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1 year ago

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The electron configuration of an element is 1s22s22p4. Describe what most likely happens when two atoms of this element move tow

ard each other.

1 answer:

ahrayia [7]1 year ago

5 0

When two atoms of this element move towards each other, they will combine in a covalent bond to form a diatomic molecule.

Looking at the electron configuration of the atoms; 1s2 2s2 2p4, we can see that these are atoms of elements in group 16. The elements in group 16 has a general outer electronic configuration of ns2np4. They have a valency of 2.

When two atoms of this element approach each other, they will combine in a covalent bond to form a compound. If this element is depicted as X, the compound formed is X2.

Learn more: brainly.com/question/1527403

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How many moles are equal to 89.23 g of calcium oxide?

adelina 88 [10]

The molecular formula of calcium oxide - CaO
The molar mass of CaO - 40 + 16 = 56 g/mol
Which means that 1 mol weighs 56 g
Therefore 56 g of CaO is - 1 mol
Then 89.23 g is equivalent to - 1/56 x 89.23 = 1.6 mol of CaO

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2 years ago

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In which 1.0 gram sample are particles arranged in a rystal structure?

lilavasa [31]

The Options are as follow,

<span> (1) CaCl</span>₂<span> (s) (3) CH</span>₃<span>OH (l)</span>

<span> (2) C</span>₂<span>H</span>₆<span> (g) (4) Cal</span>₂<span> (aq)</span>

Answer:

Option-1 is the correct answer.

Explanation:

As we know crystal formation is the property of solids. Therefore, in given options we are given with four different states of matter.

Option A, CaCl₂ is in a solid state , so it can exist in crystal form.

Option 2, C₂H₆ (Ethane) is in gas form, so it cannot form crystals.

Option 3, CH₃OH (Methanol) is present in liquid form, so it fails to form crystals.

Option 4, CaI₂, it is dissolved in water, Hence, it is in aqueous state, Therefore it also lacks crystal structure.

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1 year ago

A box has a volume of 45m3 and is filled with air held at 25∘C and 3.65atm. What will be the pressure (in atmospheres) if the sa

Marina CMI [18]

Answer:

Given:

Initial pressure: $3.65\; \rm atm$ .
Volume was reduced from $45\; \rm m^{3}$ to $5.0\; \rm m^{3}$ .
Temperature was raised from $25\; ^\circ \rm C$ to $35\; ^\circ \rm C$ .

New pressure: approximately $3.4\times 10\; \rm atm$ ( $34\; \rm atm$ .) (Assuming that the gas is an ideal gas.)

Explanation:

Both the volume and the temperature of this gas has changed. Consider the two changes in two separate steps:

Reduce the volume of the gas from $45\; \rm m^{3}$ to $5.0\; \rm m^{3}$ . Calculate the new pressure, $P_1$ .
Raise the temperature of the gas from $25\; ^\circ \rm C$ to $35\; ^\circ \rm C$ . Calculate the final pressure, $P_2$ .

By Boyle's Law, the pressure of an ideal gas is inversely proportional to the volume of this gas (assuming constant temperature and that no gas particles escaped or was added.)

For this gas, $V_0 = 45\; \rm m^{3}$ while $V_1 = 5.0\; \rm m^{3}$ .

Let $P_0$ denote the pressure of this gas before the volume change ( $P_0 = 3.65\; \rm atm$ .) Let $P_1$ denote the pressure of this gas after the volume change (but before changing the temperature.) Apply Boyle's Law to find the ratio between $P_1\!$ and $P_0\!$ :

$\displaystyle \frac{P_1}{P_0} = \frac{V_0}{V_1} = \frac{45\; \rm m^{3}}{5.0\; \rm m^{3}} = 9.0$ .

In other words, because the final volume is $(1/9)$ of the initial volume, the final pressure is $9$ times the initial pressure. Therefore:

$\displaystyle P_1 = 9.0\times P_0 = 32.85\; \rm atm$ .

On the other hand, by Amonton's Law, the pressure of an ideal gas is directly proportional to the temperature (in degrees Kelvins) of this gas (assuming constant volume and that no gas particle escaped or was added.)

Convert the unit of the temperature of this gas to degrees Kelvins:

$T_1 = (25 + 273.15)\; \rm K = 298.15\; \rm K$ .

$T_2 = (35 + 273.15)\; \rm K = 308.15\; \rm K$ .

Let $P_1$ denote the pressure of this gas before this temperature change ( $P_1 = 32.85\; \rm atm$ .) Let $P_2$ denote the pressure of this gas after the temperature change. The volume of this gas is kept constant at $V_2 = V_1 = 5.0\; \rm m^{3}$ .

Apply Amonton's Law to find the ratio between $P_2$ and $P_1$ :

$\displaystyle \frac{P_2}{P_1} = \frac{T_2}{T_1} = \frac{308.16\; \rm K}{298.15\; \rm K}$ .

Calculate $P_2$ , the final pressure of this gas:

$\begin{aligned} P_2 &= \frac{308.15\; \rm K}{298.15\; \rm K} \times P_1 \\ &= \frac{308.15\; \rm K}{298.15\; \rm K} \times 32.85\; \rm atm \approx 3.4 \times 10\; \rm atm\end{aligned}$ .

In other words, the pressure of this gas after the volume and the temperature changes would be approximately $3.4\times 10\; \rm atm$ .

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1 year ago

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