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IRINA_888 [86]
3 years ago
6

Kate gathered three boxes of the same size made of different materials: glass, clear plastic, and aluminum painted black. She pl

aced them on a window sill in the sun for an hour and then measured the warmth of the air in each box. Which of the following best describes the purpose of the experiment?
A. To relate the size of the box of the warmth of air within the box

B. To relate the type of box material to the warmth of air within the box 

C. To relate the amount of a time a box is exposed to sunlight to the warmth of air within the box 

D. To relate the type of box material to the mass of the box
Chemistry
1 answer:
zalisa [80]3 years ago
6 0
<h2>Answer:</h2>

Correct option is B.

B. To relate the type of box material to the warmth of air within the box.

<h2>Explanation:</h2>

Kate gathered three boxes of the same size made of different materials: glass, clear plastic, and aluminum painted black. She placed them on a window sill in the sun for an hour and then measured the warmth of the air in each box. She actually did this to relate the type of box material to the warmth of air within the box.

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A mixture of 0.682 mol of H2 and 0.440 mol of Br2 is combined in a reaction vessel with a volume of 2.00 L. At equilibrium at 70
ahrayia [7]

Answer: 0.294 mol of Br_2 present in the reaction vessel.

Explanation:

Initial moles of  H_2 = 0.682 mole

Initial moles of  Br_2 = 0.440 mole

Volume of container = 2.00 L

Initial concentration of H_2=\frac{moles}{volume}=\frac{0.682moles}{2.00L}=0.341M

Initial concentration of Br_2=\frac{moles}{volume}=\frac{0.440moles}{2.00L}=0.220M

equilibrium concentration of H_2=\frac{moles}{volume}=\frac{0.536mole}{2.00L}=0.268M

The given balanced equilibrium reaction is,

                            H_2(g)+Br_2(g)\rightleftharpoons 2HBr(g)

Initial conc.              0.341 M     0.220 M         0  M

At eqm. conc.    (0.341-x) M      (0.220-x) M     (2x) M

The expression for equilibrium constant for this reaction will be,

K_c=\frac{[HBr]^2}{[Br_2]\times [H_2]}

we are given : (0.341-x) = 0.268 M

x= 0.073 M

Thus equilibrium concentration of Br_2 = (0.220-x) M = (0.220-0.073) M = 0.147 M

[Br_2]=\frac{moles}{volume}\\0.147=\frac{xmole}{2.00L}\\\\x=0.294 mole

Thus there are 0.294  mol of Br_2 present in the reaction vessel.

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The answer is 0.5 moles of gold

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3 years ago
How many moles are in 1.52908x1024 molecules of Potassium phosphate (K3(PO4))?
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Answer:

<h2>2.54 moles</h2>

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

n =  \frac{N}{L} \\

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

n =  \frac{1.52908 \times  {10}^{24} }{6.02 \times  {10}^{23} }  \\

We have the final answer as

<h3>2.54 moles</h3>

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