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ivolga24 [154]
2 years ago
11

5. To keep your lunch cold at the beach, would you want a material

Chemistry
1 answer:
Leya [2.2K]2 years ago
4 0

Answer:

Minimizes heat transfer

Explanation:

To keep my lunch from getting cold at the beach, a material the minimizes heat transfer will be the most desirable.

  • Such material should be modified to prevent had transfer by convection, conduction and radiation.
  • This kind of material is desirable because heat generally flows from a place of high amount to that of low amount.
  • Since the food will mostly have a temperature higher than the ambient one, it is expected that heat will flow away from the food to the environment.
  • This will make the food cold.
  • If such heat transfer is prevented, then, the food will remain hot.
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Calculate the molarity of 1.0 mol of KCl in 750 ml of solution
Vaselesa [24]

Answer:

[KCl] = 1.33 M

Explanation:

Molarity is mol /L

Mol of solute in 1 L of solution

Volume of solution is 750 mL

750 mL / 1000 = 0.750 L

1 mol / 0.750L = M → 1.33

7 0
2 years ago
Read 2 more answers
You are presented with a mystery as part of your practical experiment. You have a solution of Pb(NO3)2 that has a worn label mak
Orlov [11]

Answer:

Minimum volume of H₂SO₄ required for H₂SO₄ to be in excess = 0.0556 mL

Explanation:

Pb(NO₃)₂ + H₂SO₄ -----> PbSO₄ + 2HNO₃

For this reaction, we know that the max concentration of Pb(NO₃) according to the bottle is 0.999M and to ensure the other reactant in the reaction is in excess, we'll do the calculation with a Pb(NO₃) that's a bit higher, that is, 1.0M.

Knowing that Concentration in mol/L = (number of moles)/(volume in L)

Number of moles of Pb(NO₃) added = concentration in mol/L × volume in L = 1 × 0.001 = 0.001 mole

According to the reaction,

1 mole of Pb(NO₃) reacts with 1 mole of H₂SO₄

0.001 mole of Pb(NO₃) will react with 0.001×1/1 mole of H₂SO₄

Therefore number of H₂SO₄ required for the reaction and for the H₂SO₄ to be in excess is 0.001 mole of H₂SO₄

So, the concentration of commercial H₂SO₄ is usually 18.0M, using this as the assumed value.

Volume of H₂SO₄ = (number of H₂SO₄ required for it to be in excess)/(concentration of H₂SO₄)

Volume of H₂SO₄ = 0.001/18 = 0.0000556 L = 0.0556 mL.

QED!!!

5 0
2 years ago
The standard free energy ( Δ G ∘ ′ ) (ΔG∘′) of the creatine kinase reaction is − 12.6 kJ ⋅ mol − 1 . −12.6 kJ⋅mol−1. The Δ G ΔG
Dmitrij [34]

Answer:

The concentration of [ADP] = 21.896*10^-6 μM

Explanation:

Given Data:

creatine + ATP -----------> ADP + creatine phosphate    

ΔG∘   = -12.6 KJ/mole  = -12600 J/mole

ΔG = -0.1 KJ/mole  =  -100 J/mole

[Creatine phosphate]  = 25 mM = 25*10^-3 M

[Creatine] = 17 mM    = 17*10^-3 M

[ATP]   =5mM = 5*10^-3M

Calculating the concentration of [ADP] using the formula;

ΔG = ΔG∘ + RTlnQc

Substituting, we have

-12600   = -100 + 8.314*298lnQc

-12600+100 = 8.314*298lnQc

-12500   = 2477.57lnQc

lnQc = -12500/2477.57

lnQc = -5.045

Qc = e^ -5.045

Qc   = 6.44*10^-3

But,

Qc    = [Creatine phosphate]*[ADP]/[creatine]*[ATP]

6.44*10^-3   = 25*10^-3*[ADP]/ (17*10^-3* 5*10^-3)

6.44*10^-3 = 25*10^-3[ADP]/8.5*10^-5

6.44*10^-3 * 8.5*10^-5 = 25*10^-3[ADP]

5.474*10^-7 = 25*10^-3[ADP]

[ADP] = 5.474*10^-7 /25*10^-3

          = 2.1896 *10^-5 M

          = 21.896*10^-6 μM

Therefore, the concentration of [ADP] = 21.896*10^-6 μM

3 0
2 years ago
Isopentyl acetate (C7H14O2), the compound responsible for the scent of bananas, can be produced commercially. Interestingly, bee
lora16 [44]

Answer:

The answer to your question is:

a) 4.64 x 10 ¹⁵ molecules

b) 9.28 x 10 ¹⁵ atoms of O2

Explanation:

MW C7H14O2 = 84 + 14 + 32 = 130 g

a)        130 g of C7H14O2 ---------------- 1 mol of C7H14O2

           1 x 10 ⁻⁶ g              ---------------      x

           x = 7.7 x 10 ⁻⁹ mol

          1 mol of C7H14O2   --------------   6 .023 x 10 ²³ molecules

          7.7 x 10⁻⁹ mol          --------------    x

          x = 4.64 x 10¹⁵ molecules

b)      130 g of C7H14O2   ----------------   1 mol of C7H14O2

         1 x 10⁻⁶  C7H14O2   -----------------     x

         x = 7.7 x 10 ⁻⁹ mol of C7H14O2

        1 mol of C7H14O2    ---------------   2 mol of O2

        7.7 x 10 ⁻⁹                 ----------------   x

         x = 1.54 x 10⁻⁸ mol of O2

       1 mol of O2 -----------------  6.023 x 10 ²³ atoms

       1.54 x 10 ⁻⁸  ----------------   x

        x = 9.28 x 10 ¹⁵ atoms of O2

8 0
2 years ago
Paige heated 3.00 g mercury (II) oxide (HgO, 216.59 g/mol) to form mercury (Hg, 200.59 g/mol) and oxygen (O2, 32.00 g/mol). She
stepan [7]

the balanced chemical equation for decomposition of HgO is as follows

2HgO --> 2Hg + O₂

stoichiometry of HgO to O₂ is 2:1

number of HgO moles heated are - 3.00 g / 216.59 g/mol = 0.0139 mol

according to stoichiometry of reaction -

number of O₂ moles formed = 0.0139 mol/ 2 = 0.00695 mol

mass of O₂ to be formed - 0.00695 mol x 32.00 g/mol = 0.2224 g

but the actual yield = 0.195 g

percent yield = actual yield / theoretical yield x 100 %

percent yield = 0.195 g / 0.2224 g x 100 % = 87.7 %

answer is 87.7 %

8 1
2 years ago
Read 2 more answers
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