Solubility rules state that group one metals and Nitrates are always soluble. Rubidium is a group one metal, so Rubidium Nitrate, RbNO3, is soluble.
Carbon dioxide and oxigen
Answer:
Explanation:
To determine the molecular formula of the compound, the empirical formula must be determined first. To determine the empirical formula, the percentage of each constituent is divided by its molar mass. This is shown below
Carbon = 60/12 = 5
Oxygen = 32/16 = 2
Hydrogen = 8/1 = 8
The next step is to divide each ratio by the smallest value. The smallest value is 2. It becomes
Carbon = 5/2 = 2.5
It is approximated to 3
Oxygen = 2/2 = 1
Hydrogen = 8/2 = 4
Therefore, the empirical formula is
C3H4O
From the given relative molecular mass of the compound, the molecular formula can be determined
Answer:
378mL
Explanation:
The following data were obtained from the question:
Pressure (P) = 99.19 kPa
Temperature (T) = 28°C
Number of mole (n) = 0.015 mole
Volume (V) =...?
Next, we shall convert the pressure and temperature to appropriate units. This is illustrated below:
For Pressure:
101.325 KPa = 1 atm
Therefore, 99.19 kPa = 99.19/101.325 = 0.98 atm
For Temperature:
T(K) = T(°C) + 273
T(°C) = 28°C
T(K) = 28°C + 273 = 301K.
Next we shall determine the volume of N2. The volume of N2 can be obtained by using the ideal gas equation as shown below:
PV = nRT
Pressure (P) = 0.98 atm
Temperature (T) = 301K
Number of mole (n) = 0.015 mole
Gas constant (R) = 0.0821atm.L/Kmol.
Volume (V) =...?
0.98 x V = 0.015 x 0.0821 x 301
Divide both side by 0.98
V = (0.015 x 0.0821 x 301) /0.98
V = 0.378 L
Finally, we shall convert 0.378 L to millilitres (mL). This is illustrated below:
1L = 1000mL
Therefore, 0.378L = 0.378 x 1000 = 378mL
Therefore, the volume of N2 collected is 378mL
The answer for that is 16, math right?
