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3 years ago

11

. If 30.0 grams of copper (II) chloride reacts with 40.0 grams of sodium nitrate, what is the limiting reactant?

1 answer:

blondinia [14]3 years ago

7 0

Answer:

The limiting reactant is copper (II) chloride

Explanation:

here you go

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How many Br atoms are in 1.98 g of Br?

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Answer:

1.5 x 10²²atoms

Explanation:

Given parameters:

Mass of Br = 1.98g

Unknown:

Number of atoms

Solution:

A mole of a substance is the amount of substance contained in the avogadro's number of particles.

To solve this problem, we must first find the number of moles present in the given mass of the Br atom:

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The activation energy (E*) for 2N2O ---> 2N2 + O2 is 250 KJ. If the k for this reaction is 0.380/M at 1001oK, what will k be

Sedbober [7]

Answer:

Explanation:

GIven that:

The activation energy = 250 kJ

k₁ = 0.380 /M

k₂ = ???

Initial temperature $T_1 =$ 1001 K

Final temperature $T_2 =$ 298 K

Applying the equation of Arrhenius theory.

$In \dfrac{k_2}{k_1 }= \dfrac{Ea}{R}( \dfrac{1}{T_1 }- \dfrac{1}{T_2})$

where ;

R gas constant = 8.314 J/K/mol

$In \dfrac{k_2}{0.380 }= \dfrac{250 * 10^3}{8,314}( \dfrac{1}{1001 }- \dfrac{1}{298})$

$In \dfrac{k_2}{0.380 }= -70.8655$

$\dfrac{k_2}{0.380 }= e^{-70.8655}$

$\dfrac{k_2}{0.380 }= 1.67303256 \times 10^{-31}$

${k_2}= 1.67303256 \times 10^{-31} \times {0.380 }$

${k_2}= 6.3575 \times 10^{-32}$ /M .sec

Half life:

At 1001 K.

$t_{1/2} = \dfrac{In_2}{k_1}$

$t_{1/2} = \dfrac{0.693}{0.38}$

$t_{1/2} =$ 1.82368 secc

At 298 K:

$t_{1/2} = \dfrac{0.693}{6.3575 \times 10^{-32}}$

$t_{1/2} =1.09 \times 10^{31} \ sec$

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