Answer:
TIMED HELP ASAP
19.11 g of MgSO₄ is placed into 100.0 mL of water. The water's temperature increases by 6.70°C. Calculate ∆H, in kJ/mol, for the dissolution of MgSO₄. (The specific heat of water is 4.184 J/g･°C and the density of the water is 1.00 g/mL). You can assume that the specific heat of the solution is the same as that of water.
1) Write the balanced chemical equation
2HCl + Na2 CO3 ----------> 2NaCl + H2CO3
2) Write the molar ratios:
2 mol HCl : 1 mol Na2CO3 : 2 mol NaCl : 1 mol H2CO3
3) Convert 0.15g of sodium carbonate to number of moles
3a) Calculate the molar mass of Na2CO3
Na: 2 * 23 g/mol = 46 g/mol
C: 12 g/mol =
O: 3 * 16 g/mol = 48 g/mol
molar mass = 46g/mol + 12g/mol + 48g/mol = 106 g/mol
3b.- Calculate the number of moles of Na2CO3
# moles = grams / molar mass = 0.15 g / 106 g/mol = 0.0014 mol Na2CO3
4) Calculate the number of moles of HCl from the molar proportion:
[0.0014 mol Na2CO3] * [2 mol HCl / 1 mol Na2CO3] = 0.0028 mol HCl
5) Calculate the volume of HCl from the definition of Molarity
Molarity, M = # moles / volume in liters
=> Volume in liters = # moles / M = 0.0028 mol / 0.1 M = 0.028 liters
0.028 liters * 1000 ml / liter = 28 ml.
Answer: 28 mililiters of 0.1 M HCl.
Answer:
It is true answer. Ozone = O3
Answer:
B. Cu + 4HNO3 → Cu(NO3)2 + 2H2O + 2NO2
Explanation:
Hello,
In this case, we should understand oxidizing agents as those substances able to increase the oxidation state of another substance, therefore, in B. reaction we notice that copper oxidation state at the beginning is zero (no bonds are formed) and once it reacts with nitric acid, its oxidation states raises to +2 in copper (II) nitrate, thus, in B. Cu + 4HNO3 → Cu(NO3)2 + 2H2O + 2NO2 nitritc acid is acting as the oxidizing agent.
Moreover, in the other reactions, copper (A.), sodium (C. and D.) remain with the same initial oxidation state, +2 and +1 respectively.
Regards.
