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4 years ago

Which chemical is produced during vigorous exercise when the supply of oxygen is limited or inadequate?

1 answer:

4 0

It is lactic acid the chemical is produced during vigorous exercise when the supply of oxygen is limited or inadequate. It forms when the body breaks down carbohydrates to use for energy during times of low oxygen levels. Hope this answers the question. Have a nice day.

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Plz help me with this question plzzzzz

LekaFEV [45]

Answer:

Explanation:

Formula: Al2O3

If we require 2Al2O3

We divide 2 by 3

Sorry for the shadow of phone and fingers.

8 0

3 years ago

When sodium metal is treated with water we get gas which is absorbed by palladium metal is ?

Scilla [17]

that gas is called hydrogen gas

8 0

4 years ago

Which of the following describes the correct order for using the scientific

larisa [96]

A. is the answer
if u have any doubts ask me
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8 0

3 years ago

Read 2 more answers

Copper metal (Cu) reacts with silver nitrate (AgNO3) in aqueous solution to form Ag and Cu(NO3)2. An excess of AgNO3 is present.

sweet [91]

Toichiometry time! Remember to look at the equation for your molar ratios in other problems.
31.75 g Cu | 1 mol Cu | 2 mol Ag | 107.9 g Ag 6851.65⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻ → ⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻ = 107.9 g Ag ∅ | 63.5 g Cu | 1 mol Cu | 1 mol Ag 63.5

There's also a shorter way to do this: Notice the molar ratio from Cu to Ag, which is 1:2. When you plug in 31.75 into your molar mass for Cu, it equals 1/2 mol. That also means that you have 1 mol Ag because of the ratio, qhich you can then plug into your molar mass, getting 107.9 as well.

3 0

3 years ago

Read 2 more answers

A radioisotope decays to give an alpha particle andPb-208.

oee [108]

Answer: The original element is $_{84}^{212}\textrm{Po}$

Explanation:

Alpha decay is defined as the process in which alpha particle is emitted. In this process, a heavier nuclei decays into a lighter nuclei. The alpha particle released carries a charge of +2 units.

The released alpha particle is also known as helium nucleus.

$_Z^A\textrm{X}\rightarrow _{Z-2}^{A-4}\textrm{Y}+_2^4\alpha$

For the given alpha decay process of an isotope:

$^{Z}_{A}\textrm{X}\rightarrow ^{208}_{82}\textrm{Pb}+_2^4\alpha$

To calculate A:

Total mass on reactant side = total mass on product side

A = 208 + 4

A = 212

To calculate Z:

Total atomic number on reactant side = total atomic number on product side

Z = 82 + 2

Z = 84

The isotopic symbol of unknown element is $_{84}^{212}\textrm{Po}$

Hence, the original element is $_{84}^{212}\textrm{Po}$

3 0

3 years ago