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tensa zangetsu [6.8K]
3 years ago
9

A boy found a solid metal box in his backyard. The box had been buried for so long, it was difficult to determine from what the

box was made. The boy measured the box and found its volume (v) to be 17.63 cubic centimeters (cm3) and its mass (m) to be 158 grams (g). The boy knows the formula for density (D) is D = m/v. This table shows the density of several metals. DENSITY OF METALS *
A. copper
B. gold
C. silver
D. tin​

Chemistry
1 answer:
Alex777 [14]3 years ago
4 0

Answer:

Box is made up of <em>copper</em>, because density is <em>8.96  g/cm³.</em>

Explanation:

Given data:

Volume of box = 17.63 cm³

Mass of box = 158 g

Which metal box is this = ?

Solution:

First we will calculate the density of box then we will compare it with the density value of given metals.

d = m/v

d = 158 g/ 17.63 cm³

d = 8.96  g/cm³

The calculated density is similar to the given density value of copper thus box is made up of copper.

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What is the new volume of a 61 L sample at STP that is moved to 183 K and 0.60 atm?
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Answer: The new volume of a 61 L sample at STP that is moved to 183 K and 0.60 atm is 54.63 L.

Explanation:

Given: V_{1} = 61 L,      T_{1} = 183 K,      P_{1} = 0.60 atm

At STP, the value of pressure is 1 atm and temperature is 273.15 K.

Now, formula used to calculate the new volume is as follows.

\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}

Substitute the values into above formula as follows.

\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\\\frac{0.60 atm \times 61 L}{183 K} = \frac{1 atm \times V_{2}}{273.15 K}\\V_{2} = 54.63 L

Thus, we can conclude that the new volume of a 61 L sample at STP that is moved to 183 K and 0.60 atm is 54.63 L.

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3 years ago
Use the periodic table to identify the element with the electron configuration 1s²2s²2p⁴. Write its orbital diagram, and give th
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Answer:

1. Orbital diagram

2p⁴   ║ ↑↓ ║  "↑"  ║   ↑

2s²    ║ ↑↓ ║

1s²     ║ ↑↓ ║

2. Quantum numbers

  • <em>n </em>= 2,
  • <em>l</em> = 1,
  • m_{l} = 0,
  • m_{s} = +1/2

Explanation:

The fill in rule is:

  • Follow shell number: from the inner most shell to the outer most shell, our case from shell 1 to 2
  • Follow the The Aufbau principle, 1s<2s<2p<3s<3p<4s<3d<4p<5s<4d<5p<6s<4f<5d<6p<7s<5f<6d<7p
  • Hunds' rule: Every orbital in a sublevel is singly occupied before any orbital is doubly occupied. All of the electrons in singly occupied orbitals have the same spin (to maximize total spin).

So, the orbital diagram of given element is as below and the sixth electron is marked between " "

2p⁴   ║ ↑↓ ║  "↑"  ║   ↑

2s²    ║ ↑↓ ║

1s²     ║ ↑↓ ║

The quantum number of an electron consists of four number:

  • <em>n </em>(shell number, - 1, 2, 3...)
  • <em>l</em> (subshell number or  orbital number, 0 - orbital <em>s</em>, 1 - orbital <em>p</em>, 2 - orbital <em>d...</em>)
  • m_{l} (orbital energy, or "which box the electron is in"). For example, orbital <em>p </em>(<em>l</em> = 1) has 3 "boxes", it was number from -1, 0, 1. Orbital <em>d</em> (<em>l </em>= 2) has 5 "boxes", numbered -2, -1, 0, 1, 2
  • m_{s} (spin of electron), either -1/2 or +1/2

In our case, the electron marked with " " has quantum number

  • <em>n </em>= 2, shell number 2,
  • <em>l</em> = 1, subshell or orbital <em>p,</em>
  • m_{l} = 0, 2nd "box" in the range -1, 0, 1
  • m_{s} = +1/2, single electron always has +1/2
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