<span>Let's </span>assume that water vapor has ideal gas
behavior. <span>
Then we can use ideal gas formula,
PV = nRT<span>
</span><span>Where, P is the pressure of the gas (Pa), V
is the volume of the gas (m³), n is the number
of moles of gas (mol), R is the universal gas constant ( 8.314 J mol</span></span>⁻¹ K⁻¹) and T is temperature in Kelvin.<span>
<span>
</span>P = 1 atm = 101325 Pa (standard pressure)
V = 13.97 L = 13.97 x 10</span>⁻³ m³<span>
n = ?
R = 8.314 J mol</span>⁻¹ K⁻¹<span>
T = 0 °C = 273 K (standard temperature)
<span>
By substitution,
</span>101325 Pa x 13.97x 10</span>⁻³
m³ = n x 8.314 J mol⁻¹ K⁻¹ x 273 K<span>
n = 0.624 mol
<span>
Hence, the moles of water vapor at STP is 0.624 mol.
According to the </span></span>Avogadro's constant, 1 mole of substance has 6.022 × 10²³ particles.
<span>
Hence, number of atoms in water vapor = 0.624 mol x </span>6.022 × 10²³ mol⁻¹
<span> = 3.758 x 10</span>²³<span>
</span>
Answer:
K3PO4
Explanation:
Recall that colligative properties depends on the number of particles present. The greater the number of particles present, the greater the degree of colligative properties of the solution. Let us look at each option individually;
SrCr2O7-------> Sr^2+ + Cr2O7^2- ( 2 particles)
C4H11N (not ionic in nature hence it can not dissociate into ions)
K3PO4-------> 3K^+ + PO4^3- (4 particles)
Rb2CO3-------> 2Rb^+ + CO3^2- (3 particles)
Hence K3PO4 has the greatest number of particles and will display the greatest colligative effect.
The answer is <span>(3) 3 × 12.4 hours
</span>
To calculate this, we will use two equations:


where:
<span>n - number of half-lives
</span>x - remained amount of the sample, in decimals
<span>

- half-life length
</span>t - total time elapsed.
First, we have to calculate x and n. x is <span>remained amount of the sample, so if at the beginning were 16 grams of potassium-42, and now it remained 2 grams, then x is:
2 grams : x % = 16 grams : 100 %
x = 2 grams </span>× 100 percent ÷ 16 grams
x = 12.5% = 0.125
Thus:
<span>

</span>




It is known that the half-life of potassium-42 is 12.36 ≈ 12.4 hours.
Thus:
<span>

</span><span>

</span>

Therefore, it must elapse 3 × 12.4 hours <span>before 16 grams of potassium-42 decays, leaving 2 grams of the original isotope</span>
<span>63.4 g/mol
First, let's determine how many atoms per unit cell in face-centered cubic.
There is 8 corners, each of which has 1 atom, and each of those atoms is shared between 8 other unit cells. So 8*1/8 = 1 atom per unit cell. Additionally, there are 6 faces, each of which has 1 atom that's shared between 2 unit cells. So 6*1/2 = 3 atoms per unit cell. So each unit cell has the mass of 1+3 = 4 atoms.
Since there is 1000 liters per cubic meter, the mass per liter is 8920 kg/1000 = 8.920 kg/L. Now the mass per unit cell is 8920 g * 4.72x10^-26 = 4.21024x10^-22 g per unit cell. The mass per atom is 4.21024x10^-22 g / 4 = 1.05256x10^-22 g/atom, Finally, multiply by Avogadro's number, getting 1.05256x10^-22 g/atom * 6.0221409x10^23 atom/mol = 63.38664625704 g/mol.
Rounding to 3 significant digits gives 63.4 g/mol.</span>