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iragen
1 month ago
14

A 10.1 g sample of NaOH is dissolved in 250.0 g of water in a coffee-cup calorimeter. The temperature increases from 23.0 °C to

________°C. Specific heat of liquid water is 4.18 J/g-K and ΔH for the dissolution of sodium hydroxide in water is 44.4 kJ/mol.
Chemistry
2 answers:
Tems11 [2.7K]1 month ago
7 0

Final Answer:

33.3 °C

Clarification:

There are two heat transfers occurring in this setup.

Heat from the NaOH solution combined with heat for warming the water equals zero.

q1 + q2 = 0

nΔH + mCΔT = 0

Information: m(NaOH) = 10.1 g

ΔH = -44.4 kJ/mol

m(H2O) = 250.0 g

C = 4.18 J/(K·mol)

Initial Temperature = 23.0 °C

Computation:

n = 10.1 g NaOH × (1 mol NaOH/40.00 g NaOH) = 0.2525 mol NaOH q1 = 0.2525 mol × (-44,400 J/mol) = -11,210 J

Total mass of solution = m(NaOH) + m(water) = 10.1 + 250.0 = 260.1 g

q2 = 260.1 × 4.18 × ΔT = 1087ΔT J

-11,210 + 1087ΔT = 0

1087ΔT = 11,210

ΔT = 11,210 / 1087 = 10.31 °C

ΔT = T2 - T1 = T2 - 23.0 = 10.31

T2 = 23.0 + 10.31 = 33.3 °C

Thus the temperature is reached at 33.3 °C.

castortr0y [3K]1 month ago
6 0

Greetings!:

Number of moles of NaOH = 10.1 / 40 = 0.2525

Energy = ΔH × moles

= 44.4 x 0.2525

= 11.21 kJ

Total mass = 10.1 + 250 = 260.1 g

Q = m Cp ΔT

11211 = 260.1 x 4.18 x ΔT

ΔT = 10.3

T2 = 10.3 + 23 = 33.3 °C

Final temperature = 33.3 ºC

Hope this information aids you!

You might be interested in
In an experiment a student mixes a 50.0 mL sample of 0.100 M AgNO₃(aq) with a 50.0 mL sample of 0.100 M NaCl(aq) at 20.0°C in a
KiRa [2933]

The enthalpy change associated with the precipitation reaction is 84 kJ/mole

Why?

The chemical equation for the reaction can be written as

AgNO₃(aq) + NaCl (aq) → AgCl(s) + NaNO₃(aq)

To determine the enthalpy change, the following equation applies

\Delta H =\frac{Q}{n}

To calculate the heat (Q):

Q=m*C*\Delta T=(100g)*(4.2 J/g*^\circ C)*(21^\circ C-20^\circ C)\\\\Q=420J

Next, we need to calculate the number of moles involved in the reaction (n):

n=[AgNO_3]*v(L)=(0.1M)*(0.05L)=0.005moles

With these two values, we can substitute them into the first equation:

\Delta H= \frac{420J}{0.005moles}=84000J/mole=84kJ/mole

Have a great day!

5 0
2 months ago
what’s the empirical and molecular formula for a compound with 20.2% sodium, 37.6% sulfur, 42.2%oxygen and a molar mass of 682.8
Anarel [2989]

Answer:

Empirical formula: Na3S4O9

Molecular formula: Na6S8O18

Explanation:

1) Upang makuha ang empirical formula, na siyang pinaka-simpleng buong bilang na ratio ng bawat elemento sa hindi kilalang compound na ito, una nating hahatiin ang dami ng bawat elemento sa kanilang sariling atomic mass.

Na= 23g/mol, S= 32g/mol, O=16g/mol

Na = 20.2/23 = 0.878mol

S = 37.6/32 = 1.175mol

O = 42.2/16 = 2.638mol

Susundan ito sa pamamagitan ng paghahati ng bawat bilang ng moles sa pinakamababang halaga (0.878mol)

Na = 0.878/0.878 = 1

S = 1.175/0.878 = 1.34

O = 2.638/0.878 = 3.004

Susunod, paramihin ang bawat decimal na bahagi ng 3 upang maging buo ang mga numero:

Na = 3

S= 4.02 ~ 4

O= 9.012 ~ 9

Kaya, ang pinaka-simpleng ratio ng Na, S, at O ay 3:4:9 kaya, ang empirical formula ay Na3S4O9

2) Para makuha ang molecular formula, gagamitin natin ang molecular weight ng bawat elemento sa empirical ratio, at ang kabuuang molar mass ng compound i.e.

Molar mass ng compound= 682.8g/mol

molecular weight ng Na3S4O9 = (3 x 23g/mol) + (4 x 32 g/mol) + (9 x 16 g/mol)

= 69 + 128 + 144

= 341 g/mol

Kaya, (Na3S4O9)n = 682.8g/mol

= (341)n = 682.8

n = 682.8/341

n~ 2

Kaya, molecular formula = (Na3S4O9)2

= Na6S8O18

8 0
2 months ago
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