Answers
Answer:
Explanation:
Hello,
In this case, since the Broglie's wavelength for bodies is defined via:
Whereas h accounts for the Planck's constant, m the mass and v the velocity, which is:
Thus, the wavelength turns out:
Best regards.
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Answers
Secondary amines react with the nitrosonium ion to generate N-Nitrosoamines.
Explanation:
Nitosonium Ion is generally utilized in the formation of Diazonium Salts which are considered excellent starting Material from synthesis point of View. Diazonium salts are formed by reacting Primary Amine or Anilines with Nitrosonium Ions. In our case, the Amine given is Secondary. So, reaction of Sec. Amines with Nitrosonium Ions stops after the formation of N-Nitrosoamine as there is no Hydrogen attached to Nitrogen atom of Amine to be eliminated and form a double and eventually triple bond with the Nitrogen atom of Nitrosonium Ion.
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Answer:
Sublimation is basically cause by the heat absorption and this process is an endothermic since it require extra energy.
It basically provide sufficient energy to molecules to control the various type of attractive forces from the neighbors and then convert it into the vapor phase.
Sublimation occur when the particle of gases become cold because some substances has high vapor pressure. Sublimation is the endothermic change and it occur below the triple point in terms of pressure and temperature.
Answers
Answer:
[K2SO4] = 4,75x10⁻⁷M ; [K⁺] = 9.50x10⁻⁷M ; [SO4⁻²] = 4,75x10⁻⁷M
SO4⁻²: 0.045ppm ; K⁺: 0.037ppm
[SO4⁻²] = 4,70x10⁻⁷ F
Explanation:
Determine the equation
K2SO4 → 2K⁺ + SO4⁻²
Each mole of potassium sulfate generates two moles of potassium cation and one mole of sulfate anion
Molar mass K2SO4: 174.26 g/m
Moles of K2SO4: grams / molar mass
2.07x10⁻⁴g / 174.26 g/m = 1.18x10⁻⁶ moles
Molarity: Moles of solute in 1 L of solution
1.18x10⁻⁶ moles / 2.5 L = 4,75x10⁻⁷M (K2SO4)
K⁺ : 4,75x10⁻⁷M . 2 = 9.50x10⁻⁷M
SO4⁻²: 4,75x10⁻⁷ M
1 mol of K2SO4 has 2 moles of K and 1 mol of SO4
1.18x10⁻⁶ moles of K2SO4 has 1.18x10⁻⁶ moles of SO4 and 2.37x10⁻⁶ moles of K.
1.18x10⁻⁶ moles of SO4⁻² are 1.13x10⁻⁴ grams (moles. molar mass)
2.37x10⁻⁶ moles of K are 9.26x10⁻⁵ grams (moles. molar mass)
These grams are in 2.5 L of water, so we need μg/mL to get ppm
2.5 L = 2500 mL
1.13x10⁻⁴ grams SO4⁻² are 113.35 μg (1 μg = 1x10⁶ g)
9.26x10⁻⁵ grams K⁺ are 92.6 μg (1 μg = 1x10⁶ g)
113.35 μg /2500 mL = 0.045ppm
92.6 μg /2500 mL = 0.037ppm
Formal concentration of SO4⁻² :
Formality = Number of formula weight of solute / Volume of solution (L)
(1.13x10⁻⁴ grams / 96.06 g ) / 2.5 L = 4,70x10⁻⁷ F
#edge2021
Answers
Answer:
it's b.
Explanation:
thank u so much for this. i appreciate it. lol.
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Answers
From the problem, you know that 1) = 2.000 g.
Now you need to find 2). You know that your crucible+anhydrous salt is 5.022 g. To find just the anhydrous salt, subtract the mass of the crucible (3.715 g).
1) = 5.022 g - 3.715 g = 1.307 g
Now you can complete our original task.
Mass H2O = 2) - 1) = 2.000 g - 1.307 g = 0.693 g.
Final answer:
The weight of the water lost is 0.693 g.
Explanation:
To calculate the weight of the water that is lost, we need to find the weight of the anhydrous salt. The anhydrous salt is the crucible with the added hydrate, minus the weight of the crucible. So, the weight of the anhydrous salt is 5.022 g - 3.715 g = 1.307 g. Since the weight of the hydrate is 2.000 g, the weight of the water that is lost is equal to the difference between the weight of the hydrate and the weight of the anhydrous salt, which is 2.000 g - 1.307 g = 0.693 g.
Learn more about Hydrates here:
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А. 14
В. 28
С. 20
D. 5
Answers
Answer:
The answer is 20
Explanation: