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3 years ago

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Who needs a higher caloric intake due to higher calorie expenditure? Person sitting on couch Person walking the dog Person playi

ng basketball

1 answer:

dimulka [17.4K]3 years ago

8 0

Person playing basketball because they are active

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Which formula is a molecular compound?<br><br> H2O2<br> CaO<br> MgS<br> NaCl

Alborosie

Answer:

H2O2

Explanation:

8 0

3 years ago

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Open the Balancing Chemical Equations interactive and select Introduction mode. Then choose Separate Water. Adjust the coefficie

aleksklad [387]

<h2> $H_2O + H_2 + O_2$ </h2>

Explanation:

1. Water decomposition

Decomposition reactions are represented by-

The general equation: AB → A + B.

Various methods used in the decomposition of water are -

Electrolysis
Photoelectrochemical water splitting
Thermal decomposition of water
Photocatalytic water splitting

Water decomposition is the chemical reaction in which water is broken down giving oxygen and hydrogen.

The chemical equation will be -

$H_2O + H_2 + O_2$

Hence, balancing the equation we need to add a coefficient of 2 in front of $H_2O$ on right-hand-side of the equation and 2 in front of $H_2$ on left-hand-side of the equation.

∴The balanced equation is -

$2 H_2O$ → $2 H_2 + O_2$

2. Formation of ammonia

The formation of ammonia is by reacting nitrogen gas and hydrogen gas.

$N_2 + H$ → $NH_3$

Hence, for balancing equation we need to add a coefficient of 3 in front of hydrogen and 2 in front of ammonia.

∴The balanced chemical equation for the formation of ammonia gas is as follows -

$N_2+3H$ → $2NH_3$ .

When 6 moles of $N_2$ react with 6 moles of $H_2$ 4 moles of ammonia are produced.

5 0

2 years ago

What is the condition of the outside air at a certain time and place?

11111nata11111 [884]

Weather... weather is the obvious answer

6 0

3 years ago

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Two common methods to generate an aldehyde is by oxidation of an alcohol and through ozonolysis.

kotykmax [81]

Answer:

<u>a</u><u>.</u><u> </u><u>True</u><u>.</u>

Explanation:

Only primary and secondary alcohols can oxidise to give an aldehyde. But a weak oxidizing agent must be used to prevent formation of a carboxylic acid or ketone.

weak oxidizing agents: Chromyl chloride, silver/oxygen/500°C

take an example of <u>e</u><u>t</u><u>h</u><u>a</u><u>n</u><u>o</u><u>l</u><u>:</u>

<u> ${ \bf{CH _{3} CH_{2}OH \: \: \frac{Ag/O_{2} }{500 \degree C} > \: \:CH _{3} CHO}}$ </u>

<u> ${ \sf{CH _{3} CHO \: \: is \: ethanal}}$ </u>

<u>B</u><u>y</u><u> </u><u>o</u><u>z</u><u>o</u><u>n</u><u>o</u><u>l</u><u>y</u><u>s</u><u>i</u><u>s</u><u>:</u>

Here, reactants are Ozone gas, Carbon tetrachloride at a temperature (<20°C), ethanoic acid, zinc and water.

take an example of propanol:

if it undergoes ozonolysis, it gives ethanal and methanal.

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3 years ago

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Consider the reaction: 2 H2O (g)-->2 H2 (g) + O2 (g). ΔH=483.6 Kj/mol. If 2 moles of H2O (g) are converted H2(g) and O2(g) ag

denis-greek [22]

DE = dH - PdV

<span>2 H2O(g) → 2 H2(g) + O2(g) </span>

<span>You can see that there are 2 moles of gas in the reactants and 3 moles of gas in the products. </span>

<span>1 moles of ideal gas occupies the same volume as 1 mole of any other ideal gas under the same conditions of temp and pressure. </span>

<span>Since it is done under constant temp and pressure that means the volume change will be equal to the volume of 1 mole of gas </span>

<span>2 moles reacts to form 3 moles </span>

<span>The gas equation is </span>

<span>PV = nRT </span>
<span>P = pressure </span>
<span>V = volume (unknown) </span>
<span>n = moles (1) </span>
<span>R = gas constant = 8.314 J K^-1 mol^-1 </span>
<span>- the gas constant is different for different units of temp and pressure (see wikki link) in this case temp and pressure are constant, and we want to put the result in an equation that has Joules in it, so we select 8.314 JK^-1mol^-1) </span>
<span>T = temp in Kelvin (kelvin = deg C + 273.15 </span>
<span>So T = 403.15 K </span>

<span>Now, you can see that PV is on one side of the equation, and we are looking to put PdV in our dE equation. So we can say </span>

<span>dE = dH -dnRT (because PV = nRT) </span>

<span>Also, since the gas constant is in the unit of Joules, we need to convert dH to Joules </span>

<span>dH = 483.6 kJ/mol = 483600 Joules/mol </span>

<span>dE = 483600 J/mol - (1.0 mol x 8.314 J mol^-1K-1 x 403.15 K) </span>
<span>dE = 483600 J/mol - 3351.77 J </span>
<span>dE = 480248.23 J/mol </span>
<span>dE = 480.2 kJ/mol </span>

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3 years ago

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