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3 years ago

HELP MEHHHHH were does the first stage of cellular respiration occur ?

Chemistry

1 answer:

Reptile [31]3 years ago

7 0

Answer:

mitochondria

Explanation:

Cellular respiration occurs primarily in the mitochondria of your cells. The mitochondria is like the powerhouse of cells .

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What is the correct answer ??

Ber [7]

Answer:

[H_3 O^+] = 1.0 ×10^-13

Explanation:

If we multiply the left side we get, 1e - 13. We add 13 to the right side while subtracting the remaining value from the left side (H3O+) than combine like terms. As you will get pH = 13.00

7 0

4 years ago

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Why does 10g Nitrogen gas contain the same number of molecules as 10 g of carbon monoxide

icang [17]

Answer:

$\fbox\colorbox{cyan}{❥i \: hope \: it \: help \: you}$

4 0

3 years ago

Read 2 more answers

How many millimeters (mm) is the length of a standard table if it is

matrenka [14]

Answer:

1,500 mm

Explanation:

if 1 meter = 1000 mm, 0.5 meters is 500 mm, so 1.50 meters is 1,500 mm

5 0

3 years ago

Two objects are brought into contact Object 1 has mass 0.76 kg, specific heat capacity 0.87) g'c and initial temperature 52.2 'C

taurus [48]

Answer:

$T_F=77.4\°C$

Explanation:

Hello there!

In this case, according to the given information, it turns out possible to set up the following energy equation for both objects 1 and 2:

$Q_1=-Q_2$

In terms of mass, specific heat and temperature change is:

$m_1C_1(T_F-T_1)=-m_2C_2(T_F-T_2)$

Now, solve for the final temperature, as follows:

$T_F=\frac{m_1C_1T_1+m_2C_2T_2}{m_1C_1+m_2C_2}$

Then, plug in the masses, specific heat and temperatures to obtain:

$T_F=\frac{760g*0.87\frac{J}{g\°C} *52.2\°C+70.7g*3.071\frac{J}{g\°C}*154\°C}{760g*0.87\frac{J}{g\°C} +70.7g*3.071\frac{J}{g\°C}} \\\\T_F=77.4\°C$

Yet, the values do not seem to have been given correctly in the problem, so it'll be convenient for you to recheck them.

Regards!

4 0

3 years ago

How many liters of O2 at 298 K and 1.00 bar are produced in 1.50 hr in an electrolytic cell operating at a current of 0.0200 A?

stellarik [79]

Answer: 0.0069L

Explanation:

2H2O(l) ---->O2(g) + 4H+(aq) + 4e-

no of moles= it/eF

NO of moles of O2 produced = (Current in Ampere x Time in second)/ (Faraday constant x Number of electrons required)

Moles of O2 produced = (0.02x (60 x 60X1.5 s)/(96485 x 4)

= 0.0002798 moles= 2.798x 10 ^-4moles

Using ideal gas equation,

P V = n R T

Where, P is the pressure,

V is the volume,

n is the number of moles,

R is the gas constant, and T is the temperature

We have, 1 bar = 0.986923 atm

Substituting the values,

V = nRT/P = (2.798 x 10-4moles x 0.08205 L atm mol K x 298 K)/ 0.986923 atm = 0.0069L

Volume of O2 produced = 0.0069L

7 0

3 years ago