Answer:
1.096g
Explanation:
You must know the atomic mass of Hydrogen, Fluorine, and Sodium before you can start:
Hydrogen: 1.008g/mol
Fluorine: 18.99g/mol
Sodium: 22.98g/mol
Next, find the composition percentage of NaF
22.98 + 18.99 = 41.97
Fluorine is 18.99/41.97 =45.25%
Sodium is 100-45.25 = 54.75%
Ultimately we want to know about HF so find how much F is in 2.3g: 2.3 * 0.4525 = 1.041g
Find comp. percentage of HF
18.99+1.008 = 19.998; H/total F/total
Hydrogen 5.041%
Fluorine 94.959%
Laws of conservation of say we have 1.041g of fluorine in our HF. We know 1.041 is 94.959% of the mass of HF so do some simple math to find the remaining: 1.041/0.94959 = 1.096g
Answer:
We need 17.2 L of Ca(OH)2
Explanation:
Step 1: Data given
Concentration of Ca(OH)2 = 1.45 M
Moles of H2SO4 = 25.0 moles
Step 2: The balanced equation
Ca(OH)2 + H2SO4 ⟶2H2O + CaSO4
Step 3: Calculate moles Ca(OH)2
For 1 mol Ca(OH)2 we need 1 mol H2SO4 to produce 2 moles H2O and 1 mol CaSO4
For 25.0 moles H2SO4 we'll need 25.0 moles Ca(OH)2 to produce 50 moles H2O and 25.0 moles CaSO4
Step 4: Calculate volume of Ca(OH)2
Volume Ca(OH)2 = moles Ca(OH)2 / concentration Ca(OH)2
Volume Ca(OH)2 = 25.0 moles / 1.45 M
Volume Ca(OH)2 = 17.2 L
We need 17.2 L of Ca(OH)2
Answer:
1.25 x 10^15Hz
Explanation:
c = frequency x wavelength
c is the speed of light, which is equal to 3.00 x 10^8 m / s
frequency = c /wavelength
= (3.00 x 10^8m /s) / (2.40 x 10^-5 cm x 1 m /100cm)
= (3.00 x 10^8 m/s) / 2.40 x 10^-7m
= 1.25 x 10^15/s 1 / s = 1Hz
So, the Frequency = 1.25 x 10^15Hz
I hope this helped :)
We can't use the nitrogen present in the atmosphere although 80% of the atmosphere consists of nitrogen only.We get this from the plants called as leguminous plants which can convert or fix atmospheric nitrogen into usable ones..
