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sdas [7]
3 years ago
13

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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Can someone please help me finish this all?
Phoenix [80]

Answer:2-liver because it doesnt belong to digestive system

3-esophagus because it doesnt belong to the circulatory system

4-heart because it doesnt belong to the nervous system

5-skull because it doesnt belong to the endocrine system

Explanation:

4 0
3 years ago
Identify the highlighted secondary lymphatic tissue located on the lateral wall of the oropharynx just inferior to the soft pala
Katena32 [7]

Answer:

The correct answer is - palatine tonsils .

Explanation:

Palatine tonsils or more normally tonsils are the pair of soft tissue masses of highlighted secondary lymphatic tissue present on the lateral wall of the oropharynx in the tonsilar fossa just below to the soft plate. It is the lymphatic tissue that is located to inferior to the soft plate.

Thus, the correct answer is - palatine tonsils.

6 0
3 years ago
The image above shows a ______ wave.
Soloha48 [4]

Answer:

longitudinal

Explanation:

the wave travels in the same direction as the movement from the person

8 0
3 years ago
If you have a 1500 g aluminum pot, how much heat energy is needed to raise its temperature by 100°C?
Nataly [62]

The heat energy required to raise the temperature of 1500 g of aluminium pot by 100°C is 135 kJ.

The heat energy required to raise the temperature of 1500 g of copper pot by 100 °C is 57.75 kJ.

Explanation:

The heat energy required to raise the temperature of any body can be obtained from the specific heat formula. As this formula states that the heat energy required to raise the temperature of the body is directly proportional to the product of mass of the body, specific heat capacity of the material and temperature change experienced by the material.

So in this problem, the mass of the aluminium is given as m = 1500 g, the specific heat of the aluminium is 0.900 J/g °C. Then as it is stated that the temperature is raised by 100 °C, so the pots are heat to increase by 100 °C from its initial temperature. This means the difference in temperature will be 100°C (ΔT = 100°C).

Then, the heat energy required to raise the temperature will be

q = m*c*del T = 1500 * 0.900 * 100 = 135000 = 135 kJ

Thus, the heat energy required to raise the temperature of 1500 g of aluminium pot by 100 °C is 135 kJ.

Similarly, the mass of copper pot is given as 1500 g, the specific heat capacity of copper is 0.385 and the difference in temperature is 100  °C.

Then, the heat energy required to raise its temperature will be

q = m*c*del T = 1500 * 0.385 * 100 = 57750 = 57.75 kJ

And the heat energy required to raise the temperature of 1500 g of copper pot by 100°C is 57.75 kJ.

So, the heat energy required to raise the temperature of 1500 g of aluminium pot by 100°C is 135 kJ. And the heat energy required to raise the temperature of 1500 g of copper pot by 100 °C is 57.75 kJ.

8 0
3 years ago
If 5.57 g of Ag2O is sealed in a 75.0-mL tube filled with 760 torr of N2 gas at 28 ∘C, and the tube is heated to 310 ∘C, the Ag2
Snezhnost [94]

Answer: Total pressure = 7293.2 torr or 9.60 atm

Explanation:

<em>Total pressure = partial pressure of nitrogen + partial pressure of oxygen</em>

The partial pressure due to nitrogen is determined using the equation of Gay-Lussac's law: <em>P₁/T₁=P₂/T₂</em>

P₁ = 760 torr = 1atm, T₁ = 28∘C = (273+28)K = 301k, P₂ = ?, T₂ = 310∘C =(310+273)K = 583K

P₂ = P₁ T₂/ T₁

P₂ = 760 * 583 / 301 = 1472.03 torr

The pressure due to Oxygen gas produced is calculated thus:

Balanced equation of the decomposition of Ag₂O at s.t.p. is as follows;

2Ag₂O ----> 4Ag + O₂(g)

2 moles of Ag₂O produces 1 mole of O₂

molar mass of Ag₂O = (2*108 + 16)g = 232g/mol

molar volume of gas at s.t.p. = 22.4L

2*232g i.e. 464g of Ag₂O produces 22.4L of O₂

5.57g of Ag₂O will produce 5.57g*22.4L/464g = 0.269L or 269mL of O₂

Using the General gas equation  P₁V₁/T₁=P₂V₂/T₂

P₁ = 1atm = 760 torr, V₁ = 269mL, T₁=273K, P₂ = ?, V₂= 75mL, T₂ = 583K

P₂ = P₁V₁T₂/V₂T₁

P₂ = 760*269*583 / 75*273

P₂ = 5821.17 torr

Total pressure = (1472.03 + 5821.17) torr

Total pressure = 7293.2 torr or 9.60 atm

7 0
3 years ago
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