This is the process that cells reproduce and replace old or damaged cells.

How can a change to another population could affect births and deaths in the parrot population.

forsale [732]

Answer:

read it first

Explanation:

Population change is governed by the balance between birth rates and death rates. If the birth rate stays the same and the death rate decreases, then population numbers will grow. If the birth rate increases and the death rate stays the same, then population will also grow.

hope it helps

6 0

2 years ago

Calculate the mass (in grams) of methylene bluecrystals that you must weigh in order to make 100.0mL of 1.25 × 10-5mol/L methyle

mario62 [17]

<u>Answer:</u> The mass of methylene blue that must be weighed is $3.99\times 10^{-4}g$

<u>Explanation:</u>

To calculate the mass of solute, we use the equation used to calculate the molarity of solution:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

We are given:

Molarity of solution = $1.25\times 10^{-5}M$

Molar mass of methylene blue = 319.85 g/mol

Volume of solution = 100.0 mL

Putting values in above equation, we get:

$1.25\times 10^{-5}M=\frac{\text{Mass of methylene blue}\times 1000}{319.85\times 100.0}\\\\\text{Mass of methylene blue}=\frac{1.25\times 10^{-5}\times 319.85\times 100.0}{1000}=3.99\times 10^{-4}g$

Hence, the mass of methylene blue that must be weighed is $3.99\times 10^{-4}g$

4 0

3 years ago

What is the concentration of OH− and pOH in a 0.00072 M solution of Ba(OH)2 at 25 ∘C? Assume complete dissociation.

Mumz [18]

Given :

0.00072 M solution of $Ba(OH)_2$ at $25^oC$ .

To Find :

The concentration of $OH^-$ and pOH .

Solution :

1 mole of $Ba(OH)_2$ gives 2 moles of $OH^-$ ions .

So , 0.00072 M mole of $Ba(OH)_2$ gives :

$[OH^-]=2 \times 0.00072\ M$

$[OH^-]=0.00144\ M$

$[OH^-]=1.44\times 10^{-3}\ M$

Now , pOH is given by :

$pOH=-log[OH^-]\\\\pOH=-log[1.44\times 10^{-3}]\\\\pOH=2.84$

Hence , this is the required solution .

3 0

3 years ago

A package contains 1.33 lb of ground round. If it contains 29% fat, how many grams of fat are in the ground round? The book is s

Effectus [21]

Answer:

To obtain the grams of fat that the ground round has, knowing that it weighs 1.33 pounds we must pass this value to grams. Since 1 pound equals 453.59 grams, 1.33 pounds equals 603.27 (453.59 x 1.33).

Now, to obtain 29 percent of 603.27, we must make the following calculation: 603.27 / 100 x 29, which gives a total of 174.94 grams.

In this way, your reasoning is correct and it is probably a mistake in the book.

6 0

3 years ago

Consider 2.4 moles of a gas contained in a 4.0 L bulb at a constant temperature of 32°C. This bulb is connected to an evacuated

Luden [163]

Complete Question

The complete question is shown on the first uploaded image

Answer

a

As the valve is opened , the gas will flow into the empty container until the both containers have the same pressure

b

$\Delta H = 0\ , \Delta E = 0 , q= 0 , w= 0$

c

The driving force for this process is the increase in entropy this is because the movement of the internal energy of the gas into a larger volume, what this does is that it increases the amount of disorder(entropy).

Explanation

In order to obtain the parameter in the part B of the question we are first obtain the initial pressure, using the ideal gas equation

$P = \frac{nRT}{V}$

$P = \frac{(2.4mol)(0.0821\frac{1 atm}{K \cdot \ mol} )}{4.0L}$

$P =15 \ atm$

The next thing is to obtain the new pressure of the gas , using boyle's law

$P_1V_1 = P_2V_2$

$P_2 = \frac{P_1 V_1}{V_2}$

$P_2 = \frac{(15 \ atm)(4.0L)}{24.0 L}$

$P_2 = 2.5 \ atm$

Since the this process is isothermal , the change in heat is equal to zero

i.e q = 0 J

The workdone to move the gas to the other container is zero because the the pressure at this second container is zero due to the fact that it is a vacuum

i.e $w = -P_{external} \Delta V$

$=-(0 \ atm) (24.0 - 4.0L)$

$= 0L \cdot atm$

Since the change in heat is zero and the workdone is zero then the change in internal energy is equal to 0

This is because the change in internal energy is equal to a summation of change in heat and the workdone

i.e $\Delta E = q + w$

$= 0J$

Generally the change in enthalpy is mathematically represented as

$\Delta H = n C_p \Delta T$

Since the temperature is zero this means that the change in temperature is zero , substituting this value for change in temperature into the equation for change in enthalpy

$\Delta H = n C_p (0)$

$= 0J$

7 0

4 years ago