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german
3 years ago
11

Which body system processes food into a useable source of energy?

Chemistry
2 answers:
anastassius [24]3 years ago
4 0

Answer:

Glucose, found in the food animals eat, is broken down during the process of cellular respiration into an energy source called ATP. When excess ATP and glucose are present, the liver converts them into a molecule called glycogen, which is stored for later use.

olchik [2.2K]3 years ago
3 0
The digestive system processes food into a usable source of energy.
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How do you test for carbon dioxide ?? Explain
g100num [7]

Answer:

Carbon dioxide reacts with calcium hydroxide solution to produce a white precipitate of calcium carbonate

Explanation:

. Limewater is a solution of calcium hydroxide. If carbon dioxide is bubbled through limewater, the limewater turns milky or cloudy white

5 0
2 years ago
What are the answers please help
LenaWriter [7]
Thank you for the free 15 points .
4 0
3 years ago
Predict the products that will be formed by the reaction below. Al(s)+Na2SO4(s)—>
Ostrovityanka [42]

Answer:

Al₂(SO₄)₃ and Na

Explanation:

Al has a charge of +3, Na has a charge of +1 and SO₄ has a charge of -2. Since cations and anions will bond we know that Al will bond with SO₄ leaving Na by itself (since this is a single replacement reaction). When Al bonds with SO₄ it makes aluminum sulfate which is Al₂(SO₄)₃ and Na will be left by itself.

3 0
3 years ago
Cytotoxic t cells can attack target cells with which chemical weapons?
Aneli [31]

Answer:

secrete cytotoxic substance which triggers apoptosis of target cell.

Explanation:

Cytotoxic T cells have cell surface receptor which recognizes the antigen present on the receptor of target cell. This interaction initiates the process of killing of target cell.

After interaction cytotoxic t cell release cytotoxic substance called granzyme and perforin. Granzyme triggers apoptosis through the activation of caspases or by making the release of cytochrome c and activation of the apoptosome.

Perforin make pores in the cell and its action is similar to complement membrane attack complex. Therefore cytotoxic substances are released by Tc cells which trigger apoptosis of target cell.

4 0
3 years ago
The reaction is proceeding at a rate of 0.0080 Ms-1 in 50.0 mL of solution in a system with unknown concentrations of A and B. W
Leokris [45]

Answer:

0.0010 mol·L⁻¹s⁻¹  

Explanation:

Assume the rate law is  

rate = k[A][B]²

If you are comparing two rates,

\dfrac{\text{rate}_{2}}{\text{rate}_{1}} = \dfrac{k_{2}\text{[A]}_2[\text{B]}_{2}^{2}}{k_{1}\text{[A]}_1[\text{B]}_{1}^{2}}= \left (\dfrac{\text{[A]}_{2}}{\text{[A]}_{1}}\right ) \left (\dfrac{\text{[B]}_{2}}{\text{[B]}_{1}}\right )^{2}

You are cutting each concentration in half, so

\dfrac{\text{[A]}_{2}}{\text{[A]}_{1}} = \dfrac{1}{2}\text{ and }\dfrac{\text{[B]}_{2}}{\text{[B]}_{1}}= \dfrac{1}{2}

Then,

\dfrac{\text{rate}_{2}}{\text{rate}_{1}} = \left (\dfrac{1}{2}\right ) \left (\dfrac{1}{2}\right )^{2} = \dfrac{1}{2}\times\dfrac{1}{4} = \dfrac{1}{8}\\\\\text{rate}_{2} = \dfrac{1}{8}\times \text{rate}_{1}= \dfrac{1}{8}\times \text{0.0080 mol$\cdot$L$^{-1}$s$^{-1}$} = \textbf{0.0010 mol$\cdot$L$^{-1}$s$^{-1}$}\\\\\text{The new rate is $\large \boxed{\textbf{0.0010 mol$\cdot$L$^\mathbf{{-1}}$s$^{\mathbf{-1}}$}}$}

8 0
3 years ago
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