Answer: 0.39 moles NaPO4
Explanation: Solution: 1 mole of NaPO4 is equal to its molar mass 118g NaPO4
45.8 g NaPO4 X 1 mole NaPO4/ 118 g NaPO4
= 0.39 moles NaPO4
Answer: (1). There are 0.0165 moles of gaseous arsine (AsH3) occupy 0.372 L at STP.
(2). The density of gaseous arsine is 3.45 g/L.
Explanation:
1). At STP the pressure is 1 atm and temperature is 273.15 K. So, using the ideal gas equation number of moles are calculated as follows.
PV = nRT
where,
P = pressure
V = volume
n = number of moles
R = gas constant = 0.0821 L atm/mol K
T = temperature
Substitute the values into above formula as follows.
![PV = nRT\\1 atm \times 0.372 L = n \times 0.0821 L atm/mol K \times 273.15 K\\n = 0.0165 mol](https://tex.z-dn.net/?f=PV%20%3D%20nRT%5C%5C1%20atm%20%5Ctimes%200.372%20L%20%3D%20n%20%5Ctimes%200.0821%20L%20atm%2Fmol%20K%20%5Ctimes%20273.15%20K%5C%5Cn%20%3D%200.0165%20mol)
2). As number of moles are also equal to mass of a substance divided by its molar mass.
So, number of moles of Arsine
(molar mass = 77.95 g/mol) is as follows.
![No. of moles = \frac{mass}{molar mass}\\0.0165 mol = \frac{mass}{77.95 g/mol}\\mass = 1.286 g](https://tex.z-dn.net/?f=No.%20of%20moles%20%3D%20%5Cfrac%7Bmass%7D%7Bmolar%20mass%7D%5C%5C0.0165%20mol%20%3D%20%5Cfrac%7Bmass%7D%7B77.95%20g%2Fmol%7D%5C%5Cmass%20%3D%201.286%20g)
Density is the mass of substance divided by its volume. Hence, density of arsine is calculated as follows.
![Density = \frac{mass}{volume}\\= \frac{1.286 g}{0.372 L}\\= 3.45 g/L](https://tex.z-dn.net/?f=Density%20%3D%20%5Cfrac%7Bmass%7D%7Bvolume%7D%5C%5C%3D%20%5Cfrac%7B1.286%20g%7D%7B0.372%20L%7D%5C%5C%3D%203.45%20g%2FL)
Thus, we can conclude that 0.0165 moles of gaseous arsine (AsH3) occupy 0.372 L at STP and the density of gaseous arsine is 3.45 g/L.
Answer:
<em>The molarity of the solution is 0,47 M</em>
Explanation:
Molarity is a concentration measurement that expresses the moles of solute (in this case NaOH) in 1 liter of solution (1000ml). First we calculate the mass of 1 mol of NaOH, to calculate the moles in 58.8 g of said compound:
Weight 1 mol NaOH= Weight Na + Weight O + Weight H=23 g + 16 g +1 g
Weight 1 mol NaOH= 40 g/mol
40 g ---1 mol NaOH
58,8g---x= (58,8g x 1 mol NaOH)/40g =1,47 mol NaOH
3,1 L solution ------1,47 mol NaOH
1 L solution --------x= (1 L solution x 1,47 mol NaOH)/3,1 L solution
<em>x= 0,47 mol NaOH ---> The molarity of the solution is 0,47 M</em>
Answer:
This reaction is an hydration of alkenes and the major product is 3-ethyl-2-heptanol
Explanation:
In this reaction in the first step water is the nuclophile and the alkene is the electrophile. Then, the carbanion is the nucleophile and attacks a proton from the water forming the respective alcohol. This carbanionic compound is a reaction intermediary. You can see the mechanism in the picture.
Is that carbanionic compound who explained the major product because the more substituted carbanion the most stable intermediary. Thus, the 3-ethyl-2-heptanol intermediary is the most stable and this product is the major one.
I hope it helps!