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maksim [4K]
3 years ago
9

Jason is preparing curry from an Indian cookbook his British brother-in-law gee him for his birthday. Unfortunately, the British

cookbook is written with units he does not typically use, so he must do some conversions before getting to work in the kitchen *) The recipe calls for 450 g of diced chicken. How many pounds of chicken should he defrost? b) The recipe calls for 125 g of diced onions. How many 4 oz onions should he dice? c) The recipe calls for 100 mL of coconut milk. How many cups should he add
Chemistry
1 answer:
Marina CMI [18]3 years ago
6 0

Answer:

0.99 pounds

2,000 oz

0.423 cups

Explanation:

In order  to convert this units we need to look up their equivalences.

1 g equals 0,0022 pounds approximately

Then we need to cross-multiply:

\frac{1 g}{0.0022 pounds} = \frac{450 g}{ Xpounds} \\x = \frac{450 g x 0.0022 pounds}{1 g}

450 g equals 0.99 pounds approximately

We can do the same calculation for the other 2 ingredients

1 g equals 16 oz

Then (125 g x 16 oz) / 1 oz = 2,000 oz  (or 500 4 oz)

1 cup equals 236.59 ml

Then (100 ml x 1 cup)/ 236.59 = 0.423 cups

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Given the data calculated in Parts A, B, C, and D, determine the initial rate for a reaction that starts with 0.45 M of reagent
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Answer:

\large\boxed{\large\boxed{0.0014M/s}}

Explanation:

From the table, first find the order of reaction, then find the rate constant, write the rate equation, and, finally, subsititue the data for the reaction that starts with 0.45M of reagent A and 0.90 M of reagents B and C.

<u>1. Table</u>

Trial  [A] (M)    [B] (M)   [C] (M)    Initial rate (M/s)

 1        0.20      0.20       0.20         6.0×10⁻⁵

2        0.20      0.20       0.60         1.8×10⁻⁴

3        0.40      0.20        0.20        2.4×10⁻⁴

4        0.40      0.40        0.20        2.4×10⁻⁴

<u>2. Orders</u>

a) From trials 3 and 4 you learn that the initial concentration of B does not change the change teh rate of the reaction. Hence the order with respect to [B] is 0.

b) From trials 1 and 2 you learn that when the concentration of C is tripled the rate of reaction is also tripled:

  • 0.60 / 0.2 = 3, and
  • 1.8×10⁻⁴ / 6.0×10⁻⁵ = 3

Hence, the order with respect to C is 1.

c) From trials 1 and 3 you get:

  • 0.40/0.2 = 2
  • 2.4×10⁻⁴ /  6.0×10⁻⁵ = 4

Which means that when the concentration of A is doubled, the rate of the reaction is quadruplicated. Hence, the order of the reaction with respect to A is 2.

<u>3. Rate equation</u>

Ther orders are:

              a=2\\\\b=0\\\\c=1

Hence the rate is:

            rate=k[A]^a{B}^b[C]^c\\ \\ rate=k[A]^2[B]^0[C]^1=k[A]^2C

<u>4. Rate constant, k</u>

<u />

You can use any trial to find the value of the constant, k

Using trial 1:

            6.0\times 10^{-5}M/s=k(0.20M)^2(0.20M)\\ \\ k=\frac{6.0\times 10^{-5}M/s}{(0.20M)^2(0.20M)}=0.0075M^{-2}s^{-1}

<u>5. Rate law:</u>

       rate=k[A]^2C=0.0075[A]^2[C]

<u>6. Substitute</u>

Subsititue the data for the reaction that starts with 0.45M of reagent A and 0.90 M of reagents B and C.

        rate=0.0075M^{-2}s^{-1}[A]^2[C]=0.0075M^{-2}s^{-1}[0.45M]^2[0.9M]

        r=0.00136688M/s\approx 0.0014M/s

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