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4 years ago

1. All of the following are correct comparisons of anaerobic and aerobic respiration except

Chemistry

2 answers:

Lisa [10]4 years ago

8 0

Answer:

1. AEROBIC RESPIRATION IS DONE BY ANIMAL CELLS WHILE ANAEROBIC RESPIRATION IS DONE BY PLANT AND ANIMAL CELLS

2. CELLULAR RESPIRATION OCCURS IN BOTH PLANTS AND ANIMALS

3. CELLULAR RESPIRATION GIVES CARBON DIOXIDE, WATER AND ENERGY AS BY PRODUCTS.

4. LACTIC ACID AND ENERGY

5. MITOCHODRION

Explanation:

Aerobic and anaerobic respiration occur in both plant and animal cells depending on the absence or presence of oxygen. in the presence of oxygen, plants and animal cells use aerobic pathway to produce their energy during metabolism. While in the absence of oxygen, the cells undergo anaerobic respiration and lactic acid is formed as byproduct.

2. This is true because both plant and animal cell perform cellular respiration. This is the process of manufacturing food substances needed by the cells.

3. The byproducts of cellular respiration are carbon dioxide, water and energy. Energy is used by the cells for its activities and some are stored as glycogen in the presence of excess energy. Carbon dioxide is excreted from the cells by the lungs into the atmosphere, water is excreted by the kidneys.

4. The byproducts of anaerobic respiration is lactic acid and energy.

5. Mitochondrion is the site for cellular respiration as it serves as the power house of the cell.

xxMikexx [17]4 years ago

3 0

1. All of the following are correct comparisons of anaerobic and aerobic respiration except
- Answer: aerobic respiration is only done by animal cells but anaerobic is done by plant and animal cells

2. Which of the following is a true statement?

- Answer: Cellular respiration occurs in both plant and animal cells.

3. Carbon dioxide, water, and energy are the products of
- Answer: cellular respiration

4.When anaerobic respiration occurs in muscle cells, it produces
-Answer: lactic acid and energy

5. What is the location within a cell where cellular respiration occurs?
- Answer: The mitochondrion

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Now, Johnny the pool guy must add enough Na2CO3 to get the pH back up to 7.60. How many grams of Na2CO3 does Johnny need to add

vaieri [72.5K]

Answer:

m = 3.4126 g

Explanation:

First, the question is incomplete but I already put in the comments the rest of the question.

Let's solve the first two questions, and then the actual question you are asking here to give you a better explanation of how to do it.

1) We need the volume of the pool, in this case is easy. Assuming the pool is rectangular, we use the volume of a parallelepiped which is the following:

V = h * d * w

We have the data, but first we will convert the feet to centimeter. This is because is easier to work the volume in cm³ than in feet.

So the height, width and depth of the pool in centimeter are:

h = 32 * 30.48 = 975.36 cm

w = 18 * 30.48 = 548.64 cm

d = 5.3 * 30.48 = 161.54 cm

Now the volume:

V = 975.36 * 548.64 * 161.54

V = 86,443,528.79 cm³ or 86,443,528.79 mL or 86,443.5 L

2) If the pool has a pH of 6.4, the concentration of H+ can be calculated with the following expression:

[H+] = antlog(-pH) or 10^(-pH)

Replacing we have:

[H+] = 10^(-6.4)

[H+] = 3.98x10^-7 M

3) Finally the question you are asking for.

According to the reaction:

Na2CO3 + H+ → 2Na+ + HCO3−

We can see that there is ratio of 1:1 between the H+ and the Na2CO3, so, if we have initially a concentration of 3.98x10^-7 M, the difference between the new concentration of H+ and the innitial, will give the concentration to be added to the pool to raise the pH. Then, with the molecular weight of Na2CO3 (105.98 g/mol) we can know the mass needed.

The new concentration of [H+] is:

[H+] = 10^(-7.6) = 2.58x10^-8 M

The difference of both [H+] will give concentration of Na2CO3 used:

3.98x10^-7 - 2.58x10^-8 = 3.73x10^-7 M

The moles:

moles = 3.73x10^-7 * 86,443.5 = 0.0322 moles

Finally the mass:

m = 0.0322 * 105.98

m = 3.4126 g

5 0

3 years ago

Question 2(Multiple Choice Worth 3 points)

viktelen [127]

Answer:

2.0 x 10⁻³ M/s.

