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Natalija [7]
3 years ago
15

If a sheet of aluminum foil (d = 2.70 g/cm3) weighs 5.31 lbs and is 0.0130 mm thick, the surface area of the foil is closest toâ

¦?
Physics
1 answer:
Salsk061 [2.6K]3 years ago
5 0
Given:
d = 2.70 g/cm³, the density
M = 5.31 lbs, the mass
t = 0.0130 mm, thickness

Convert all data into SI units.
d = (2.70 g/cm³)*(1/1000 kg/g)*(10² cm/m)³
   = 2.7 x 10³ kg/m³
M = (5.31 lb)*(0.543592 kg/lb)
    = 2.4086 kg
t = (0.013 mm)*(10⁻³ m/mm)
  = 1.3 x 10⁻⁴ m

Let the surface area f the foil be A m².
Then the mass of the foil is 
M = d*A*t
    = (2.7 x 10³ kg/m³)*(A m²)*(1.3 x 10⁻⁴ m) 
    = 0.351A kg

Because the mass is 2.4086 kg, therefore
A = 2.4086/0.351
   = 6.862 m²

Answer: The surface area is 6.862 m².

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If 2.40 g of KNO3 reacts with sufficient sulfur (S8) and carbon (C), how much P-V work will the gases do against an external pre
creativ13 [48]

Answer:

-112.876J

Explanation:

In order to solve this question, we would need to incorporate Stoichiometry, which involves using relationships between reactants and/or products in a chemical reaction to determine desired quantitative data.

Here's a balanced equation for the reaction:

16KNO_3(s) + 24C(s) + S_8(s)    \to 24CO_2(g) + 8N_2(g) + 8K_2S(s)

Let us define P - V work as;

w_{pv} = - P_{external}  \triangle Volume

where  \triangle (Volume) = (V_{final} - V_{initial})

External pressure is given as  1.00atm , therefore the work solely depends on the change in volume and since the reactants are solids, none of the reactants contribute to the volume. Hence,  V_i = 0.

To find the volume of the products, we need to first find the amount of moles of the product made from  2.40_gKNO_3, using the molar mass of  KNO_3  which is 101.1032 g/mol  

2.40_gKNO_3 . {\frac{1molKNO_3}{101.1032_g}} = 0.0237molKNO_3

Now let us convert moles of  KNO_3  into moles of CO_2 and N_2  using the stoichiometric ratios from our balanced equation of the reaction.

0.0237molKNO_3 . {\frac{24molCO_2}{16molKNO_3}} = 0.0356molCO_2

0.0237molKNO_3 . {\frac{8molN_2}{16molKNO_3}} = 0.01185molN_2

K_2S is not factored into the volume calculation because it is a solid.

Now let us also convert the moles of  CO_2  and  N_2 into grams using their respective molar masses.

0.0356molCO_2 . {\frac{44.01_g}{1molCO_2}} = 1.567_gCO_2

0.01185molN_2 . {\frac{28.014_g}{1molN_2}} = 0.332_gN_2

We will now proceed to convert grams into volume using the density values provided.

1.567_gCO_2 . {\frac{1L}{1.830_g}} = 0.856LCO_2

0.332_gN_2 . {\frac{1L}{1.165_g}} = 0.285LN_2

Summing up the two volumes, we get the final volume

0.856L + 0.258L = 1.114L = V_f

Plugging everything into the w_{pv} equation, we get:

w_{pv} = -1atm(1.114L - 0L) = -1.114L.atm

Finally, let us convert L.atm into joules using the conversion rate of;

1L.atm = 101.325J\\-1.114L.atm. {\frac{101.325J}{1L.atm}} = -112.876J

7 0
3 years ago
What assumption can be made about the densities of two different spheres that have
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Answer:

Their densities are different as well

Explanation:

Density is worked out by dividing mass by volume. If the mass was the same then the densities would be as well. But it's not.

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Answer:

This conveyor belt should be connected to fixed pulleys in case their total effort is less than the load. But if it's not that, the belt should be connected to movable pulleys.

Explanation:

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Answer:60000 joules

Explanation:

power=500watts

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