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Pepsi [2]
2 years ago
7

If you want 2 miles of Silver (Ag) from the reaction below, how many grams of copper should you start with? Reaction: Cu + 2AgNO

3 → 2Ag + Cu(NO3)2
Chemistry
1 answer:
RSB [31]2 years ago
8 0

Answer:

You should start with 63.54 grams of copper.

Explanation:

The chemical reactions are processes in which the nature of the substances changes, that is, from some initial substances called reactants, totally different ones called products are obtained.

In the chemical reaction, the formulas of reagents and products appear preceded by numbers (the stoichiometric coefficients) that indicate the proportions according to which the transformation occurs. So you can say that stoichiometry establishes relationships between the molecules or elements that make up the reactants of a chemical equation with the products of said reaction. The relationships that are established are MOLAR relationships between the compounds or elements that make up the chemical equation: always in MOLES.

The stoichiometric coefficients of a chemical equation are due to the fact that the atoms present before the reaction must be the same after the reaction, although they will have been rearranged to produce new substances.

If you want 2 moles of silver (Ag), for stoichiometry of the reaction you need a moles of copper Cu. Being the molar mass of copper Cu 63.54 g / mole, then:

1 mole*63.54 g/mole= 63.54 g

<u><em> You should start with 63.54 grams of copper.</em></u>

<u><em></em></u>

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B. begin with a hypothesis

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Answer is: K <span>be for the reaction at 375 K is 326.
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T₁<span><span> = 298 K
</span>T</span>₂<span><span> = 375 K
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R = 8,314 J/K</span></span></span>·mol.<span>
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K</span>₂ = ?The van’t Hoff equation: ln(K₂/K₁) = -ΔH/R(1/T₂ - 1/T₁).
ln(K₂/6,8·10⁵) = 92220 J/mol / 8,314 J/K·mol (1/375K - 1/298K).
ln(K₂/6,8·10⁵) = 11092,13 · (0,00266 - 0,00335).
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3 years ago
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7 0
3 years ago
N2O5 decomposes to form NO2 and O2 with first-order kinetics. The initial concentration of N2O5 is 3.0 M and the reaction runs f
kow [346]

Answer : The final concentration of N_2O_5 is, 2.9 M

Explanation :

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant  = 5.89\times 10^{-3}\text{ min}^{-1}

t = time passed by the sample  = 3.5 min

a = initial concentration of the reactant  = 3.0 M

a - x = concentration left after decay process = ?

Now put all the given values in above equation, we get

3.5=\frac{2.303}{5.89\times 10^{-3}}\log\frac{3.0}{a-x}

a-x=2.9M

Thus, the final concentration of N_2O_5 is, 2.9 M

3 0
2 years ago
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