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stira [4]
3 years ago
10

Rosa is breathing normally and eats a peanut butter sandwich for lunch. Peanut butter contains a lot of protein, and bread is mo

stly starch. Rosa plans to go for a run later this afternoon.
What does she need from the food she ate and the air she breathes so that she can go on her run? How do Rosa’s body systems work together to get the molecules she needs into her cells? How do her cells use these molecules to make energy for her body to run?
Chemistry
2 answers:
jeka57 [31]3 years ago
8 0

Answer:

She mainly needs the Peanut Butter because it has a lot of protein and her cells worked together to help her run.

FrozenT [24]3 years ago
3 0

Answer:

mostly PB for proteins

Explanation:

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2.3 Zinc has five naturally occurring isotopes: 48.63% of 64 Zn with an atomic weight of 63.929 amu; 27.90% of 66Zn with an atom
lakkis [162]

<u>Answer:</u> The average atomic mass of element Zinc is 65.40 amu.  

<u>Explanation:</u>

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i .....(1)

  • <u>For _{30}^{64}\textrm{Zn} isotope:</u>

Mass of _{30}^{64}\textrm{Zn} isotope = 63.929 amu

Percentage abundance of _{30}^{64}\textrm{Zn} isotope = 48.63 %

Fractional abundance of _{30}^{64}\textrm{Zn} isotope = 0.4863

  • <u>For _{30}^{66}\textrm{Zn} isotope:</u>

Mass of _{30}^{66}\textrm{Zn} isotope = 65.926 amu

Percentage abundance of _{30}^{66}\textrm{Zn} isotope = 27.90 %

Fractional abundance of _{30}^{66}\textrm{Zn} isotope = 0.2790

  • <u>For _{30}^{67}\textrm{Zn} isotope:</u>

Mass of _{30}^{67}\textrm{Zn} isotope = 66.927 amu

Percentage abundance of _{30}^{67}\textrm{Zn} isotope = 4.10 %

Fractional abundance of _{30}^{67}\textrm{Zn} isotope = 0.0410

  • <u>For _{30}^{68}\textrm{Zn} isotope:</u>

Mass of _{30}^{68}\textrm{Zn} isotope = 67.925 amu

Percentage abundance of _{30}^{68}\textrm{Zn} isotope = 18.75 %

Fractional abundance of _{30}^{68}\textrm{Zn} isotope = 0.1875

  • <u>For _{30}^{70}\textrm{Zn} isotope:</u>

Mass of _{30}^{70}\textrm{Zn} isotope = 69.925 amu

Percentage abundance of _{30}^{70}\textrm{Zn} isotope = 0.62 %

Fractional abundance of _{30}^{70}\textrm{Zn} isotope = 0.0062

Putting values in equation 1, we get:

\text{Average atomic mass of Zinc}=[(63.929\times 0.4863)+(65.926\times 0.2790)+(66.927\times 0.0410)+(67.925\times 0.1875)+(69.925\times 0.0062)]

\text{Average atomic mass of Zinc}=65.40amu

Hence, the average atomic mass of element Zinc is 65.40 amu.

7 0
3 years ago
How does the burning of fossil fuels increase the greenhouse effect answers?
miskamm [114]
The burning of fossil fuels like coal, natural gas etc... leads to the emmission of carbondioxide which is a green house gas and leads to the increase of heat in the atmosphere. This results in greenhouse effect.
8 0
3 years ago
Balance each of the following examples of heterogeneous equilibria and write each Kc expression. Then calculate the value of Kc
marysya [2.9K]

Explanation:

1) 2 Al(s) + 2 NaOH(aq) + 6 H_2O(l) \longleftrightarrow 2 Na[Al(OH)_4](aq) + 7 H_2(g)

Kc=\frac{[Na[Al(OH)_4]]^2*[H_2]^7}{[NaOH]^2}

The Kc for the reverse reaction is the inverse of the Kc of the reaction:

Kc_{reverse}=\frac{1}{Kc}=\frac{1}{11}=0.091

2) H_2O(l) + SO_3(g) \longleftrightarrow H_2SO_4 (aq)

Kc=\frac{[H_2SO_4]}{[SO_3]^2}

The Kc for the reverse reaction is the inverse of the Kc of the reaction:

Kc_{reverse}=\frac{1}{Kc}=\frac{1}{0.0123}=81.3

3)  P_4(s) + 3 O_2(g) \longleftrightarrow P_4O_6(s)

Kc=\frac{1}{[O_2]^3}

The Kc for the reverse reaction is the inverse of the Kc of the reaction:

Kc_{reverse}=\frac{1}{Kc}=\frac{1}{1.56}=0.641

5 0
3 years ago
How would the addition of protons affect the concentration of CH3COOH? How would the addition of OH– affect the amount of CH3COO
Deffense [45]

Answer:

1) increase concentration

2) decrease the amount

3) decrease the concentration

4) it would increase

Explanation: edge 2021

8 0
3 years ago
Read 2 more answers
g, Assuming the precipitate is totally insoluble in water, which aqueous ions will be present in the solution (collected in the
Allushta [10]

Answer:

Cl⁻, Na⁺, OH⁻

Explanation:

The titration is:

CuCl₂(aq) + 2 NaOH(aq) → Cu(OH)₂(s) + 2 NaCl(aq)

In solution, before the reaction, the ions are Cu²⁺ and Cl⁻. The addition of NaOH (Na⁺ + OH⁻) produce the precipitation of Cu²⁺ forming Cu(OH)₂(s). When you reach the equivalence point, there is no Cu²⁺ because precipitates completely. All OH⁻ ions reacts when are added but when Cu²⁺ is finished, excess OH⁻ ions still in solution helping to detect the equivalence point.

Thus, ions present after the equivalence point are:<em> Cl⁻, Na⁺</em> (Don't react, spectator ions), and <em>OH⁻</em>.

3 0
3 years ago
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