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Pani-rosa [81]
3 years ago
6

What attracts heat more? black or aluminum foil?​

Chemistry
2 answers:
Inessa05 [86]3 years ago
8 0

Answer:

"<em><u>A</u></em><em><u>L</u></em><em><u>U</u></em><em><u>M</u></em><em><u>I</u></em><em><u>N</u></em><em><u>U</u></em><em><u>M</u></em><em><u> </u></em>FOIL" will conduct heat much more effectively than black construction paper.

Svetach [21]3 years ago
6 0

Answer:

\pink{\mid{\fbox{\tt{What attracts heat more?}}\mid}}

Black – the color that absorbs all visible wavelengths of light – attracts the most heat, followed by violet, indigo, blue, green, yellow, orange and red, in descending order.

\pink{\mid{\fbox{\tt{ black or aluminum foil? }}\mid}}

Black Foil, BlackWrap, Lindcraft Foil, Cinefoil and Shadowfoils are all names for black aluminum foil that's primarily used to block out light and is attached to light fixtures and or their barn doors. Used in any situation where extreme heat might otherwise burn flags or cutters.

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5 0
4 years ago
You want to make 500 ml of a 1 N solution of sulfuric acid (H2SO4, MW: 98.1). How many grams of sulfuric acid do you need?
umka21 [38]

Answer:

24.525 g of sulfuric acid.

Explanation:

Hello,

Normality (units of eq/L) is defined as:

N=\frac{eq_{solute}}{V_{solution}}

Since the sulfuric acid is the solute, and we already have the volume of the solution (500 mL) but we need it in liters (0.5 L, just divide into 1000), the equivalent grams of solute are given by:

eq_{solute}=N*V_{solution}=1\frac{eq}{L}*0.5L=0.5 eq

Now, since the sulfuric acid is diprotic (2 hydrogen atoms in its formula) 1 mole of sulfuric acid has 2 equivalent grams of sulfuric acid, so the mole-mass relationship is developed to find its required mass as follows:

m_{H_2SO_4}=0.5eqH_2SO_4(\frac{1molH_2SO_4}{2 eqH_2SO_4}) (\frac{98.1 g H_2SO_4}{1 mol H_2SO_4} )\\m_{H_2SO_4}=24.525 g H_2SO_4

Best regards.

4 0
4 years ago
6. How many moles of water would require 92.048 kJ of heat to raise its temperature from 34.0 °C to 100.0 °C? (3 marks)​
scoray [572]

Taking into account the definition of calorimetry, 0.0185 moles of water are required.

<h3>Calorimetry</h3>

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

Sensible heat is defined as the amount of heat that a body absorbs or releases without any changes in its physical state (phase change).

So, the equation that allows to calculate heat exchanges is:

Q = c× m× ΔT

where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation.

<h3>Mass of water required</h3>

In this case, you know:

  • Heat= 92.048 kJ
  • Mass of water = ?
  • Initial temperature of water= 34 ºC
  • Final temperature of water= 100 ºC
  • Specific heat of water = 4.186 \frac{J}{gC}

Replacing in the expression to calculate heat exchanges:

92.048 kJ = 4.186 \frac{J}{gC}× m× (100 °C -34 °C)

92.048 kJ = 4.186 \frac{J}{gC}× m× 66 °C

m= 92.048 kJ ÷ (4.186 \frac{J}{gC}× 66 °C)

<u><em>m= 0.333 grams</em></u>

<h3>Moles of water required</h3>

Being the molar mass of water 18 \frac{g}{mole}, that is, the amount of mass that a substance contains in one mole, the moles of water required can be calculated as:

amount of moles=0.333 gramsx\frac{1 mole}{18 grams}

<u><em>amount of moles= 0.0185 moles</em></u>

Finally, 0.0185 moles of water are required.

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8 0
3 years ago
1) If 0.193 grams of toluene is dissolved in 2.532 grams of p-xylene, what is the molality of toluene in the solution?2) If a fr
AveGali [126]

Answer:

The value of K_f for xylene is 4.309°C/m.

The molar mass of pentane using this data is 73.82 g/mol.

Explanation:

\Delta T_f=K_f\times \frac{\text{Amount of solute}}{\text{Molar mass of solute}\times \text{Mass of solvent(kg)}}

where,

\Delta T_f =depression in freezing point

K_f = freezing point constant  

we have :

1) freezing point constant  for xylene = K_f =?

Mass of toluene = 0.193 g

Mass of xylene = 2.532 kg = 0.002532 kg ( 1 g =0.001 kg)

\Delta T_f=3.57^oC

3.57^oC=K_f\times \frac{0.193 g}{92 g/mol\times 0.002532 kg}

K_f=4.309^oC/m

The value of K_f for xylene is 4.309°C/m.

2)

Mass of pentane = 0.123 g

molar mass of pentame= M

Mass of xylene = 2.493 g =  0.002493 kg

Freezing point Constant of xylene = K_f=4.309^oC/m

2.88^oC=4.309^oC/m\times \frac{0.123g}{M\times 0.002493 kg}

M = 73.82 g/mol

The molar mass of pentane using this data is 73.82 g/mol.

5 0
3 years ago
Suppose that you add 26.7 g of an unknown molecular compound to 0.250 kg of benzene, which has a K f of 5.12 oC/m. With the adde
nadya68 [22]

Answer: The molar mass of the unknown compound is 200 g/mol

Explanation:

Depression in freezing point is given by:

\Delta T_f=i\times K_f\times m

\Delta T_f=2.74^0C = Depression in freezing point

i= vant hoff factor = 1 (for molecular compound)

K_f = freezing point constant = 5.12^0C/m

m= molality

\Delta T_f=i\times K_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}

Weight of solvent (benzene)= 0.250 kg  

Molar mass of solute = M g/mol

Mass of solute  = 26.7 g

2.74^0C=1\times 5.12\times \frac{26.7g}{Mg/mol\times 0.250kg}

M=200g/mol

Thus the molar mass of the unknown compound is 200 g/mol

4 0
4 years ago
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