Answer:
<em> 14, 508J/K</em>
ΔHrxn =q/n
where q = heat absorbed and n = moles
Explanation:
<em>m = mass of substance (g) = 0.1184g</em>
1 mole of Mg - 24g
<em>n</em> moles - 0.1184g
<em>n = 0.0049 moles.</em>
Also, q = m × c × ΔT
<em> Heat Capacity, C of MgCl2 = 71.09 J/(mol K)</em>
<em>∴ specific heat c of MgCL2 = 71.09/0.0049 (from the formula c = C/n)</em>
<em>= 14, 508 J/K/kg</em>
ΔT= (final - initial) temp = 38.3 - 27.2
= 11.1 °C.
mass of MgCl2 = 95.211 × 0.1184 = 11.27
⇒ q = 11.27g × 11.1 °C × <em>14, 508 j/K/kg </em>
<em>= 1,7117.7472 J °C-1 g-1</em>
<em />
<em>∴ ΔHrxn = q/n</em>
<em>=1,7117.7472 ÷ 0.1184 </em>
<em>= 14, 508J/K</em>
Answer:
the lighter fresh water rises up and over the salt water
Explanation:
this is because the salt water is denser
Answer:
0.067 M
Explanation:
Given data:
Volume of solution = 1.00 L
Mass of BaCl₂ = 14 g
Molarity of solution = ?
Solution:
Molarity is used to describe the concentration of solution. It tells how many moles are dissolve in per litter of solution.
Formula:
Molarity = number of moles of solute / L of solution
Now we will calculate the number of moles first.
Number of moles = mass/ molar mass
Number of moles = 14 g/ 208.23 g/mol
Number of moles = 0.067 mol
Now we will calculate the molarity.
Molarity = 0.067 mol / 1.00 L
Molarity = 0.067 M ( mol/L=M)
Answer:
a) 0.925 mol Na2CO3 can be theoretically produced
b) 0.075 moles of the excess starting material remains
Explanation:
balaced chemical:
2 NaOH(s) + CO2(g) ↔ Na2CO3(s) + H2O(aq)
1.85n 1.00n Xn
moles theor. Na2CO3:
⇒ nNaCO3 = 1.85nNaOH * ( nNa2CO3 / 2nNaOH)
⇒ nNa2CO3 = 0.925nNa2CO3
moles of the excess:
⇒moles CO2 react = 1.85nNaOH * nCO2 / 2nNaOH = 0.925n CO2
⇒moles CO2 excess = 1.00n - 0.925n = 0.075n excess CO2