Explanation:

The rate of the reaction is the change of the concentration of the reactants or the products with time.

The rate of the reaction = - Δ[reactants]/Δt = - [(0.6 M - 1.8 M)]/(580 s - 0 s) = 2.069 x 10⁻³ M/s.

5 0

3 years ago

Balance the equation: Fe + O2 ----> Fe2O3

LenaWriter [7]

Fe + O2 ----> Fe2O3
4Fe + 3O2 ----> 2Fe2O3
Left Hand Side;
Fe=4
O2=6
Right Hand Side:
Fe=4
O2=6

7 0

3 years ago

When metallic sodium is dissolved in liquid sodium chloride, electrons are released into the liquid. These dissolved electrons a

qaws [65]

Answer:

The edge of the length is $\mathbf{L = 8.54 \times 10^{-10} \ m}$

Explanation:

From the given information:

The associated energy for a particle in three - dimensional box can be expressed as:

$E_n = \dfrac{h^2}{8mL^2}(n_x^2+n_y^2+n_z^2)$

here;

h = planck's constant = $6.626 \times 10^{-34} \ Js$

$n_i$ = the quantum no in a specified direction

m = mass (of particle)

L = length of the box

At the ground state $n_x = n_y = n_z=1$

The energy at the ground state can be calculated by using the formula:

$E_1 =\dfrac{3h^2}{8mL^2}$

At first excited energy level, one of the quantum values will be 2 and the others will be 1.

Thus, the first excited energy will be: 2,1,1

∴

$E_2 =\dfrac{(2^2+1^2+1^2)h^2}{8mL^2}$

$E_2 =\dfrac{(4+1+1)h^2}{8mL^2}$

$E_2 =\dfrac{(6)h^2}{8mL^2}$

The transition energy needed to move from the ground to the excited state is:

$\Delta E= E_2 - E_1$

$\Delta E= \dfrac{6h^2}{8mL^2}- \dfrac{3h^2}{8mL^2}$

$\Delta E= \dfrac{3h^2}{8mL^2}}$ ----- (1)

Recall that:

the wavelength identified with the electronic transition is: 800 nm

800 nm = 8.0 × 10⁻⁷ m

However, the energy-related with the electronic transition is:

$\Delta E =\dfrac{hc}{\lambda}$

$\Delta E =\dfrac{6.626 \times 10^{-34} \times 2.99 \times 10^8}{8.0 \times 10^{-7} }$

$\Delta E =2.48 \times 10^{-19} \ J$

Replacing the value of $\Delta E$ in (1); then:

$2.48 \times 10^{-19}= \dfrac{3h^2}{8mL^2}}$

Making the edge length L the subject of the formula; we have:

$L = \sqrt{\dfrac{3h^2}{8m \times2.48 \times 10^{-19}} }$

$L = \sqrt{\dfrac{3\times (6.626 \times 10^{-34})^2}{8(9.1 \times 10^{-31} ) \times2.48 \times 10^{-19}} }$

$\mathbf{L = 8.54 \times 10^{-10} \ m}$

Thus, the edge of the length is $\mathbf{L = 8.54 \times 10^{-10} \ m}$

5 0

3 years ago

When zn(oh)2(s) was added to 1.00 l of a basic solution, 1.09×10−2 mol of the solid dissolved. what is the concentration of oh−

uysha [10]

The concentration of the basic solution is determined by:

N = (number of moles / volume of solution)

number of moles = 1.09 x 10^-2 mol
volume of solution = 1 liter
N of basic solution = 1.09 x 10^-2 mol / 1 liter
N = 1.09 x 10^-2 mol/L

The initial concentration of Zn (OH)2 is 0; the basic solution is 1.09x10^-2 M, then the concentration of OH in the final solution is 1.09x10^-2 M

7 0

3 years ago

Read 2 more answers