Answer:
28,125 g
Explanation:
m(O₂) = 100 g
n(O₂) =M(O₂)/m(O₂)=32 gram per mole / 100 g = 0,32 mole
O₂ + 2H₂ ⇒ 2H₂O , n(H₂O) = 2 × n(O₂)
m(H₂O) = M(H₂O)/2 × n(O₂) = 18 grams per mole / 2 × 0,32 mole = 28,125 g
Answer:
6.52×10⁴ GHz
Explanation:
From the question given above, the following data were obtained:
Wavelength (λ) = 4.6 μm
Velocity of light (v) = 2.998×10⁸ m/s
Frequency (f) =?
Next we shall convert 4.6 μm to metre (m). This can be obtained as follow:
1 μm = 1×10¯⁶ m
Therefore,
4.6 μm = 4.6 μm × 1×10¯⁶ m / 1 μm
4.6 μm = 4.6×10¯⁶ m
Next, we shall determine frequency of the light. This can be obtained as follow:
Wavelength (λ) = 4.6×10¯⁶ m
Velocity of light (v) = 2.998×10⁸ m/s
Frequency (f) =?
v = λf
2.998×10⁸ = 4.6×10¯⁶ × f
Divide both side by 4.6×10¯⁶
f = 2.998×10⁸ / 4.6×10¯⁶
f = 6.52×10¹³ Hz
Finally, we shall convert 6.52×10¹³ Hz to gigahertz. This can be obtained as follow:
1 Hz = 1×10¯⁹ GHz
Therefore,
6.52×10¹³ Hz = 6.52×10¹³ Hz × 1×10¯⁹ GHz / 1Hz
6.52×10¹³ Hz = 6.52×10⁴ GHz
Thus, the frequency of the light is 6.52×10⁴ GHz
Answer:
All of the above.
Explanation:
In positive deviation from Raoult's Law occur when the vapour pressure of components is greater than what is expected value in Raoult's law.
When a solution is non ideal then it shows positive or negative deviation.
Let two solutions A and B to form non- ideal solutions.let the vapour pressure of component A is 
and vapour pressure of component B is 
.
= Vapour pressure of component A in pure form
= Vapour pressure of component B in pure form
=Mole fraction of component A
=Mole fraction of component B
The interaction between A- B is less than the interaction A- A and B-B interaction.Therefore, the escaping tendency of liquid molecules in mixture is greater than the escaping tendency in pure form.Hence, the vapour pressure of a mixture is greater than the initial value of vapour pressure.
,
Therefore, 
Therefore, the enthalpy of mixing is greater than zero and change in volume is greater than zero.
Hence, option a,b,c and d are true.
The easiest way to approximate ph level is through the ph paper measurement system because using pH paper would be best for doing a quick measurement in the field. It does not essentially need to be calibrated or standardized and the papers make available an instant estimate.
