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3 years ago

While Cameron is intensely working out, his muscles begin to get sore and start to burn. What type of cellular

Chemistry

1 answer:

max2010maxim [7]3 years ago

4 0

Answer:

1) Anaerobic respiration

2) Lactic acid

3) Stop the work out

Explanation:

1) Anaerobic respiration

Muscular anaerobic respiration to produce energy during exercises or workout, takes place when the level of oxygen available for the amount of energy being generated is not sufficient, and the balance generated energy is produced along with lactic acid by anaerobic respiration

2) Lactic acid

The produced lactic acid from the anaerobic respiration at the muscle sites results in a burning sensation being felt during strenuous muscular activity in the presence of low oxygen supply

3) Stop the work out

In order to stop the burning, the muscular activity is to be stopped, to allow the resumption of the aerobic process of energy production, and to allow the produced lactic acid to be cleared

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Copper is commonly mined as an ore with a variable percent composition of copper (II) sulfide. This ore is also sometimes referr

pantera1 [17]

Answer:

$m_{CuO}=93.6gCuO$

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

$CuS+\frac{3}{2} O_2\rightarrow CuO+SO_2$

Thus, given the 1.00-kg of 12.5% ore, we can compute the theoretical yield of copper (II) oxide via stoichiometry:

$m_{CuO}^{theoretical}=1.00kgCuS*\frac{1000gCuS}{1kgCuS} *\frac{12.5gCuS}{100gCuS} *\frac{1molCuS}{95.6gCuS} *\frac{1molCuO}{1molCuS} *\frac{79.5gCuO}{1molCuO} \\\\m_{CuO}^{theoretical}=103.95gCuO$

Whereas the third factor accounts for the percent purity of the covellite. Then, given the percent yield, we can compute the actual yield by:

$m_{CuO}=103.95gCuO*0.9\\\\m_{CuO}=93.6gCuO$

Regards.

6 0

3 years ago

What is the break down of food into energy

AnnyKZ [126]

it is nutrients that's it

3 0

3 years ago

A gas occupies a volume of 2.4 L at 14.1 kPa. What volume will the gas occupy at 84.6 kPa?

Naddik [55]

Answer:

<u>0.40 L</u>

Explanation:

Boyle's law for gases states that, at constant temperature, the volume and pressure of a fixed amount of gas are inversely related.

Mathematically, that is:

PV = constant

P₁V₁ = P₂V₂

Here, you have:

V₁ = 2.4 L

P₁ = 14.1 Kpa

P₂ = 84.6 KPa

V₂ = ?

Then, you can solve for V₂:

V₂ = P₁V₁ / P₂

Substitute and compute:

V₂ = 14.1 KPa × 2.4L / 84.6 KPa = 0.40 L ← answer

3 0

3 years ago

The titration of a 20.0-mLmL sample of an H2SO4H2SO4 solution of unknown concentration requires 22.87 mLmL of a 0.158 M KOHM KOH

kkurt [141]

Answer:

0.0905 M

Explanation:

Let's consider the neutralization reaction between H2SO4 and KOH.

H₂SO₄ + 2 KOH → K₂SO₄ + 2 H₂O

22.87 mL of 0.158 M KOH react. The reacting moles of KOH are:

0.02287 L × 0.158 mol/L = 3.61 × 10⁻³ mol

The molar ratio of H₂SO₄ to KOH is 1:2. The reacting moles of H₂SO₄ are 1/2 × 3.61 × 10⁻³ mol = 1.81 × 10⁻³ mol

1.81 × 10⁻³ moles of H₂SO₄ are in 20.0 mL. The molarity of H₂SO₄ is:

M = 1.81 × 10⁻³ mol / 0.0200 L = 0.0905 M

3 0

2 years ago

A hypothetical element has an atomic weight of 48.68 amu. It consists of three isotopes having masses of 47.00 amu, 48.00 amu, a

Morgarella [4.7K]

Answer : The percent abundance of the heaviest isotope is, 78 %

Explanation :

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:

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

As we are given that,

Average atomic mass = 48.68 amu

Mass of heaviest-weight isotope = 49.00 amu

Let the percentage abundance of heaviest-weight isotope = x %

Fractional abundance of heaviest-weight isotope = $\frac{x}{100}$

Mass of lightest-weight isotope = 47.00 amu

Percentage abundance of lightest-weight isotope = 10 %

Fractional abundance of lightest-weight isotope = $\frac{10}{100}$

Mass of middle-weight isotope = 48.00 amu

Percentage abundance of middle-weight isotope = [100 - (x + 10)] % = (90 - x) %

Fractional abundance of middle-weight isotope = $\frac{(90-x)}{100}$

Now put all the given values in above formula, we get:

$48.68=[(47.0\times \frac{10}{100})+(48.0\times \frac{(90-x)}{100})+(49.0\times \frac{x}{100})]$

$x=78\%$

Therefore, the percent abundance of the heaviest isotope is, 78 %

5 0

3 years ago

Read 2 more answers