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Vaselesa [24]
3 years ago
7

A weather balloon has a volume of 750ml when filled with helium at 8c at a pressure of 380torr. What is the new volume of the ba

lloon, where the pressure is 0.2torr and the temperature of 75c

Chemistry
2 answers:
Lubov Fominskaja [6]3 years ago
6 0

Explanation:

Below is an attachment containing the solution

DerKrebs [107]3 years ago
3 0

Answer:

Using general gas laws

P1V1/T1 = P2V2/T2

Where P1=380torr = 380mmHg

Vi = 750ml = 0.75dm3

T1= 8c = 281k

P2=0.2torr= 0.2mmHg

T2= 75c = 348k

V2=?

(380*0.75)/281 = (0.2*V2)/348

V2= 1764.77

Explanation:

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Choose the options below that are true.
Assoli18 [71]

Answer:

The options <u>(A) -</u>The rate law for a given reaction can be determined from a knowledge of the rate-determining step in that reaction's mechanism.  and <u>(C) </u>-The rate laws of bimolecular elementary reactions are second order overall ,<u>is true.</u>

Explanation:

(A) -The rate law can only be calculated from the reaction's slowest or rate-determining phase, according to the first sentence.

(B) -The second statement is not entirely right, since we cannot evaluate an accurate rate law by simply looking at the net equation. It must be decided by experimentation.

(C) -Since there are two reactants, the third statement is correct: most bimolecular reactions are second order overall.

(D)-The fourth argument is incorrect. We must track the rates of and elementary phase that is following the reaction in order to determine the rate.

<u>Therefore , the first and third statement is true.</u>

5 0
2 years ago
A 40.0 mL sample of 0.18 M HCI is titrated with 0.36 M CoHsNH2. Dctermine the pH at these points: At the beginning (before base
Whitepunk [10]

Answer:

at the beginning:

pH = 0.745

Explanation:

HCl is a strong acid, so:

  • HCl + H2O  → H3O+  +  Cl-

       0.18 M             0.18        0.18.....equilibrium

before base is added:

∴ [ H3O+ ] ≅ <em>C </em>HCl = 0.18 M

⇒ pH = - Log [ H3O+ ] = - Log ( 0.18 )

⇒ pH = 0.745

8 0
3 years ago
Calculate the wavelength, in nanometers, of the spectral line produced when an electron in a hydrogen atom undergoes the transit
Lapatulllka [165]

<u>Answer:</u> The wavelength of spectral line is 656 nm

<u>Explanation:</u>

To calculate the wavelength of light, we use Rydberg's Equation:

\frac{1}{\lambda}=R_H\left(\frac{1}{n_f^2}-\frac{1}{n_i^2} \right )

Where,

\lambda = Wavelength of radiation

R_H = Rydberg's Constant  = 1.097\times 10^7m^{-1}

n_f = Final energy level = 2

n_i= Initial energy level = 3

Putting the values in above equation, we get:

\frac{1}{\lambda }=1.097\times 10^7m^{-1}\left(\frac{1}{2^2}-\frac{1}{3^2} \right )\\\\\lambda =\frac{1}{1.524\times 10^6m^{-1}}=6.56\times 10^{-7}m

Converting this into nanometers, we use the conversion factor:

1m=10^9nm

So, 6.56\times 10^{-7}m\times (\frac{10^9nm}{1m})=656nm

Hence, the wavelength of spectral line is 656 nm

6 0
3 years ago
A closed container holds 2.0 moles of CO2 gas at STP. How many moles of oxygen can be placed in a container of the same size at
PilotLPTM [1.2K]

Answer: 2 moles

Explanation:

STP is Standard Temperature and Pressure. That means the pressure is 1.00 atm and the temperature is 273K. Since the oxygen is placed in the same container, we can use Ideal Gas Law to figure out what container the CO₂ used.

Ideal Gas Law: PV=nRT

P=1.00 atm

n=moles

R=0.08206 Latm/Kmol

T=273K

CO₂

V=\frac{nRT}{P}

V=\frac{(2.0 mol)(0.08206Latm/Kmol)(273K)}{1.00atm}

V=44.8L

Since we know that CO₂ has a 44.8 L container, we can use that to find the moles of oxygen.

n=\frac{PV}{RT}

n=\frac{(1.00atm)(44.8L)}{(0.08206Latm/Kmol)(273K)}

n=1.99=2mol

There are 2 mol of oxygen.

5 0
3 years ago
How many grams are in 44.8 liters of nitrogen gas, n2?<br> a) 56g<br> b) 27g<br> c) 36g<br> d) 112g
kkurt [141]

a) 56g

<h3>Calculation:</h3>

At STP,

22.4 L of N₂ = 1 mol

We have given 44.8 L of N₂, therefore,

44.8 L of N₂ = \frac{44.8}{22.4}

                    = 2 mol

We know that,

1 mol of N₂ = 28 g

Hence,

2 mol of N₂ = 28 × 2

                   = 56g

Hence, there are 56 g of N₂ in 44.8 L of nitrogen gas.

Learn more about calculation at STP here:

brainly.com/question/9509278

#SPJ4

3 0
2 years ago
Read 2 more answers
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