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tamaranim1 [39]
4 years ago
5

What is the balanced equation for: Ammonium Carbonate and Magnesium Sulfate react to yield Ammonium Sulfate plus Magnesium Carbo

nate?
Chemistry
1 answer:
Debora [2.8K]4 years ago
5 0

Answer:

(NH₄)₂CO₃ (aq) + MgSO₄(aq)  →  (NH₄)₂SO₄(aq) + MgCO₃ (s) ↓  

Explanation:

We determine the reactants:

(NH₄)₂CO₃ → Ammonium carbonate

MgSO₄ → Magnesium sulfate

We determine the products:

(NH₄)₂SO₄ ; MgCO₃

All the salts from carbonates are precipitates:

(NH₄)₂CO₃ (aq) + MgSO₄(aq)  →  (NH₄)₂SO₄(aq) + MgCO₃ (s) ↓  

Ratio is all 1:1. 1 mol of ammonium carbonate reacts to 1 mol of magnessium sulfate in order to produce 1 mol of ammonium sulfate and 1 mol of magnessium carbonate.

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Option C: Sulfur Dioxide is the answer  

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5.0 x 10^24 molecules equal
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<span>4.999999999999999e</span>+<span>24 this is what i got on the calculator but i dont know if its right.</span>
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3 years ago
Stannum has a body centered tetragonal with lattice constant, a = b = 5.83A and c = 3.18A. If the atomic radius is 0.145 nm, det
Fed [463]

Answer:

the atomic packing factor of Sn is 0.24

Explanation:

a = b = 5.83A and c = 3.18A.

Volume of unit cell = a²c

= (5.83)² *  3.18 * 10⁻²⁴ cm³

= 1.08 * 10⁻²²cm³

Volume of atoms =

2 \times  \frac{4}{3} \pi r^3

(∴ BCC, effective number of atom is 2)

Volume of atoms =

2 * \frac{4}{3} *3.14*(0.145*10^-^7cm)^3

= 2.55*10⁻²³cm³

\text {Atomic packing factor}=\frac{\text {volume occupied by atom}}{\text {volume of unit cell }}

=\frac{2.55*10^-^2^3}{1.08*10^-^2^2} \\\\=0.24

<h3>therefore, the atomic packing factor of Sn is 0.24</h3>
4 0
4 years ago
Lead ions and magnesium ions both have a charge of 2+. predict the ionic equation for this reaction.
FinnZ [79.3K]
Answer:
PbMg

Explanation:
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4 0
2 years ago
What is the total mass of D-glucose dissolved in a 2-μL aliquot of the solution used for this experiment?
RideAnS [48]

Answer:

The total mass of D-Glucose dissolved in a 2μL aliquot is 1 E-4 g

Explanation:

providing a solution to 5% weight-volume as found in commerce:

⇒ % 5 = (5g d-glucose/ 100 mL sln)×100

⇒ 0.05 =  g C6H12O6/mL sln

⇒ g C6H12O6 = (2 μL sln)×(0.001 mL/μL)×(0.05 g C6H12O6/mL sln)

⇒ g C6H12O6 = 1 E-4 g C6H12O6

5 0
4 years ago
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