Answer : The mass of sodium bromide added should be, 18.3 grams.
Explanation :
Molality : It is defined as the number of moles of solute present in kilograms of solvent.
Formula used :
Solute is, NaBr and solvent is, water.
Given:
Molality of NaBr = 0.565 mol/kg
Molar mass of NaBr = 103 g/mole
Mass of water = 315 g
Now put all the given values in the above formula, we get:
Thus, the mass of sodium bromide added should be, 18.3 grams.
The answer should be D hope this helps
Answer:
ΔH = 2.68kJ/mol
Explanation:
The ΔH of dissolution of a reaction is defined as the heat produced per mole of reaction. We have 3.15 moles of the solid, to find the heat produced we need to use the equation:
q = m*S*ΔT
<em>Where q is heat of reaction in J,</em>
<em>m is the mass of the solution in g,</em>
<em>S is specific heat of the solution = 4.184J/g°C</em>
<em>ΔT is change in temperature = 11.21°C</em>
The mass of the solution is obtained from the volume and the density as follows:
150.0mL * (1.20g/mL) = 180.0g
Replacing:
q = 180.0g*4.184J/g°C*11.21°C
q = 8442J
q = 8.44kJ when 3.15 moles of the solid react.
The ΔH of the reaction is:
8.44kJ/3.15 mol
= 2.68kJ/mol
Answer: the pressure releases gas. The two most abundant gases are sulfur dioxide and carbon dioxide, and if levels of these gases increase,
Explanation:
Answer:
4.42x10⁻¹⁹ J/molecule
Explanation:
At a double bond, there's sigma and a pi bond, and at a single bond, there's only a sigma bond. Thus, if the energy to break both sigma and pi is 614 kJ/mol, and the energy to break only the sigma bond is 348 kJ/mol, the energy to break only the pi bond is:
E = 614 - 348 = 266 kJ/mol
Knowing that 1 kJ = 1000 J, E = 266,000 J/mol
By Avogadro's number, 1 mol = 6.02x10²³ molecules, thus:
E = 266,000 J/mol * 1mol/6.02x10²³ molecules
E = 4.42x10⁻¹⁹ J/molecule
