Answer:
The value of the partial pressure of the oxygen 
= 690 torr
Explanation:
Total pressure of the mixture of gases = 736 torr
The partial pressure of water vapor = 46 torr
From the law of pressure we know that
Total pressure = The partial pressure of water vapor + The partial pressure of oxygen 
Put the values of pressures in above equation we get,
⇒ 736 = 46 +
⇒ = 736 - 46
⇒ = 690 torr
This is the value of the partial pressure of the oxygen.
Answer:
Here's what I get
Explanation:
Assume the initial concentrations of H₂ and I₂ are 0.030 and 0.015 mol·L⁻¹, respectively.
We must calculate the initial concentration of HI.
1. We will need a chemical equation with concentrations, so let's gather all the information in one place.
H₂ + I₂ ⇌ 2HI
I/mol·L⁻¹: 0.30 0.15 x
2. Calculate the concentration of HI
![Q_{\text{c}} = \dfrac{\text{[HI]}^{2}} {\text{[H$_{2}$][I$_{2}$]}} =\dfrac{x^{2}}{0.30 \times 0.15} = 5.56\\\\x^{2} = 0.30 \times 0.15 \times 5.56 = 0.250\\x = \sqrt{0.250} = \textbf{0.50 mol/L}\\\text{The initial concentration of HI is $\large \boxed{\textbf{0.50 mol/L}}$}](https://tex.z-dn.net/?f=Q_%7B%5Ctext%7Bc%7D%7D%20%3D%20%5Cdfrac%7B%5Ctext%7B%5BHI%5D%7D%5E%7B2%7D%7D%20%7B%5Ctext%7B%5BH%24_%7B2%7D%24%5D%5BI%24_%7B2%7D%24%5D%7D%7D%20%3D%5Cdfrac%7Bx%5E%7B2%7D%7D%7B0.30%20%5Ctimes%200.15%7D%20%3D%20%205.56%5C%5C%5C%5Cx%5E%7B2%7D%20%3D%200.30%20%5Ctimes%200.15%20%5Ctimes%205.56%20%3D%200.250%5C%5Cx%20%3D%20%5Csqrt%7B0.250%7D%20%3D%20%5Ctextbf%7B0.50%20mol%2FL%7D%5C%5C%5Ctext%7BThe%20initial%20concentration%20of%20HI%20is%20%24%5Clarge%20%5Cboxed%7B%5Ctextbf%7B0.50%20mol%2FL%7D%7D%24%7D)
3. Plot the initial points
The graph below shows the initial concentrations plotted on the vertical axis.
Answer:they will out live us himans
Explanation:
<span>Stating that collisions of gas particles are perfectly elastic means that no kinetic energy is lost when the gases collide, and no kinetic energy is gained. Some of the properties of gases include volume, pressure, thermal conductivity, pressure, and viscosity.</span>
Answer:
18.84 g of silver.
Explanation:
We'll begin by calculating the number atoms present in 5.59 g of sulphur. This can be obtained as follow:
From Avogadro's hypothesis,
1 mole of sulphur contains 6.02×10²³ atoms.
1 mole of sulphur = 32 g
Thus,
32 g of sulphur contains 6.02×10²³ atoms.
Therefore, 5.59 g of sulphur will contain = (5.59 × 6.02×10²³) / 32 = 1.05×10²³ atoms.
From the calculations made above, 5.59 g of sulphur contains 1.05×10²³ atoms.
Finally, we shall determine the mass of silver that contains 1.05×10²³ atoms.
This is illustrated below:
1 mole of silver = 6.02×10²³ atoms.
1 mole of silver = 108 g
108 g of silver contains 6.02×10²³ atoms.
Therefore, Xg of silver will contain 1.05×10²³ atoms i.e
Xg of silver = (108 × 1.05×10²³)/6.02×10²³
Xg of silver = 18.84 g
Thus, 18.84 g of silver contains the same number of atoms (i.e 1.05×10²³ atoms) as 5.59 g of sulfur
