ; The charges of the magnesium (2+) are balanced with the oxygen charge (2-).

$Al^{3+}: O^{2-}=2:3$ ; This is correct because if charges are like this, you must find the least common factor in order to know the ratio. The LCF is 6, therefore, for the atom with a 3+ charge, you need 2 of them, and for the atom with a 2- charge, you need 3 of them. This keeps the charge neutral.

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3 years ago

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Ammonia NH3 may react with oxygen to form nitrogen gas and water.4NH3 (aq) + 3O2 (g) \rightarrow 2 N2 (g) + 6H2O (l)If 2.15g of

bagirrra123 [75]

Answer:

NH3 is the limiting reactant

The % yield is 36.1 %

Explanation:

Step 1: Data given

Mass of NH3 = 2.15 grams

Mass of O2 = 3.23 grams

Molar mass of NH3 = 17.03 g/mol

Molar mass of O2 = 32 g/mol

volume of N2 produced = 0.550 L

Temperature = 295 K

Pressure = 1.00 atm

Step 2: The balanced equation:

4NH3 (aq) + 3O2 (g) → 2 N2 (g) + 6H2O (l)

Step 3: Calculate moles of NH3

Moles NH3 = Mass NH3 / Molar Mass NH3

Moles NH3 = 2.15 grams / 17.03 g/mol

Moles NH3 = 0.126 moles

Step 4: Calculate moles of O2

Moles O2 = 3.23 grams / 32 g/mol

Moles O2 = 0.101 moles

Step 5: Calculate the limiting reactant

For 4 moles NH3 consumed, we need 3 moles of O2 to produce, 2 moles of N2 and 6 moles of H2O

NH3 is the limiting reactant. It will completely be consumed ( 0.126 moles).

O2 is in excess, there will be 3/4 * 0.126 = 0.0945 moles consumed

There will remain 0.101 - 0.945 = 0.0065 moles of O2

Step 6: Calculate moles of N2

For 4 moles NH3 consumed, we need 3 moles of O2 to produce, 2 moles of N2 and 6 moles of H2O

For 4 moles NH3 , we'll have 2 moles of N2 produced

For 0.126 moles NH3 consumed, we'll have 0.063 moles of N2 produced.

Step 7: Calculate volume of N2 produced

p*V = n*R*T

⇒ with p = the pressure of the gas = 1.00 atm

⇒ with V = the volume = TO BE DETERMINED

⇒ with n = the number of moles N2 = 0.063 moles

⇒ with R = the gasconstant = 0.08206 L*atm/K*mol

⇒ with T = the temperature = 295

V = (nRT)/p

V = (0.063*0.08206*295)/1

V = 1.525 L = theoretical yield

Step 8: Calculate the % yield

% yield = actual yield / theoretical yield

% yield = (0.550 L / 1.525 L)*100%

% yield = 36.1 %

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2 years ago

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