How is an ionic compound made up of charged particles can be electrically neutral
1 answer:
You might be interested in
1) Chemical equation
<span>4C3H5(NO3)3(l) 12CO2(g) + 6N2(g) + 10H2O(g) + O2(g)
2) Molar ratios
4 mol C3H5(NO3)3 : 10 mol H2O
3) Convert grams of water into number of moles:
numer of moles = mass in grams / molar mass = 120 g / 18.0158 g/mol =
number of moles = 6.66 moles H2O.
4) use proportion and solve for x
4 mol C3H5(NO3)3 / 10 mol H2O = x / 6.66 mol H2O
=> x = 6.66 mol H2O * 4 mol C3H5(NO3)3 / 10 mol H2O = 2.66 mol C3H5(NO3)3
5) Convert </span><span>2.66 mol C3H5(NO3)3 into grams
mass in grams = molar mass * number of moles = 227.0995 g * 2.66 mol = 605.07 g
Answer: 605.07 g
</span>
<span>4C3H5(NO3)3(l) 12CO2(g) + 6N2(g) + 10H2O(g) + O2(g)
2) Molar ratios
4 mol C3H5(NO3)3 : 10 mol H2O
3) Convert grams of water into number of moles:
numer of moles = mass in grams / molar mass = 120 g / 18.0158 g/mol =
number of moles = 6.66 moles H2O.
4) use proportion and solve for x
4 mol C3H5(NO3)3 / 10 mol H2O = x / 6.66 mol H2O
=> x = 6.66 mol H2O * 4 mol C3H5(NO3)3 / 10 mol H2O = 2.66 mol C3H5(NO3)3
5) Convert </span><span>2.66 mol C3H5(NO3)3 into grams
mass in grams = molar mass * number of moles = 227.0995 g * 2.66 mol = 605.07 g
Answer: 605.07 g
</span>
Answer:
Explanation:
We can use Gay-Lussac's Law of Combining Volumes to solve this problem.
Gases at the same temperature and pressure react in the same ratios as their coefficients in the balanced equation.
1. Write the chemical equation.
Ratio: 1 L 3 L
N₂ + 3H₂ ⟶ 2NH₃
V/L: 2
2. Calculate the volume of H₂.
According to Gay-Lussac, 3 L of H₂ react with 1 L of N₂.
Then, the conversion factor is (3 L H₂/1 L N₂).
Answer:
Explanation:
The measure of R is always the same, but the numbers may differ depending on the units you use.
For example, in SI units, R = 8.314 Pa·m³K⁻¹mol
If your measurement uses different units, you must either convert your units to SI or use a value of R consistent with your units.
If you use bars and litres, R = 0.083 14 bar·L·K⁻¹mol⁻¹.
If you use kilopascals and litres, R = 8.314 kPa·L·K⁻¹mol
If you use atmospheres and litres, R = 0.082 06 L·atm·K⁻¹mol⁻¹.
If you use Torr and cubic centimetres, R = 62 368 Torr·cm³ K⁻¹mol⁻¹.
The only units that don't change are "K⁻¹mol⁻¹".
Answer:
here you go buddy. good luck
