The enthalpy change of the reaction when sodium hydroxide and sulfuric acid react can be calculated using the mass of solution, temperature change, and specific heat of water.

The balanced chemical equation for the reaction can be represented as,

$H_{2}SO_{4}(aq) + 2NaOH (aq) ----> Na_{2}SO_{4}(aq) + 2H_{2}O(l)$

Given volume of the solution = 101.2 mL + 50.6 mL = 151.8 mL

Heat of the reaction, q = $m C .$ Δ $T$

m is mass of the solution = 151.8 mL * $\frac{1 g}{1 mL} = 151.8 g$

C is the specific heat of solution = 4.18 $\frac{J}{g. ^{0}C}$

ΔT is the temperature change = $31.50^{0}C - 21.45^{0}C = 10.05^{0}C$

q = $151.8 g (4.18 \frac{J}{g ^{0}C})(10.05^{0}C) = 6377 J$

Moles of NaOH = $101.2 mL * \frac{1L}{1000 mL}*\frac{1.00 mol}{L} = 0.1012 mol$ NaOH

Moles of $H_{2}SO_{4}$ = $50.6 mL * \frac{1 L}{1000 mL} * \frac{1.0 mol}{1 L} = 0.0506 mol H_{2}SO_{4}$

Enthalpy of the reaction = $\frac{6377 J*\frac{1kJ}{1000J}}{0.0506 mol} = 126 kJ/mol$

5 0

3 years ago

What is the mass (grams) of salt in 10.0 m' of ocean water? ball park-4x10's (1.000 molsalt -58.44 g salt, 1.0 L ocean water -0.

koban [17]

Answer:

3.5 × 10⁵ g of salt

Explanation:

<em>What is the mass (grams) of salt in 10.0 m³ of ocean water?</em>

We have this data:

1.000 mol salt is equal to 58.44 g salt
1.0 L of ocean water contains 0.60 mol of salt

We will need the following relations:

1 L = 1dm³
1 m³ = 10³ dm³

We can use proportions:

$10.0m^{3} .\frac{10^{3}dm^{3} }{1m^{3} } .\frac{1L}{1dm^{3} } .\frac{0.60molSalt}{1.0L} .\frac{58.44gSalt}{1molSalt} =3.5 \times 10^{5} gSalt$

8 0

3 years ago

Nick hypothesizes that wax has a higher melting point than chocolate how can he test his hypothesis

LUCKY_DIMON [66]

Melt chocolate and measure the temperature and then melt wax and measure the temperature and compare his results.

5 0

3 years ago

Is xenon (xe) more or less reactive than bromine (br)? why? ​

Is xenon (xe) more or less reactive than bromine (br)? why?