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hodyreva [135]
3 years ago
12

If the length, width, and height of a box are 10.00 cm, 7.25 cm and 3.00 cm, respectively, what is the volume of the box in unit

s of milliliters and liters? (a) How many mL will the box contain? (b) How many L will the box contain?
Chemistry
1 answer:
Blizzard [7]3 years ago
3 0

<u>Answer:</u>

<u>For a:</u> The volume of the box is 217.5 mL

<u>For b:</u> The volume of the box is 0.2175 L

<u>Explanation:</u>

The box is a type of cuboid.

To calculate the volume of cuboid, we use the equation:

V=lbh

where,

V = volume of cuboid

l = length of cuboid = 10.00 cm

b = breadth of cuboid = 7.25 cm

h = height of cuboid = 3.00 cm

Putting values in above equation, we get:

V=10.00\times 7.25\times 3.00=217.5cm^3

  • <u>For a:</u>

To convert the volume of cuboid into milliliters, we use the conversion factor:

1mL=1cm^3

So,

\Rightarrow 217.5cm^3\times (\frac{1mL}{1cm^3})\\\\\Rightarrow 217.5mL

Hence, the volume of the box is 217.5 mL

  • <u>For b:</u>

To convert the volume of cuboid into liters, we use the conversion factor:

1L=1000cm^3

So,

\Rightarrow 217.5cm^3\times (\frac{1L}{1000cm^3})\\\\\Rightarrow 0.2175L

Hence, the volume of the box is 0.2175 L

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Which tool was most likely used in a procedure if the lab report shows that approximately 300 mL of water was used?
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What is the mass of 3.0 x x times 10^23 atoms of neon?
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Answer:

10.09 grams

Explanation:

First you need to know the number of moles you are dealing with.

If you know that each mole has 6.022x10²³ of something (in this case of atoms), you can divide 3x10²³ atoms of neons by 6.022x10²³ to obtain the number of moles.

You have 0.5 moles of Neon, so then by the periodic table, you see that the molar mass of neon is 20.18g/mol, so by each mole you have 20.18 grams of neon. Multiply 20.18 grams by 0.5 moles and you got 10.09 grams of Neon

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Why do solids maintain a specific shape and volume?
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Answer:

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<em>Hope it helps...</em>

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3 years ago
The reaction described by H2(g)+I2(g)⟶2HI(g) has an experimentally determined rate law of rate=k[H2][I2] Some proposed mechanism
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Answer:

Mechanism A and B are consistent with observed rate law

Mechanism A is consistent with the observation of J. H. Sullivan

Explanation:

In a mechanism of a reaction, the rate is determinated by the slow step of the mechanism.

In the proposed mechanisms:

Mechanism A

(1) H2(g)+I2(g)→2HI(g)(one-step reaction)

Mechanism B

(1) I2(g)⇄2I(g)(fast, equilibrium)

(2) H2(g)+2I(g)→2HI(g) (slow)

Mechanism C

(1) I2(g) ⇄ 2I(g)(fast, equilibrium)

(2) I(g)+H2(g) ⇄ HI(g)+H(g) (slow)

(3) H(g)+I(g)→HI(g) (fast)

The rate laws are:

A: rate = k₁ [H2] [I2]

B: rate = k₂ [H2] [I]²

As:

K-1 [I]² = K1 [I2]:

rate = k' [H2] [I2]

<em>Where K' = K1 * K2</em>

C: rate = k₁ [H2] [I]

As:

K-1 [I]² = K1 [I2]:

rate = k' [H2] [I2]^1/2

Thus, just <em>mechanism A and B are consistent with observed rate law</em>

In the equilibrium of B, you can see the I-I bond is broken in a fast equilibrium (That means the rupture of the bond is not a determinating step in the reaction), but in mechanism A, the fast rupture of I-I bond could increase in a big way the rate of the reaction. Thus, just <em>mechanism A is consistent with the observation of J. H. Sullivan</em>

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