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3 years ago

After examining a cell with a microscope, a student concludes that it is a plant cell.

Chemistry

1 answer:

Scorpion4ik [409]3 years ago

7 0

Explanation:

The student most likely observed both (A) Vacuole and (C) Chloroplast that led him to the conclusion, that the cell is a plan cell.

Reason :

In a plant cell, Vacuole occupies 70% to 90% of the total volume of the cell, making it easier to identify.

Chloroplast contains photosynthetic pigment called chlorophyll (green in colour), which is only found in plant cell.

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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

sladkih [1.3K]

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.

1. Table

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

2. Orders

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.

3. Rate equation

Ther orders are:

$a=2\\\\b=0\\\\c=1$

Hence the rate is:

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

4. Rate constant, k

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

Using trial 1:

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

5. Rate law:

$rate=k[A]^2C=0.0075[A]^2[C]$

6. Substitute

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

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

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

7 0

3 years ago

A geologist is attempting to date a layer of sedimentary rock. He performs radioactive dating using uranium-235, which has a hal

Bad White [126]

Answer:

2.11 billion years.

Explanation:

I hope this helps! Have a great rest of your day!

4 0

3 years ago

Read 2 more answers

Assume that all of this fossil fuel is in the form of octane (C8H18) and calculate how much CO2 in kilograms is produced by worl

AnnZ [28]

The given question is incomplete. the complete question is:

The world burns the fossil fuel equivalent of approximately $9.50\times 10^{12}$ kg of petroleum per year. Assume that all of this petroleum is in the form of octane. Calculate how much CO2 in kilograms is produced by world fossil fuel combustion per year.( Hint: Begin by writing a balanced equation for the combustion of octane.)

Answer: $29\times 10^{12}kg$

Explanation:

Combustion is a chemical reaction in which hydrocarbons are burnt in the presence of oxygen to give carbon dioxide and water.

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

$2_C8H_{18}+17O_2\rightarrow 16CO_2+18H_2O$