what is the volume at standard temperature and pressure for 28.4g of N2H6SO3?
That compound is called ammonium sulfamate and is a white crystalline powder used as a weed killer. Its density is about 1.8 g/cm^3 at STP
You can't use the perfect gas law on powders. The volume will be 28.4/1.8 = 16 cm^3
See http://en.wikipedia.org/wiki/Ammonium_sulfamate
for more information about the compound
thank you. (:
To find the volume of a given amount of a substance at standard temperature and pressure (STP), you need to use the ideal gas law equation. The ideal gas law equation is PV = nRT, where:
P is the pressure (at STP, it is 1 atmosphere or 101.3 kilopascals)
V is the volume
n is the number of moles of the substance
R is the ideal gas constant (0.0821 L·atm/(mol·K) or 8.314 J/(mol·K))
T is the temperature in Kelvin
To solve the equation, we need three of the variables. In this case, we have the mass of the substance (28.4g) and we need to find the volume at STP.
First, we need to calculate the number of moles (n) of N2H6SO3 using its molar mass. The molar mass of N2H6SO3 is calculated by adding the molar masses of nitrogen (N), hydrogen (H), sulfur (S), and oxygen (O).
The atomic masses are:
N: 14.01 g/mol
H: 1.01 g/mol
S: 32.07 g/mol
O: 16.00 g/mol
N2H6SO3:
(2 * N) + (6 * H) + S + (3 * O)
(2 * 14.01) + (6 * 1.01) + 32.07 + (3 * 16.00)
Calculating the molar mass:
28.02 + 6.06 + 32.07 + 48.00
= 114.15 g/mol
Now that we have the molar mass (114.15 g/mol) and the mass (28.4 g), we can calculate the number of moles (n) using the formula:
n = mass / molar mass
n = 28.4 g / 114.15 g/mol
Calculating the number of moles:
n = 0.249 mol
Now that we know the number of moles (n), we can calculate the volume using the ideal gas law equation. At STP, the temperature is 273.15 K.
PV = nRT
(1 atm) * V = (0.249 mol) * (0.0821 L·atm/(mol·K)) * (273.15 K)
Solving for V:
V = (0.249 * 0.0821 * 273.15) / 1
V = 5.69 L (rounded to two decimal places)
Therefore, the volume at standard temperature and pressure (STP) for 28.4g of N2H6SO3 is approximately 5.69 liters.