Explanation:
Given
The enthalpy of formation of RbF (s) is –557.7kJ/mol
The standard enthalpy of formation of RbF (aq, 1 m) is –583.8 kJ/mol
The enthalpy of solution of RbF = Enthalpy of RbF (aq) - Enthalpy of formation of RbF (s)
= -583.8 - (-557.7) kJ/mol
= -26.1 kJ/mol
The enthalpy is negative which means that the temperature will rise when RbF is dissolved.
1.4mL
Explanation:
Given parameters:
Mass of object = 3.5g
density = 2.5g/mL
Unknown:
Volume of the object = ?
Solution:
Density is the mass per unit volume of a body. It is an intensive property that is the same for the same substances.
Density = 
since the volume is unknown; we make it the subject of the formula;
Volume = 
Volume = 
Volume of the object = 1.4mL
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Answer:
1.72x10⁻⁵ g
Explanation:
To solve this problem we use the PV=nRT equation, where:
- R = 0.082 atm·L·mol⁻¹·K⁻¹
- T = 25 °C ⇒ (25+273.16) = 298.16 K
And we <u>solve for n</u>:
- 1 atm * 5.7x10⁶ L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 298.16 K
Finally we <u>convert moles of helium to grams</u>, using its <em>molar mass</em>:
- 4.29x10⁻⁶ mol * 4 g/mol = 1.72x10⁻⁵ g
<span>NaCl
First calculate the molar mass of NaCl and AgNO3 by looking up the atomic weights of each element used in either compound
Sodium = 22.989769
Chlorine = 35.453
Silver = 107.8682
Nitrogen = 14.0067
Oxygen = 15.999
Now multiply the atomic weight of each element by the number of times that element is in each compound and sum the results
For NaCl
22.989769 + 35.453 = 58.44277
For AgNO3
107.8682 + 14.0067 + 3 * 15.999 = 169.8719
Now calculate how many moles of each substance by dividing the total mass by the molar mass
For NaCl
4.00 g / 58.44277 g/mol = 0.068443 mol
For AgNO3
10.00 g / 169.8719 g/mol = 0.058868
Looking at the balanced equation for the reaction, there is a 1 to 1 ratio in molecules for the reaction. Since there is a smaller number of moles of AgNO3 than there is of NaCl, that means that there will be some NaCl unreacted, so the excess reactant is NaCl</span>