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liubo4ka [24]
3 years ago
5

The energy released in cellular respiration was originally present in ____________

Chemistry
1 answer:
marusya05 [52]3 years ago
5 0

Answer:

ATP

Explanation:

During cellular respiration, glucose is broken down in the presence of oxygen to produce carbon dioxide and water. Energy released during the reaction is captured by the energy-carrying molecule ATP (adenosine triphosphate).

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TIMED PLS HELP AND WILL GIVE BRAINLIST
Strike441 [17]

Answer:

19.4 g of alum, will be its theoretical yield

Explanation:

The reaction is:

2 Al + 2 KOH + 4 H₂SO₄ + 22H₂O → 3H₂ + 2KAl(SO₄)₂•12H₂O

Let's determine the amount of acid.

M are the moles contained in 1 L of solution or it can be mmoles that are contained in 1 mL of solution

M = mmol /mL

M . mL = mmol

We replace: 8.3 mL . 9.9 M = 82.17 mmoles

We convert to moles: 82.17 mmol . 1 mol / 1000mmol = 0.082 moles

Ratio is 4:2

4 moles of sulfuric acid can make 2 moles of alum

By the way, 0.082 moles of acid may produce ( 0.082 . 2) /4 = 0.041085 moles.

We convert moles to mass:

Molar mass of alum is: 473.52 g/mol.

0.041085 moles . 473.52 g/mol = 19.4 g

6 0
3 years ago
What are the nine major plates of the lithosphere
artcher [175]
North American plate
Pacific plate
Nazca plate
Antarctic plate
south American plate
African plate
Eurasian plate
indo-austrailian plate

6 0
3 years ago
A) Find the gas speed of sulfur dioxide at 100.0 degrees Celsius? ______________
gtnhenbr [62]

a. 381.27 m/s

b. the rate of effusion of sulfur dioxide = 2.5 faster than nitrogen triiodide

<h3>Further explanation</h3>

Given

T = 100 + 273 = 373 K

Required

a. the gas speedi

b. The rate of effusion comparison

Solution

a.

Average velocities of gases can be expressed as root-mean-square averages. (V rms)  

\large {\boxed {\bold {v_ {rms} = \sqrt {\dfrac {3RT} {Mm}}}}

R = gas constant, T = temperature, Mm = molar mass of the gas particles  

From the question  

R = 8,314 J / mol K  

T = temperature  

Mm = molar mass, kg / mol  

Molar mass of Sulfur dioxide = 64 g/mol = 0.064 kg/mol

\tt v=\sqrt{\dfrac{3\times 8.314\times 373}{0.064} }\\\\v=381.27~m/s

b. the effusion rates of two gases = the square root of the inverse of their molar masses:  

\rm \dfrac{r_1}{r_2}=\sqrt{\dfrac{M_2}{M_1} }

M₁ = molar mass sulfur dioxide = 64

M₂ =  molar mass nitrogen triodide = 395

\tt \dfrac{r_1}{r_2}=\sqrt{\dfrac{395}{64} }=\dfrac{20}{8}=2.5

the rate of effusion of sulfur dioxide = 2.5 faster than nitrogen triodide

4 0
3 years ago
G.P.E. on Earth is determined by an object's _______and _______
lozanna [386]

Answer:

mass and height of the object.

5 0
2 years ago
The study of the interaction between animals and plants within a specific environment is called
sergij07 [2.7K]

Answer:

The study is called ecology

Explanation:

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8 0
3 years ago
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