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ale4655 [162]
3 years ago
10

You’re at the zoo and have a big red 1.80 L helium balloon. The barometric pressure today is 785 mmHg. Then you hear the roar of

a lion. Startled, you accidentally release the balloon. It flies away. By the time it reaches the clouds, the atmospheric pressure that high is only 3.00 atmospheres. What would the volume of the balloon up there? (Temperature is constant)
Chemistry
2 answers:
Marianna [84]3 years ago
8 0

Answer:

The volume of the balloon up there will be 0.618 L

Explanation:

Given:

V₁ = 1.8 L

P₁ = 785 mmHg = 1.03 atm

P₂ = 3 atm

Question: What would the volume of the balloon up there, V₂ = ?

According the ideal gas law:

PV = nRT

However, in this case, the number of moles of helium balloon is constant. The temperature will also be assumed to be constant. Therefore, the expression of the ideal gases is as follows:

P₁V₁ = P₂V₂

Solving for V₂

V_{2} =\frac{P_{1}V_{1}  }{P_{2} } =\frac{1.03*1.8}{3} =0.618L

nadezda [96]3 years ago
5 0

Answer:

0.62 L

Explanation:

Step 1:

Data obtained from the question.

Initial Volume (V1) = 1.80 L

Initial pressure (P1) = 785 mmHg

Final pressure (P2) = 3.00 atm

Final volume (V2) =?

Step 2:

Conversion of the pressure in mmHg to atm.

It is important to express the initial and the final pressure in the same unit. Either express both in atm or in mmHg. What ever the case is, we'll still arrive at same answer. Here, we shall be converting from mmHg to atm. This is illustrated below:

760mmHg = 1atm

Therefore, 785 mmHg = 785/760 = 1.03 atm

Step 3:

Determination of the final volume. This is illustrated below.

We shall be applying the Boyle's law equation since the temperature is constant.

P1V1 = P2V2

Initial Volume (V1) = 1.80 L

Initial pressure (P1) = 1.03 atm

Final pressure (P2) = 3.00 atm

Final volume (V2) =?

P1V1 = P2V2

1.03 x 1.8 = 3 x V2

Divide both side by 3

V2 = (1.03 x 1.8) /3

V2 = 0.62 L

Therefore, the new volume of the balloon is 0.62 L

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2. Fe(s) + O2(g) → Fe3O4(s) a. When 13.54 g of O2 is mixed with 12.21 g of Fe, which is the limiting reactant? b. What mass in g
julsineya [31]

The percent yield shows the extent to which the reactants are converetd into products . The limiting reactant is used up in the reaction.

<h3>What is a limiting reactant?</h3>

A limiting reactant is the reactant that is in the least amount in the system.  Now;

Number of moles of Fe =  12.21 g/56 g/mol = 0.22 moles

Number of moles of O2 = 13.54 g/32 g/mol = 0.42 moles

Balanced reaction equation;

3Fe(s) + 2 O2(g) = Fe3O4(s)

If 3 moles of Fe reacts with 2 mole of O2

0.22 moles of Fe reacts with  0.22 moles * 2 mole/3 moles = 0.15 moles

Hence, Fe is the limiting reactant

If 3 mole of Fe produces 1 mole of  Fe3O4(s)

 0.22 moles  of O2   produces   0.22 moles * 1 mole/3 moles of  Fe3O4(s) = 0.1073 moles

Mass of  Fe3O4(s) =0.1073 moles  * 232 g/mol =16.9 g

Number of moles of excess reactant = 0.42 moles - 0.15 moles = 0.27 moles

Mass of excess reactant = 0.27 moles * 32 g/mol = 8.64 g

percent yield = 15.88 g/16.9 g * 100/1

= 93.4%

Learn more about percent yiled: brainly.com/question/13463225

8 0
2 years ago
calcualte pressure at STP in 10.0 L vessel after reaction of 1.0 L hydrochloride acid (concentration 35% and density 1.28 g/cub.
sattari [20]

Answer:

The pressure in the vessel is 13,3 atm.

