Answer is 10 units of water molecules.
Use the formula
first step:
Use the formula
molarity= mole/liter
change ml to l
plug in data
to get .1=mole/.25 or .1M*.25liter
which =.025 moles
then divide .025 moles by two because there are two OH in Sr(OH)2
then multiply that by 265.76 (the molar mass of water)
.0125*265.76
which is 3.32grams this is your answer
Answer:
31395 J
Explanation:
Given data:
mass of water = 150 g
Initial temperature = 25 °C
Final temperature = 75 °C
Energy absorbed = ?
Solution:
Formula:
q = m . c . ΔT
we know that specific heat of water is 4.186 J/g.°C
ΔT = final temperature - initial temperature
ΔT = 75 °C - 25 °C
ΔT = 50 °C
now we will put the values in formula
q = m . c . ΔT
q = 150 g × 4.186 J/g.°C × 50 °C
q = 31395 J
so, 150 g of water need to absorb 31395 J of energy to raise the temperature from 25°C to 75 °C .
Answer: Th enthalpy of combustion for the given reaction is 594.244 kJ/mol
Explanation: Enthalpy of combustion is defined as the decomposition of a substance in the presence of oxygen gas.
W are given a chemical reaction:
To calculate the enthalpy change, we use the formula:
This is the amount of energy released when 0.1326 grams of sample was burned.
So, energy released when 1 gram of sample was burned is = 
Energy 1 mole of magnesium is being combusted, so to calculate the energy released when 1 mole of magnesium ( that is 24 g/mol of magnesium) is being combusted will be:
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
