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FinnZ [79.3K]
3 years ago
10

Based on the balanced chemical equation for the preparation of malachite, what is the composition of the bubbles formed when sod

ium carbonate is added to the solution of copper sulfate?
Chemistry
1 answer:
spin [16.1K]3 years ago
3 0

Answer:

Carbon Dioxide = CO2

Explanation:

The synthesis of Malachite is seen in the chemical formula:

CuSO 4 . 5H2O(aq) + 2NaCO3(aq) --> CuCO 3 Cu(OH) 2 (s) + 2Na 2 SO 4 (aq) + CO 2 (g) + 9H 2 O(l)

The bubbles mentioned in the question hints that our interest is the compounds in their gseous phase  (g).

Upon examining the chemical equation, only CO2 is in the gaseous state and hence the only one that can be formed as bubbles,

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The composition of a liquid-phase reaction 2A - B was monitored spectrophotometrically. The following data was obtained: t/min 0
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Answer:

1) The order of the reaction is of FIRST ORDER

2)   Rate constant k = 5.667 × 10 ⁻⁴

Explanation:

From the given information:

The composition of a liquid-phase reaction 2A - B was monitored spectrophotometrically.

liquid-phase reaction 2A - B signifies that the reaction is of FIRST ORDER where the rate of this reaction is directly proportional to the concentration of A.

The following data was obtained:

t/min                    0         10         20          30             40          ∞

conc B/(mol/L)    0       0.089    0.153     0.200       0.230    0.312

For  a first order reaction:

K = \dfrac{1}{t} \ In ( \dfrac{C_{\infty} - C_o}{C_{\infty} - C_t})

where :

K = proportionality  constant or the rate constant for the specific reaction rate

t = time of reaction

C_o = initial concentration at time t

C _{\infty} = final concentration at time t

C_t = concentration at time t

To start with the value of t when t = 10 mins

K_1 = \dfrac{1}{10} \ In ( \dfrac{0.312 - 0}{0.312 - 0.089})

K_1 = \dfrac{1}{10} \ In ( \dfrac{0.312 }{0.223})

K_1 =0.03358 \  min^{-1}

K_1 \simeq 0.034 \  min^{-1}

When t = 20

K_2= \dfrac{1}{20} \ In ( \dfrac{0.312 - 0}{0.312 - 0.153})

K_2= 0.05 \times  \ In ( 1.9623)

K_2=0.03371 \ min^{-1}

K_2 \simeq 0.034 \ min^{-1}

When t = 30

K_3= \dfrac{1}{30} \ In ( \dfrac{0.312 - 0}{0.312 - 0.200})

K_3= 0.0333 \times  \ In ( \dfrac{0.312}{0.112})

K_3= 0.0333 \times  \ 1.0245

K_3 = 0.03412 \ min^{-1}

K_3 = 0.034 \ min^{-1}

When t = 40

K_4= \dfrac{1}{40} \ In ( \dfrac{0.312 - 0}{0.312 - 0.230})

K_4=0.025 \times  \ In ( \dfrac{0.312}{0.082})

K_4=0.025 \times  \ In ( 3.8048)

K_4=0.03340 \ min^{-1}

We can see that at the different time rates, the rate constant of k_1, k_2, k_3, and k_4 all have similar constant values

As such :

Rate constant k = 0.034 min⁻¹

Converting it to seconds ; we have :

60 seconds = 1 min

∴

0.034 min⁻¹ =(0.034/60) seconds

= 5.667 × 10 ⁻⁴ seconds

Rate constant k = 5.667 × 10 ⁻⁴

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