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liraira [26]
3 years ago
11

Sometimes engineers increase output through crossbreeding. This is when they:

Chemistry
1 answer:
kondaur [170]3 years ago
5 0
Combine two plants so that each plant's strength compensates for the other's weakness apek
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The half-life of a pesticide determines its persistence in the environment. A common pesticide degrades in a first-order process
denis23 [38]

Answer:

0.1066 hours

Explanation:

A common pesticide degrades in a first-order process with a rate constant (k) of 6.5 1/hours. We can calculate its half-life (t1/2), that is, the times that it takes for its concentration to be halved, using the following expression.

t1/2 = ln2/k

t1/2 = ln2/6.5 h⁻¹

t1/2 = 0.1066 h

The half-life of the pesticide is 0.1066 hours.

7 0
3 years ago
A particular reaction, A- products, has a rate that slows down as the reaction proceeds. The half-life of the reaction is found
Thepotemich [5.8K]

Explanation:

Half life of zero order and second order depends on the initial concentration. But as the given reaction slows down as the reaction proceeds, therefore, it must be second order reaction. This is because rate of reaction does not depend upon the initial concentration of the reactant.

a. As it is a second order reaction, therefore, doubling reactant concentration, will increase the rate of reaction 4 times. Therefore, the statement  a is wrong.

b. Expression for second order reaction is as follows:

\frac{1}{[A]} =\frac{1}{[A]_0} +kt

the above equation can be written in the form of Y = mx + C

so, the plot between 1/[A] and t is linear. So the statement b is true.

c.

Expression for half life is as follows:

t_{1/2}=\frac{1}{k[A]_0}

As half-life is inversely proportional to initial concentration, therefore, increase in concentration will decrease the half life. Therefore statement c is wrong.

d.

Plot between A and t is exponential, therefore there is no constant slope. Therefore, the statement d is wrong

8 0
3 years ago
Calculate the change in energy when 75.0 grams of water drops from<br> 31.0C to 21.6.
zysi [14]

Answer: Step 1: Calculate qsur (the surrounding is

usually the water)

qsur = ? J

m = 75.0 g water

c = 4.184 J/g

oC

ΔT = (Tfinal- Tinitial)= (21.6 – 31.0) = -9.4 oC

qsur = m · c · (ΔT)

qsur = (75.0g) (4.184 J/g

oC) (-9.4 oC)

qsur = - 2949.72 J

First, using the information we know that we

must solve for qsur, which is the water. We know

the mass for water, 75.0g, the specific heat of

the water, 4.184 j/g

o

c, and the change in

temperature, 21.6-31.0 = -9.4 oC. Plugging it

into the equation, we solve for qsur.

Step 2: Calculate qsys qsys = - (qsur)

qsys = - (- 2949.72 J)

qsys = + 2949.72

In this case, the qsur is negative, which means

that the water lost energy. Where did it go? It

went to the system. Thus, the energy of the

system is negative, opposite, the energy of the

surrounding.

Step 3: Calculate moles of the substance

that is the system

Given: 12.8 g KCl

Mol system = (g system given)

(molar mass of system)

Mol system = (12.8 g KCl)

(39.10g + 35.45g)

Mol system = 12.8 g KCl

74.55 g

Mol system = 0.172

Here, we solve for the mol in the system by

using the molar mass of the material in the

system.

Step 4: Calculate ΔH ΔH = q sys .

Mol system

ΔH= + 2949.72 J

0.172 mol

ΔH= +17179.81 J/mol or +1.72 x 104

J/mol

i hope this helps

5 0
3 years ago
Read 2 more answers
Which is the molar mass of H2O?
Svetllana [295]
H=(2x1.008)=2.016
O= 15.999

15.999
+ 2.016
______
18.015 g/mol :)

3 0
3 years ago
Read 2 more answers
The bonding found in calcium chloride is
photoshop1234 [79]
<span>The bonding found in calcium chloride is i</span>onic bonds. 

I hope this helps!
4 0
3 years ago
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