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Alborosie
2 years ago
9

Which of the following choices represents hydrogen peroxide

Chemistry
2 answers:
GaryK [48]2 years ago
8 0
A molocule i think sorry if im wrong


worty [1.4K]2 years ago
8 0
Oxygen atom i might be wrong 
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glycerol, pyruvate, glycerol -3-phosphate, dihydroxyacetone phosphate and glucose♡ hope this helps♡

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3 years ago
Describe the sequence of events in the formation of an evaporite
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- Water enter from creeks, bringing whatever material it has dissolved in its way

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3 0
3 years ago
When a piece of metal is irradiated with UV radiation (λ = 162 nm), electrons are ejected with a kinetic energy of 3.54×10-19 J.
dsp73

We have that the work function of the metal

\phi=1.227*10^{-18}J

From the Question we are told that

UV radiation (λ = 162 nm)

Kinetic energy K.E =3.54*10-19 J.

Generally the equation for Kinetic energy    is mathematically given as

KE =\frac{hc}{\pi-\phi} \\\\\phi =\frac{ 6.626*10^{-34} * 3*10^8}{162*10^{-9} -3.54*10^{-19}}

\phi=1.227*10^{-18}J

For more information on this visit

brainly.com/question/12669551?referrer=searchResults

8 0
2 years ago
If the sample contained 2.0 moles of KClO3 at a temperature of 214.0 °C, determine the mass of the oxygen gas produced in grams
Westkost [7]

Answer : The mass of the oxygen gas produced in grams and the pressure exerted by the gas against the container walls is, 96 grams and 1.78 atm respectively.

Explanation : Given,

Moles of KCl_3 = 2.0 moles

Molar mass of O_2 = 32 g/mole

Now we have to calculate the moles of MgO

The balanced chemical reaction is,

2KClO_3\rightarrow 2KCl+3O_2

From the balanced reaction we conclude that

As, 2 mole of KClO_3 react to give 3 mole of O_2

So, 2.0 moles of KClO_3 react to give \frac{2.0}{2}\times 3=3.0 moles of O_2

Now we have to calculate the mass of O_2

\text{ Mass of }O_2=\text{ Moles of }O_2\times \text{ Molar mass of }O_2

\text{ Mass of }O_2=(3.0moles)\times (32g/mole)=96g

Therefore, the mass of oxygen gas produced is, 96 grams.

Now we have to determine the pressure exerted by the gas against the container walls.

Using ideal gas equation:

PV=nRT\\\\PV=\frac{w}{M}RT\\\\P=\frac{w}{V}\times \frac{RT}{M}\\\\P=\rho\times \frac{RT}{M}

where,

P = pressure of oxygen gas = ?

V = volume of oxygen gas

T = temperature of oxygen gas = 214.0^oC=273+214.0=487K

R = gas constant = 0.0821 L.atm/mole.K

w = mass of oxygen gas

\rho = density of oxygen gas = 1.429 g/L

M = molar mass of oxygen gas = 32 g/mole

Now put all the given values in the ideal gas equation, we get:

P=1.429g/L\times \frac{(0.0821L.atm/mole.K)\times (487K)}{32g/mol}

P=1.78atm

Thus, the pressure exerted by the gas against the container walls is, 1.78 atm.

7 0
3 years ago
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