What words do you use in a hypothesis

Solutions of hydrogen in palladium may be formed by exposing Pd metal to H gas. The concentration of hydrogen in the palladium d

Juliette [100K]

Explanation:

Mass of solution , m= 0.94 g + 215 g = 215.94 g

Volume of the solution = V

Density of solution = d = $10.8 g/cm^3$

$V=\frac{m}{d}=\frac{215.94 g}{10.8 g/cm^3}=19.99 cm^3=19.99 mL$

$1 cm^3=1 mL$

$1 mL=0.001 L$

$V=0.01999 L$

Moles of hydrogen = $\frac{0.94 g}{2 g/mol}=0.47 mol$

Molarity of hydrogen gas:

$Molarity=\frac{Moles}{\text{Volume of solution (L)}}$

$=\frac{0.47 mol}{0.01999 L}=0.02351 mol/L$

Molality of hydrogen gas :

Mass of solvent or here it is palladium = 215 g =0.215 kg (1 g = 0.001 kg)

$Molality=\frac{Moles}{\text{Mass of solvent(kg)}}$

$=\frac{0.47}{0.215 kg}=2.1860 mol/kg$

Mass percentage of hydrogen is solution :

$\frac{\text{Mass of hydrogen}}{\text{Mass of solution}}\times 100$

$=\frac{0.94 g}{215.94 g}\times 100=0.435\%$

5 0

4 years ago

17.

yanalaym [24]

Answer:

ammonia gas and hydrogen chloride gas.

3 0

3 years ago

lord [1]

Answer:

$\boxed{6.08 \times 10^{23}}$

Explanation:

Data:

I = 2.15 A

t = 8 min 24 s

T = 26.0 °C

V = 65.4 mL

p = 774.2 To

1. Write the equation for the half-reaction

2H₂O ⟶ O₂ + 4H⁺ + 4e⁻

2. Calculate the moles of oxygen

$p = \text{774.2 To} \times \dfrac{\text{1 atm}}{\text{760 To}} = \text{1.0187 atm}$

V = 0.0654 L

T = (26.0 + 273.15) K = 299.15 K

$\begin{array}{rcl}pV& = & nRT\\1.0189 \times 0.0654 & = & n \times 0.08206 \times 299.15\\0.06662 & = & 24.55n\\\\n & = & \dfrac{0.06662}{24.55}\\\\n & = & 2.714 \times 10^{-3}\\\end{array}$

3. Calculate the moles of electrons

$\text{n} = \text{2.714 $\times 10^{-3}$ mol oxygen} \times \dfrac{\text{4 mol electrons}}{\text{ 1 mol ozygen}} = \text{ 0.01086 mol electrons}$

4. Calculate the number of coulombs

t = 8 min 24 s =504 s

Q = It = 504 s × 2.10 C·s⁻¹= 1058 C

5. Calculate the number of electrons

$\text{No. of electrons} = \text{1058 C} \times \dfrac{\text{1 electron}}{1.602 \times 10^{-19}\text{ C}} = 6.607 \times 10^{21}\text{ electrons}$

6. Calculate Avogadro's number

$N_{\text{A}} = \dfrac{6.607 \times 10^{21}}{0.01086} = \mathbf{6.08 \times 10^{23}}\\\\\text{The experimental value of Avogadro's number is } \boxed{\mathbf{6.08 \times 10^{23}}}$

3 0

4 years ago

2. Which of the following is not a physical property of water? It is a colorless liquid. It is composed of hydrogen and oxygen.

S_A_V [24]

The correct answer is C. Sugar dissolves in it

Hope this helps

-AaronWiseIsBae

7 0

3 years ago

in a torricell barometer a pressure of one atmospheric pressure supports q 760 mm coloumn of mercury.if the original tube contai

dybincka [34]

Answer:

The answer to the question is

The height of the mercury fluid column remain the same.

Explanation:

The pressure, P in a column of fluid of height, h is given by

P = (Density p)×(height of fluid column h)×(gravity g)

Therefore, when the diameter is doubled we have

Density of the mercury in the tube with twice the diameter = (Mass of mercury)/(volume of mercury) where the volume of mercury = h×pi×(Diameter×2)^2/4 = h×pi×Diameter^2. Therefore the volume increases by a factor of 4 and therefore the mass increases by a factor of 4 which means that the density remains the same hence

P = p×h1×g = p×h2×g Therefore h1 = h2

The height of the fluid column remain the same

5 0

3 years ago