Answer: 3-nitropyridine
Explanation:
You have a sealed 2 liter flask that contains nothing but water and carbon dioxide. The flask is half-filled with liquid water, has a temperature of 25°c, and the overall pressure within the flask is 0. 1 atm. How many moles of co2 are in the flask? at this temperature, you may take the kh value for co2 as 0. 033 m / atm.
In this exercise we want to calculate the amount of moles, so this is going to be:
[tex](4.6)(10^{-3}) \ mols \ CO_2[/tex]
Knowing that Henry's law is given by:
[tex]C = KHP[/tex]
Where constants are given by:
C = Concentration KH = Henry's law constant = [tex]0.033 m/atm[/tex]P = partial pressure = [tex]0.07 atm[/tex]Before we can find the concentration of CO2 (and hence the moles of CO2), we first need to find its partial pressure. We look up the vapor pressure of water at 25º and find it to be 0.03 atm. Since the total pressure is equal to 0.1 atm, this mean the partial pressure of:
[tex]CO_2 = 0.1 \ atm - 0.03 \ atm = 0.07 \ atm[/tex]
Now using Henry's law, we find the concentration:
[tex]C = (0.033)*( 0.07) = (2.31)*(10^{-3})[/tex]
Converting to moles of CO2, we have:
[tex](2.31)*(10^{-3})*( 2) = (4.6)*(10^{-3})[/tex]
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Solid calcium hydroxide is dissolved in water until the pH of the solution is 11.44. The hydroxide ion concentration [OH–] of the solution is:______.A. 1.1 * 10-11 M.B. 3.06 M.C. 8.7 * 10-4 M.D. 1.0 * 10-14 M.E. None of these.
Answer:
c the answer is c that is the answer
I NEED THE ANSWER ASAP! Its science btw
Which statement describes conditions in which a mineral can form?
A. Crystals of minerals dissolve in the groundwater in caves.
B. Materials dissolved in seawater crystallize on an ocean bottom.
C. Molten materials are cooled in a netalworks factory
D. Materials are mined from deposits deep underground.
Answer:
b
Explanation:
22) Calculate the energy of a photon of radiation with a frequency of 9.50 x 1013 Hz.
The energy of the photon with a frequency of 9.50×10¹³ Hz is 6.29×10¯²⁰ J
From the question given above, the following data were obtained:
Frequency (f) = 9.50×10¹³ Hz
Planck's constant (h) = 6.626×10¯³⁴ Js
Energy (E) =?The energy of the photon can be obtained as follow:
E = hf
E = 6.626×10¯³⁴ × 9.50×10¹³
E = 6.29×10¯²⁰ JTherefore, the energy of the photon is 6.29×10¯²⁰ J
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