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IgorC [24]
3 years ago
9

A balloon contains 269.7 L of helium at 6.12ºC and 1.00 atm. What is the temperature (in ºC) of the gas if the volume has increa

sed by 320.4 L at the same pressure?
Chemistry
1 answer:
MatroZZZ [7]3 years ago
3 0

Answer:

T2= 7.3°C

Explanation:

To solve this problem we will use Charles law equation i.e,

V1/T1 = V2/T2    

Given data

V1 = 269.7 L

T1 = 6.12 °C

V2= 320.4 L

T2=?

Solution:

Now we will put the values in equation

269.7 L / 6.12°C  = 320.4 L / T2

T2=  320.4 L × 6.12°C/ 269.7 L

T2= 1960.85 °C. L /269.7 L

T2= 7.3°C

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Atoms of hydrogen can become helium atoms during a chemical reaction.<br> True<br> False
marishachu [46]

Answer:

false

in chemical reaction element doesnot change into another

Explanation:

3 0
3 years ago
Read 2 more answers
2.40 g of NH4Cl is added to 19.4 g of water. Calculate the molality of the solution.
Sholpan [36]
Molality is the moles of solute per kg of solvent.

Moles of NH₄Cl = 2.4 / (14 + 4 x 1 + 35.5)
= 0.0448 mole

Molality = 0.0448 / (19.4 / 1000)
= 2.31 m
3 0
3 years ago
At 20.°C, a 1.2-gram sample of Mg ribbon reacts rapidly with 10.0 milliliters of 1.0 M HCl(aq). Which change in conditions would
Pavlova-9 [17]

Answer: (2) decreasing the concentration of HCl(aq) to 0.1 M

Explanation: Rate of a reaction depends on following factors:

1. Size of the solute particles: If the reactant molecules are present in smaller size, surface of particles and decreasing the size increases the surface area of the solute particles. Hence, increasing the rate of a reaction.  

2. Reactant concentration: The rate of the reaction is directly proportional to the concentration of reactants.

3. Temperature: Increasing the temperature increases the energy of the molecules and thus more molecules can react to give products and rate increases.

(1) Increasing the initial temperature to 25°C will increase the reaction rate.

(2) Decreasing the concentration of HCl(aq) to 0.1 M will decrease the reaction rate due to lesser concentration.

(3) Using 1.2 g of powdered Mg will increase the reaction rate due to large surface area.

(4) Using 2.4 g of Mg ribbon will increase the reaction rate due to high concentration of reactants.

7 0
3 years ago
Read 2 more answers
How many ug of nickel (Ni) are required to make 25.00 nanoliters of a 1.25 mol/L solution? Be sure to report your answer to the
devlian [24]

volume of Ni = 25 nL = 25 x 10⁻⁹ L

mol Ni = 25 x 10⁻⁹ L x 1.25 mol/L = 3.125 x 10⁻⁸

mass = mol x Ar Ni

mass = 3.125 x 10⁻⁸ x 59 g/mol

mass = 1.84 x 10⁻⁶ g = 1.84 μg

4 0
2 years ago
Read 2 more answers
THIS IS A THREE PART QUESTION IF YOU CAN HELP IT WOULD BE REALLY APPRECIATED SO I DONT FAIL.
Nostrana [21]

Answer:

1. B = 1.13M

2. A. 8.46%

3. D = 0.0199

Explanation:

1. Molarity of a a solution = number of moles of solute/ volume of solution in L

Number of moles of solute = mass of solute/molar mass of solute

Molar mass of NaC₂H₃O₂ = 82 g/mol, mass of NaC₂H₃O₂ = 245 g

Number of moles of NaC₂H₃O₂ = 245 g / 82 g/mol = 2.988 moles

Molarity of solution = 2.988 mols/ 2.65 L = 1.13 M

2. Percentage by mass of a substance = mass of substance /mass of solution × 100%

Mass of 2.65 L of water = density × volume

Density of water = 1 Kg/L = 1000 g/L; volume of waterb= 2.65 L

Mass of water = 1000 g/L × 2.65 L = 2650 g

Mass of solution = mass of water + mass of solute = 245 + 2650 =2895 g

Percentage by mass of NaC₂H₃O₂ = 245/2895 × 100% = 8.46%

3. Mole fraction of NaC₂H₃O₂ = moles of NaC₂H₃O₂/moles of solution

Moles of water = mass /molar mass

Mass of water = 2650 g; molar mass of water = 18 g/mol

Moles of water = 2650 g / 18 g/mol = 147.222 moles

Moles of solution = moles of solute + moles of water = 147.222 + 2.988 = 150.21

Moles of NaC₂H₃O₂ =2.988 moles

Moles fraction of NaC₂H₃O₂ =2.988/150.21 = 0.0199

6 0
3 years ago
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