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zysi [14]
3 years ago
12

What is the largest river basin in North Carolina?

Chemistry
2 answers:
user100 [1]3 years ago
8 0

Answer:

The Mackenzie River

Explanation:

Liula [17]3 years ago
6 0
The french broad? idk sorry
You might be interested in
Ethylene oxide (EO) is prepared by the vapor-phase oxidation of ethylene. Its main uses are in the preparation of the antifreeze
Rashid [163]

Answer:

a. ΔH^0_{rxn} = -108.0\frac{kJ}{mol}

b. 320.76° C

Explanation:

a.)

we can solve this type of question (i.e calculate ΔH^0_{rxn} , for the gas-phase reaction )  using the Hess's Law.

ΔH^0_{rxn} =  E_{product} deltaH^0_{t}-E_{reactant} deltaH^0_{t}

Given from the question, the table below shows the corresponding  ΔH^0_{t}(kJ/mol) for each compound.

Compound                    H^0_{t}(kJ/mol)

Liquid EO                       -77.4

CH_4_(g_)                            -74.9                

CO_(g_)                              -110.5

If we incorporate our data into the above previous equation; we have:

ΔH^0_{rxn} = (-110.5 kJ/mol + (-74.9 kJ/mol) ) - (-77.4 kJ/mol)

          =   -108.0 \frac{kJ}{mol}

b.)

We are to find the final temperature if the average specific heat capacity of the products is 2.5 J/g°C

Given that:

the specific heat capacity (c) = 2.5 J/g°C

T_{initial} = 93.0°C   &

the  enthalpy of vaporization  (ΔH^0_{vap}) = 569.4 J/g

If, we recall; we will remember that; Specific Heat Capacity is the amount of heat needed to raise the temperature of one gram of a substance by one kelvin.

∴ the specific heat capacity (c) is given as =  \frac{Heat(q)}{mass*changeintemperature(T_{initial}-T_{final})}

Let's not forget as well, that  ΔH^0_{vap} = \frac{q}{mass}

If we substitute  ΔH^0_{vap}  for  \frac{q}{mass} in the above equation, we have;

specific heat capacity (c) = \frac{deltaH^0_{vap}}{T_{final}-T_{initial}}

Making (T_{final}- T_{initial}) the subject of the formula; we have:

T_{final}- T_{initial}  = \frac{delat H^0_{vap}}{specificheat capacity}

(T_{final}-93.0^0C)=\frac{569.4J/g}{2.5J/g^0C}

T_{final}=\frac{569.4J/g}{2.5J/g^0C}+93.0^0C

         = 227.76°C +93.0°C

          = 320.76°C

∴ we can thereby conclude that the final temperature = 320.76°C                

7 0
3 years ago
If gas is initially at 350L and 500k then changes to 295K what is the new volume
Dimas [21]
<h2>Hello!</h2>

The answer is:

The new volume is equal to 206.5 L.

<h2>Why?</h2>

To solve this problem, we need to assume that the pressure is constant, and use the Charle's Law equation, so, solving we have:

\frac{V_1}{T_1}=\frac{V_2}{T_2}

We are given:

V_1=350L\\T_1=500K\\T_2=295K

Then, using the Charle's Law equation, we have:

\frac{350L}{500K}=\frac{V_2}{295K}

\frac{350L}{500K}=\frac{V_2}{295K}\\\\V_2=\frac{350L}{500K}*295K=206.5L

Hence, we have that the new volume is equal to 206.5 L.

Have a nice day!

5 0
3 years ago
Water boils at 212 degrees
max2010maxim [7]

Answer:

Fahrenheit

Explanation:

Bc i said so LOL JKJK ABAHGTRDSED

5 0
3 years ago
How do you do this?
Alona [7]
It would be Atom 2 because the proton and neutron are both nine and the election is 8 which is -1 to nine <span />
7 0
3 years ago
calculate the volume occupied by 10g of propane gas, under normal conditions of temperature and pressure
andriy [413]

Answer:

5.5 L

Explanation:

First we <u>convert 10 g of propane gas</u> (C₃H₈) to moles, using its <em>molar mass</em>:

  • 10 g ÷ 44 g/mol = 0.23 mol

Then we <u>use the PV=nRT formula</u>, where:

  • P = 1 atm & T = 293 K (This are normal conditions of T and P)
  • n = 0.23 mol
  • R = 0.082 atm·L·mol⁻¹·K⁻¹
  • V = ?

1 atm * V = 0.23 mol * 0.082 atm·L·mol⁻¹·K⁻¹ * 293 K

  • V = 5.5 L
3 0
3 years ago
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