Nvm i got the answer

Calculate the enthalpy for this reaction: 2C(s) + H2(g) ---> C2H2(g) ΔH° = ??? kJ Given the following thermochemical equation

nordsb [41]

Answer:

The enthalpy for given reaction is 232 kilo Joules.

Explanation:

$C_2H_2(g) + \frac{5}{2}O_2(g)\rightarrow 2CO_2(g) + H_2O(l), \Delta H^o_{1} = -1,123 kJ$ ...[1]

$C(s) + O_2(g)\rightarrow CO2(g), \Delta H^o_{2} = -340 kJ$ ..[2]

$H_2(g) + \frac{1}{2}O_2(g)\rightarrow H_2O(l) ,\Delta H^o_{3} = -211 kJ$ ..[3]

$2C(s) + H_2(g)\rightarrow C_2H_2(g),\Delta H^o_{4} =?$ ..[4]

2 × [2] + [3] - [1] ( Using Hess's law)

$\Delta H^o_{4}=2\times \Delta H^o_{2}+\Delta H^o_{3} - \Delta H^o_{1}$

$\Delta H^o_{4}=2\times (-340 kJ) + (-211 kJ) - (-1,123 kJ)$

$\Delta H^o_{4}=232 kJ$

The enthalpy for given reaction is 232 kilo Joules.

5 0

3 years ago

What two things must all matter have?

OleMash [197]

1.) Mass

2.) Can occupy space (Volume)

6 0

3 years ago

Read 2 more answers

What is the percentage error of length measurement of 0.229cm if the correct value is 0.225cm

Mars2501 [29]

Percent error is the difference between the measured and known value, divided by the known value, multiplied by 100%.

So first, we take our measured value, .299 cm, minus our known value, .225 cm.

.299 cm - .225 cm=.004 cm

Next, we divide that by our known value

$\frac{.004}{.225}=.0177777778$

Finally, multiply your answer by 100

.0177777778 x 100= 1.77777778 %

Round to three significant figures, and you're done.

=1.78 % error

5 0

3 years ago

Read 2 more answers

Using the Bohr model, determine the energy, in joules, necessary to ionize a ground-state hydrogen atom. Show your calculations.

lord [1]

Answer:

The energy required to ionize the ground-state hydrogen atom is 2.18 x 10^-18 J or 13.6 eV.

Explanation:

To find the energy required to ionize ground-state hydrogen atom first we calculate the wavelength of photon required for this operation.

It is given by Bohr's Theory as:

1/λ = Rh (1/n1² - 1/n2²)

where,

λ = wavelength of photon

n1 = initial state = 1 (ground-state of hydrogen)

n2 = final state = ∞ (since, electron goes far away from atom after ionization)

Rh = Rhydberg's Constant = 1.097 x 10^7 /m

Therefore,

1/λ = (1.097 x 10^7 /m)(1/1² - 1/∞²)

λ = 9.115 x 10^-8 m = 91.15 nm

Now, for energy (E) we know that:

E = hc/λ

where,

h = Plank's Constant = 6.625 x 10^-34 J.s

c = speed of light = 3 x 10^8 m/s

Therefore,

E = (6.625 x 10^-34 J.s)(3 x 10^8 m/s)/(9.115 x 10^-8 m)

E = (2.18 x 10^-18 J)(1 eV/1.6 x 10^-19 J)

5 0

3 years ago

Uranium – 235 has a half-life of 713 million years. Would uranium – 235 or carbon – 14 be more useful for dating Cambrian time e

kenny6666 [7]

Uranium-235 would be more useful for dating in Cambrian time because Cambrian time was 540 million years ago while the half life of carbon-14 is only 5,730 years
Hope this helps

6 0

3 years ago