This might be right. Im not quite sure. This is what my 5th grade science teacher told me. 'Look at the 2 LR's and add them together. Then look at the total amount which is 32. When you add the 2 LR's you get 17. So subtract 17 from 32 and you get 15. So: C:15 is your answer." (LR's stands for liquid reactants)
10+17=17
Total amount is 32
32-17=15
15 is you mass
Hope this Helps
Hey there !
Molecular Weight: 262.86 g/mol
Volume in liters : 1,000 mL => 1,000 / 1000 => 1 L
Number of moles of solution :
number of moles = volume x molarity
number of moles = 1 x 0.3 => 0.3 moles
Therefore:
1 mole Mg₃(PO₄)₂ ------------------- 262.86 g
0.3 moles ----------------------------- mass of Mg₃(PO₄)₂ ??
mass of Mg₃(PO₄)₂ = 0.3 x 262.86 / 1
mass of Mg(PO₄)₂ => 78.858 g
Hope this helps!
Answer:
To find the mass percent of hydrogen in hydrogen chloride, we must divide the weight of the hydrogen atom alone by the weight of the entire molecule. Then we multiply by 100% to find the percentage. Thus, 2.77% of the mass of hydrogen chloride is hydrogen.
Explanation:
i hope you understand better
Answer:
B)
Explanation:
The total ionic equation contains all the species in the reaction in the ionic form.
Spectator ions are defined as the ions which does not get involved in a chemical equation or they are ions which are found on both the sides of the chemical reaction present in ionic form.
Spectator ions are also written in total ionic equations.
The chemical equation is:
The total ionic equation is -
Answer:
The enthalpy change during the reaction is -7020.09 kJ/mole.
Explanation:
First we have to calculate the heat gained by the calorimeter.

where,
q = heat gained = ?
c = specific heat = 
= final temperature = 
= initial temperature = 
Now put all the given values in the above formula, we get:


The heat gained by water present in calorimeter. = q'

where,
q' = heat gained = ?
m = mass of water = 
c' = specific heat of water = 
= final temperature = 
= initial temperature = 

q ' = 12,060.38 J
Now we have to calculate the enthalpy change during the reaction.

where,
= enthalpy change = ?
Q = heat gained = -(q+q') = -(2,485.25 J + 12,060.38 J)= -14,545.63 J
Q = -14.54563 kJ
n = number of moles fructose = 

Therefore, the enthalpy change during the reaction is -7020.09 kJ/mole.