The first compound to precipitate will be AgI.
Explanation: To form a precipitate from a salt solution, the ionic product must surpass the solubility product. Given that AgI has a significantly low Ksp, it will precipitate before PbI2 does. The concentration of AgI solution also affects precipitation speed; the highest concentration of AgI in the choices provided is...
Ca3(PO4)2 is the correct formula.
Answer:
9.69g
Explanation:
To find the needed outcome, we first need to determine the number of moles of N2 present in 7.744L of the gas.
1 mole of gas takes up 22.4L at STP.
Thus, X moles of nitrogen gas (N2) will fill 7.744L, meaning
X moles of N2 = 7.744/22.4 = 0.346 moles
Next, we will convert 0.346 moles of N2 to grams to achieve the result sought. The calculation goes as follows:
Molar Mass of N2 = 2x14 = 28g/mol
Number of moles N2 = 0.346 moles
Find the mass of N2 =?
Mass = number of moles × molar mass
Mass of N2 = 0.346 × 28
Mass of N2 = 9.69g
Hence, 7.744L of N2 consists of 9.69g of N2
Solution:
The gas's new temperature is 604K
Justification:
Assuming standard temperature and pressure, we can determine the gas's temperature using the ideal gas law;
Step 1: Formulate the general gas law equation
P1V1/T1 = P2V2/T2
Step 2: Insert the values, converting as needed to standard units.
P1 = 0.800 atm
V1 = 0.180 L
T1 = 29°C = 273 + 29 = 302K
P2 = 3.20 atm
V2 = 90 mL = 90 * 10^-3 L = 0.09 L
Step 3: Solve for T2
The new gas temperature T2 is calculated as:
T2 = P2V2T1/(P1V1)
T2 = 3.20 * 0.09 * 302 / (0.800 * 0.180)
T2 = 86.976 / 0.144
T2 = 604K
The gas's new temperature is 604K.
To calculate the moles of MgSO4.7H2O, we find the molar mass equals 246, thus moles = 32 / 246 = 0.13 moles. Upon heating, all 7 H2O from one molecule will evaporate. The total moles of H2O present amount to 7 x 0.13 = 0.91, and the mass of that H2O is 0.91 x 18 = 16.38g. Therefore, the mass of the anhydrous MgSO4 that remains is 32 - 16.38 = 15.62 g.
