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boyakko [2]
3 years ago
10

In a campfire, carbon from wood reacts with oxygen in a combustion reaction. If you’ve ever started a campfire, you know you mus

t begin by lighting small pieces of wood (called kindling) instead of logs. Likewise, as this video shows, iron filings burn in the presence of air to make iron oxide. Why do iron filings burn while a heavy iron pipe does not? Why do twigs ignite better than logs? Give your answer in terms of atoms.
Chemistry
2 answers:
hichkok12 [17]3 years ago
7 0
Increased surface area means more<span> places where </span>wood can<span> turn into flame - where heat, fuel, and oxygen </span>are<span> all in the same place at the same time. </span>The mass of a small particle is such that the entire particle can be heated<span> to ignition temperature. An </span>iron<span> rod will </span>not burn<span> as it has relatively very less surface area per unit mass </span>but iron filings<span> with more surface area per unit mass can </span>burn<span> in air and get converted to </span>iron oxide.<span>Kindling </span>or<span> smaller pieces </span>or wood <span>have a lot of surface area and therefore have </span>more<span> contact with oxygen, making them easier to burn.</span>
Alenkasestr [34]3 years ago
5 0
One thing I can tell is the fact that more surface area means more wood to burn into flames. When it starts to burn Heat, fuel, and oxygen occur in the same place at the same time to put it into flames.




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A freezer compartment is covered with a 2-mm-thick layer of frost at the time it malfunctions. If the compartment is in ambient
ruslelena [56]

Answer:

The time required to melt the frost is 3.25 hours.

Explanation:

The time required to melt the frost dependes on the latent heat of the frost and the amount of heat it is transfered by convection to the air .

The heat transferred per unit area can be expressed as:

q=h_c*A*\Delta T\\\\q/A=h_c*\Delta T

being hc the convective heat transfer coefficient (2 Wm^-2K^-1) and ΔT the difference of temperature (20-0=20 °C or K).

q/A=h_c*\Delta T=2\frac{W}{m^2K}*20K=40\frac{W}{m^2}

If we take 1 m^2 of ice, with 2 mm of thickness, we have this volume

V=T*A = 0.002 m * 1 m^2=0.002m^3

The mass of the frost can be estimated as

M=\rho * V=700\frac{kg}{m^3}*0.002m^3= 1.4 kg

Then,  the amount of heat needed to melt this surface (1 m²) of frost is

Q=L*M=334\frac{kJ}{kg}*1.4kg= 467.6kJ

The time needed to melt the frost can be calculated as

t=\frac{Q}{(q/A)}=\frac{467.6kJ/m2}{40W/m2} = 11.69\frac{kJ}{W}*\frac{1W*s}{1J}*\frac{1000J}{1kJ}=   11690s=3.25h

7 0
3 years ago
A Brønsted-Lowry acid Question 9 options: A) has a lower pH than vinegar. B) is any species that donates a proton. C) is any spe
rusak2 [61]

Answer:

B) is any species that donates a proton.

7 0
3 years ago
Cuantos moles de CO2 se requieren para reaccionar con 2 moles de Ba (OH)2
EastWind [94]

Answer:

hola soy jess, tu respuesta esta aqui

¿cuantos moles de CO2 se requiere para reaccionar 2 moles de Ba(OH)2

2 mol Ba(OH)₂ × \frac{1molCO_{2} }{1molBa (OH)_{2}}

1molBa(OH)

2

1molCO

2

= 2 moles CO₂

Explanation:

espero que pueda ayudarte

hermana/hermano

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3 0
3 years ago
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Answer:

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5 0
2 years ago
For Na2HPO4:(( (Note that for H3PO4, ka1= 6.9x10-3, ka2 = 6.4x10-8, ka3 = 4.8x10-13 ) a) The active anion is H2PO4- b) The activ
Komok [63]

Answer:

Check the explanation

Explanation:

Answer – Given, H_3PO_4 acid and there are three Ka values

K_{a1}=6.9x10^8, K_{a2} = 6.2X10^8, and K_{a3}=4.8X10^{13}

The transformation of H_2PO_4- (aq) to HPO_4^2-(aq)is the second dissociation, so we need to use the Ka2 = 6.2x10-8 in the Henderson-Hasselbalch equation.

Mass of KH2PO4 = 22.0 g , mass of Na2HPO4 = 32.0 g , volume = 1.00 L

First we need to calculate moles of each

Moles of KH2PO4 = 22.0 g / 136.08 g.mol-1

                             = 0.162 moles

Moles of Na2HPO4 = 32.0 g /141.96 g.mol-1

                             = 0.225 moles

[H2PO4-] = 0.162 moles / 1.00 L = 0.162 M

[HPO42-] = 0.225 moles / 1.00 L = 0.225 M

Now we need to calculate the pKa2

pKa2 = -log Ka

       = -log 6.2x10-8

       = 7.21

We know Henderson-Hasselbalch equation

pH = pKa + log [conjugate base] / [acid]

pH = 7.21 + log 0.225 / 0.162

     = 7.35

The pH of a buffer solution obtained by dissolving 22.0 g of KH2PO4 and 32.0 g of Na2HPO4 in water and then diluting to 1.00 L is 7.35

6 0
3 years ago
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