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mamaluj [8]
3 years ago
7

Which of the following will occur soon after a population of prey increases slightly in size?

Chemistry
2 answers:
jeka943 years ago
5 0
It is E. because the predators will have slightly more meals with their prey slightly increasing population.
Salsk061 [2.6K]3 years ago
4 0

Answer: Option E

Explanation:

The role of nature is that predators eat prey. They serve as food for the predators. If there is a increase in the size of the prey population then there will be increase in the size of predators.

Example: The lions feed on the rabbit for the its energy needs. If the size of rabbits will increase then the there will be a slight increase in the population of the lions as there will be a proper supply of food for them.

hence, the correct answer is option E

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Consider the rate law below.
Serga [27]

C. quadruples the rate

<h3>Further explanation</h3>

Given

The rate law :

R=k[A]²

Required

The rate

Solution

There are several factors that influence reaction kinetics :  

  • 1. Concentration  
  • 2. Surface area  
  • 3. Temperature  
  • 4. Catalyst  
  • 5. Pressure  
  • 6. Stirring  

The rate is proportional to the concentration.

If the concentration increased, the reaction rate will increase

The reaction is second-order overall(The exponent is 2)

The concentration of A is doubled, the reaction rate will increase :

r = k[A]² ⇒ r= k[2A]²⇒r=4k[A]²

<em>The reaction rate will quadruple.</em>

5 0
3 years ago
Is centigrams or grams bigger.<br> example: which unit is larger.<br> 35cg or 35 g
sp2606 [1]

Answer:

35g

Explanation:

5 0
3 years ago
Read 2 more answers
What does C=o represent
Novay_Z [31]

Carbon monoxide (CO) is a colorless, odorless, and tasteless gas that is slightly less dense than air. ... Carbon monoxide consists of one carbon atom and one oxygen atom, connected by a triple bond that consists of two covalent bonds as well as one dative covalent bond.  


I think it's probably not right.

7 0
3 years ago
A sample of 0.3283 g of an ionic compound containing the bromide ion (Br−) is dissolved in water and treated with an excess of A
Pavel [41]

Answer:

92.49 %

Explanation:

We first calculate the number of moles n of AgBr in 0.7127 g

n = m/M where M = molar mass of AgBr = 187.77 g/mol and m = mass of AgBr formed = 0.7127 g

n = m/M = 0.7127g/187.77 g/mol = 0.0038 mol

Since 1 mol of Bromide ion Br⁻ forms 1 mol AgBr, number of moles of Br⁻ formed = 0.0038 mol and

From n = m/M

m = nM . Where m = mass of Bromide ion precipitate and M = Molar mass of Bromine = 79.904 g/mol

m = 0.0038 mol × 79.904 g/mol = 0.3036 g

% Br in compound = m₁/m₂ × 100%

m₁ = mass of Br in compound = m = 0.3036 g (Since the same amount of Br in the compound is the same amount in the precipitate.)

m₂ = mass of compound = 0.3283 g

% Br in compound = m₁/m₂ × 100% = 0.3036/0.3283 × 100% = 0.9249 × 100% = 92.49 %

4 0
3 years ago
What is the final temperature of the solution formed when 1.52 g of NaOH is added to 35.5 g of water at 20.1 °C in a calorimeter
Inessa [10]

Answer : The final temperature of the solution in the calorimeter is, 31.0^oC

Explanation :

First we have to calculate the heat produced.

\Delta H=\frac{q}{n}

where,

\Delta H = enthalpy change = -44.5 kJ/mol

q = heat released = ?

m = mass of NaOH = 1.52 g

Molar mass of NaOH = 40 g/mol

\text{Moles of }NaOH=\frac{\text{Mass of }NaOH}{\text{Molar mass of }NaOH}=\frac{1.52g}{40g/mole}=0.038mole

Now put all the given values in the above formula, we get:

44.5kJ/mol=\frac{q}{0.038mol}

q=1.691kJ

Now we have to calculate the final temperature of solution in the calorimeter.

q=m\times c\times (T_2-T_1)

where,

q = heat produced = 1.691 kJ = 1691 J

m = mass of solution = 1.52 + 35.5 = 37.02 g

c = specific heat capacity of water = 4.18J/g^oC

T_1 = initial temperature = 20.1^oC

T_2 = final temperature = ?

Now put all the given values in the above formula, we get:

1691J=37.02g\times 4.18J/g^oC\times (T_2-20.1)

T_2=31.0^oC

Thus, the final temperature of the solution in the calorimeter is, 31.0^oC

4 0
3 years ago
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