Explanation:

The reaction that occurs in vessel (where limestone is 96% of CaCO₃) is:

2 HCl (aq)+ CaCO₃ (s) → CaCl₂(aq)+ H₂O(l)+ CO₂(g)

The increase in the pressure of the vessel after the reaction is by formation of a gas (CO₂). So we have to find the produced moles of this gas and apply the gas ideal law to find the pressure.

We have to find the limit reactant, to do so, we have to calculate the moles of each reactant in the reaction, the one that have the less moles will be the limit reactant:

HCl:

1,0L × (35/100) × (1000 cm³/1L) × (1,28 g/ 1cm³) × (1mol HCl/ 36,46 g) ÷ 2mol

(Concentration)      (L to cm³)         (cm³ to g)      (g to mol)  (moles of reaction)

moles of HCl= 6,14 mol

CaCo₃:

   1,0 kg     ×       (96/100)                ×   (1000 g/1kg) × (1 mol/100,09g)

(Limestone) (CaCo₃ in limestone)          (kg to g)            (g to mol)

moles of CaCo₃= 9,59 mol

So, <em>reactant limit is HCl</em>

This reaction have a yield of 97%. So, the CO₂ moles are:

6,14 mol × 97÷ = 5,96 mol CO₂

The ideal gas formula to obtain pressure is:

P = nRT/V

Where: n = 5,96mol; R= 0,082 atm×L/mol×K; T = 273,15 (until STP conditions) and V= 10,0 L

Replacing this values in the equation the pressure is

P = 13,3 atm

I hope it helps!

7 0
3 years ago
I still need a bit of help.
Ostrovityanka [42]
H. The atom would no longer be aluminum
If you added a proton to an atom of aluminum it would become a silicone ion
5 0
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In this photograph what rock formation is presented by the blue lines?
Effectus [21]

It is representing syncline rock formation

there are two rock formation based on fold formation : syncline and anticline

In syncline rock formation there is fold like trough unlike anticline where it is like crust

In syncline the fold is downward as shown in photo and the new rock is outer fold and old at inner side

3 0
3 years ago
Read 2 more answers
a sample of 3.00 g of so2 (g)originally in a 5.00 L vesselat 21 degee Celsius is transferred to a 10.0 L vessel at 26 degree Cel
eimsori [14]

Answer:

1) The partial pressure of SO₂ gas in the larger container = 0.115 atm.

2) The partial pressure of N₂ gas in the larger container = 0.206 atm.

3) The total pressure in the vessel = 0.321 atm.

Explanation:

  • To calculate the partial pressure of each gas, we can use the general law of ideal gas: PV = nRT.

where, P is the partial pressure of the gas in atm,

V is the volume of the vessel in L,

n is the no. of moles of the gas,

R is the general gas constant (R = 0.082 L.atm/mol.K),

T is the temperature of the gas in K.

<u><em>1) What is the partial pressure of SO₂ gas in the larger container?</em></u>

<em>∵ P = nRT/V.</em>

n = mass/molar mass = (3.0 g)/(64.066 g/mol) = 0.047 mol.

R = 0.082 L.atm/mol.K.

T = 26 °C + 273.15 = 299.15 K.

V = 10.0 L. (The volume of the new container)

∴ P = nRT/V = (0.047 mol)(0.082 L.atm/mol.K)(299.15 K)/(10.0 L) = 0.115 atm.

<u><em>2) What is the partial pressure of N₂ gas in the larger container?</em></u>

<em>∵ P = nRT/V.</em>

n = mass/molar mass = (2.35 g)/(28.0 g/mol) = 0.084 mol.

R = 0.082 L.atm/mol.K.

T = 26 °C + 273.15 = 299.15 K.

V = 10.0 L. (The volume of the new container)

∴ P = nRT/V = (0.084 mol)(0.082 L.atm/mol.K)(299.15 K)/(10.0 L) = 0.206 atm.

<u><em>3) What is the total pressure in the vessel?</em></u>

  • According to Dalton's law the total pressure exerted is equal to the sum of the partial pressures of the individual gases.

<em>∵ The total pressure in the vessel = the partial pressure of SO₂ + the partial pressure of N₂.</em>

∴ The total pressure in the vessel = 0.115 + 0.206 = 0.321 atm.

5 0
3 years ago
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