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victus00 [196]
3 years ago
11

some places such as caves and sinkholes have lower elevations than surrounding areas infer how a sink might be shown on a topogr

aphic map

Chemistry
1 answer:
Andreyy893 years ago
5 0

Answer:

By closed contours with hachures

Explanation:

A sinkhole is a depression in the ground that forms when water erodes an underlying rock layer.

Sinkholes are denoted by closed contours with tick marks (hachures) that point towards lower ground.

The topographical map below shows the difference between the contours of a hill and those of a sinkhole.

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Given the following at 25C calculate delta Hf for HCN (g) at 25C. 2NH3 (g) +3O2 (g) + 2CH4 (g) ---> 2HCN (g) + 6H2O (g) delta
AysviL [449]

<u>Answer:</u> The \Delta H_f for HCN (g) in the reaction is 135.1 kJ/mol.

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. The equation used to calculate enthalpy change is of a reaction is:

\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]

For the given chemical reaction:

2NH_3(g)+3O_2(g)+2CH_4(g)\rightarrow 2HCN(g)+6H_2O(g)

The equation for the enthalpy change of the above reaction is:

\Delta H_{rxn}=[(2\times \Delta H_f_{(HCN)})+(6\times \Delta H_f_{(H_2O)})]-[(2\times \Delta H_f_{(NH_3)})+(3\times \Delta H_f_{(O_2)})+(2\times \Delta H_f_{(CH_4)})]

We are given:

\Delta H_f_{(H_2O)}=-241.8kJ/mol\\\Delta H_f_{(NH_3)}=-80.3kJ/mol\\\Delta H_f_{(CH_4)}=-74.6kJ/mol\\\Delta H_f_{(O_2)}=0kJ/mol\\\Delta H_{rxn}=-870.8kJ

Putting values in above equation, we get:

-870.8=[(2\times \Delta H_f_{(HCN)})+(6\times (-241.8))]-[(2\times (-80.3))+(3\times (0))+(2\times (-74.6))]\\\\\Delta H_f_{(HCN)}=135.1kJ

Hence, the \Delta H_f for HCN (g) in the reaction is 135.1 kJ/mol.

8 0
3 years ago
A balloon vendor at a street fair is using a tank of helium to fill her balloons. The tank has a volume of 124.0 L and a pressur
amm1812

Answer:

She lost 50.88 moles

Explanation:

Step 1: Data given

The volume of the tank = 124.0 L

The initial pressure = 104.0 atm

The temperature = 24.0 °C = 297 K

The pressure drops to 94.0 atm

The temperature stays constant at 297 K

Step 2: Calculate the initial number of moles

p*V = n*R*T

n = (p*V)/(R*T)

⇒with p = the initial pressure = 104.0 atm

⇒with V = the initial volume = 124.0 L

⇒with n = the initial number of moles = TO BE DETERMINED

⇒with R = the gas constant = 0.08206 L*atm/mol*K

⇒ with T = the temperature= 297 K

n = (104.0*124.0)/(0.08206*297)

n = 529.14 moles

Step 3: Calculate final number of moles

p*V = n*R*T

n = (p*V)/(R*T)

⇒with p = the initial pressure = 94.0 atm

⇒with V = the initial volume = 124.0 L

⇒with n = the initial number of moles = TO BE DETERMINED

⇒with R = the gas constant = 0.08206 L*atm/mol*K

⇒ with T = the temperature= 297 K

n = (94.0*124.0)/(0.08206*297)

n = 478.26 moles

Step 4: Calculate the difference of moles

529.14 moles - 478.26 moles = 50.88 moles

She lost 50.88 moles

4 0
4 years ago
You are on the Titanic and want to find something that will float because you didn't
damaskus [11]

Find volume of pillow

L=78cm

B=55cm

H=25cm

\\ \bull\tt\longrightarrow V=LBH

\\ \bull\tt\longrightarrow V=25(78)(55)

\\ \bull\tt\longrightarrow V=107250cm^3

\\ \bull\tt\longrightarrow V=10.72m^3

Now

Mass=5.5kg

\\ \bull\tt\longrightarrow Density=\dfrac{Mass}{Volume}

\\ \bull\tt\longrightarrow Density=\dfrac{5.5}{10.72}

\\ \bull\tt\longrightarrow Density=0.5kg/m^3

Density of water=1000kg/m^3

As it is less than density of water it will float on water

8 0
3 years ago
Why does aluminium oxide (Al2O3) have a higher melting point than sodium oxide (Na2O)?
Lilit [14]

Al2O3 has a higher melting point than Na2O. This is because the ionic bond between Al3+ ions and O2- ions is stronger than that between Na+ and O2-. The charge on the Al3+ ion is larger than that of the Na+ ion

5 0
4 years ago
In aqueous solution amino acids are rarely found in the neutral, unionized form.
True [87]

a. True.

There is always an equilibrium of the type

NH₃⁺CHRCOOH ⇌ NH₃⁺CHRCOO⁻ ⇌ NH₂CHRCOO⁻

The compound is <em>always in an ionized form</em>.

There are no unionized NH₂CHRCOOH molecules in the solution.

3 0
3 years ago
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