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photoshop1234 [79]
3 years ago
15

Calculate the number of moles of aluminum, sulfur, and oxygen atoms in 4.00 moles of aluminum sulfate, al2(so4)3. express the nu

mber of moles of al, s, and o atoms numerically, separated by commas.
Chemistry
2 answers:
earnstyle [38]3 years ago
8 0
<span>Answer: 8, 12, 48.
</span>

<span>Explanation:
</span>

<span>1) Aluminum:
</span>

<span>There are 2 atoms of Al in the unit formula Al₂(SO₄)₃, so there are 4x2 = 8 atoms of Al in 4 unit formulas of the same compound.
</span>

<span>2) Sulfur:
</span>

There are 3 atoms of S in the unit formula <span>Al₂(SO₄)₃, so there are 4x3 = 12 atoms of S in 4 unit formulas of the same compound
</span>

3) Oxygen:

<span>
</span><span>There are 4x3 = 12 atoms of O in the unit formula </span><span>Al₂(SO₄)₃, so there are 4x12 = 48 atoms of O in 4 unit formulas of the same compound.</span>


So, the answer is 8, 12, 48
bagirrra123 [75]3 years ago
3 0
<span>To calculate the number of moles of aluminum, sulfur, and oxygen atoms in 4.00 moles of aluminum sulfate, al2(so4)3. We will simply inspect the "number" of aluminum, sulfur, and oxygen atoms available per one mole of the compound. Here we have Al2(SO4)3, which means that for every mole of aluminum sulfate, there are 2 moles of aluminum, 3 (1 times 3) moles of sulfur, and 12 (4x3) moles of oxygen. Since we have four moles of Al2(SO4)3 given, we simply multiply 4 times the moles present per 1 mole of the compound. So we have 4x2 = 8 moles of Al, 4x3 = 12 moles of sulfur, and 4x12 = 48 moles of oxygen.

So the answer is:
8,12,48


</span>
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How many molecules are in 69 grams of Na?
12345 [234]
Molar mass Na = 23.0 g/mol

1 mol ---- 23.0 g
n mol ---- 69 g

n = 69 / 23.0

n = 3.0 moles

1 mole -------- 6.02x10²³ molecules
3.0 moles ---- ?

3.0 *  6.02x10²³ / 1

= 1.806x10²⁴ molecules

hope this helps!




7 0
2 years ago
Which will dissolve faster in water one gram cube of sugar or one gram of powder sugar?
krek1111 [17]
The  powder sugar because has more contact area 
8 0
3 years ago
What is the molality of sodium chloride in solution that is 13.0% by mass sodium chloride and that has a density of 1.10 g/ml?
8090 [49]
Answer is: molality od sodium chloride is 2,55 mol/kg.
V(solution) = 100 ml.
m(solution) = d(solution) · V(solution).
m(solution) = 1,10 g/ml · 100 ml.
m(solution) = 110 g.
ω(NaCl) = 13,0% = 0,13.
m(NaCl) = ω(NaCl) · m(solution).
m(NaCl) = 0,13 · 110 g.
m(NaCl) = 14,3 g.
n(NaCl) = m(NaCl) ÷ M(NaCl).
n(NaCl) = 14,3 g ÷ 58,5 g/mol.
n(NaCl) = 0,244 mol.
m(H₂O) = 110 g - 14,3 g.
m(H₂O) = 95,7 g = 0,0957 kg.
b(NaCl) = n(NaCl) ÷ m(H₂O).
b(NaCl) = 0,244 mol ÷ 0,0957 kg.
b(NaCl) = 2,55 mol/kg.
3 0
3 years ago
If an experiment with 10.2 g barium chloride produced 14.5 g silver chloride, calculate thr experimental mole ratio of silver ch
Leni [432]
The  experimental  mole ratio  of silver chloride  to  barium chloride  is calculated as below

fin the mole of each compound

mole= mass/molar  mass
moles of AgCl = 14.5g/142.5 g/mol = 0.102  moles of AgCl
moles of BaCl2 = 10.2 g/208 g/mol =  0.049  moles of BaCl2

find the  mole ratio  by dividing each mole with the smallest  mole(0.049)

AgCl= 0.102/0.049 =2
BaCl2 = 0.049/0.049 =1
therefore  the mole   ratio   AgCl to  BaCl2  is  2 :1
3 0
2 years ago
Read 2 more answers
A soft silvery metal has two naturally occurring isotopes: mass 84.9118, accounting for 72.15% and mass 86.9092, accounting for
zloy xaker [14]

Answer: The atomic weight of the metal would be 85.47.

Explanation:

Mass of isotope 1 of metal = 84.9118

% abundance of isotope 1 of metal = 72.15% = \frac{72.15}{100}

Mass of isotope 2 of metal= 86.9092

% abundance of isotope 2 of metal = 27.85% = \frac{27.85}{100}

Formula used for average atomic mass of an element :

\text{ Average atomic mass of an element}=\sum(\text{atomic mass of an isotopes}\times {{\text { fractional abundance}})

A=\sum[(84.9118)\times \frac{72.15}{100})+(86.9092)\times \frac{27.85}{100}]]

A=85.47

Therefore, the atomic weight of the metal would be 85.47.

6 0
3 years ago
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