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nikdorinn [45]
3 years ago
9

Balancing chemical equations

Mathematics
1 answer:
sashaice [31]3 years ago
7 0
. ...
Example 2. Zn + HCl ---> ZnCl2 + H2 ...
Example 3. Ca(OH)2 + H3PO4 ---> Ca3(PO4)2 + H2O. ...
Example 4. FeCl3 + NH4OH ---> Fe(OH)3 + NH4Cl. ...
Example 5. S8 + F2 ---> SF6 ...
Example 6. C2H6 + O2 ---> CO2 + H2O. ...
Example 7. Al2(CO3)3 + H3PO4 ---> AlPO4 + CO2 + H2O. Summary
To be useful, chemical equations must always be balanced. Balanced chemical equations have the same number and type of each atom on both sides of the equation.
The coefficients in a balanced equation must be the simplest whole number ratio. Mass is always conserved in chemical reactions. mber of atoms of each element in reactants is not equal to the number of atoms of each element present in product, then the chemical equation is called unbalanced chemical equation. An equation is balanced when the same number of each element is represented on the reactant and product sides. Equations must be balanced to accurately reflect the law of conservation of matter.

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3 0
3 years ago
A=8, b=5, C=90 degree; Find c, A, and B<br><br> This is for the law of sine and cosine
brilliants [131]

Answer:

• c = √89 ≈ 9.434

• A = arcsin(8/√89) ≈ 57.995°

• B = arcsin(5/√89) ≈ 32.005°

Step-by-step explanation:

By the law of cosines, ...

c² = a² + b² -2ab·cos(C)

Since c=90°, cos(C) = 0 and this reduces to the Pythagorean theorem for this right triangle.

c = √(8² +5²) = √89 ≈ 9.434

Then by the law of sines (or the definition of the sine of an angle), ...

sin(A) = a/c·sin(C) = a/c = 8/√89

A = arcsin(8/√89) ≈ 57.995°

sin(B) = b/c·sin(C) = b/c = 5/√89

B = arcsin(5/√89) ≈ 32.005°

6 0
3 years ago
Factorise the following expressions 5x^2-25xy​
Natali [406]

Answer: 5x ( x - 5y )

Step-by-step explanation:

3 0
3 years ago
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RUDIKE [14]

Answer:

27 = 3 \times 3 \times 3\\\\\frac{1}{27}  = \frac{1}{3 \times 3 \times 3} = \frac{1}{3^{3}}  = 3^{-3}

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3 years ago
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Fynjy0 [20]

Answer:

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Step-by-step explanation:

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3 years ago
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