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3 years ago

Which step will decrease the pressure of a gas inside a closed cubical container

2 answers:

7 0

Decreasing the temperature will decrease the pressure of a gas inside a closed cubical container. In a closed container, the Volume (V) is fixed and the no of mole of the gas(n) are also fixed. So, we write the gas equation : PV = nRT Here, V,R and n are constants. So, Pressure is directly proportional to the temperature. Thus, on increasing the temperature, pressure will increase and on decreasing the temperature, the pressure will decrease.

Nadya [2.5K]3 years ago

7 0

decreasing the temperature inside the container

hope this helps!

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What is the name of the element with a valence electron configuration of 4s24p3?

Gnesinka [82]

The element Arsenic, or As in the periodic table, possesses the valence electron configuration of 4s2 4p3. It has an atomic number of 33 and an atomic mass of 74.92, and is classified as a metalloid.

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4 years ago

Two or more atoms chemically bonded together are called

BARSIC [14]

Answer:

A compound is "a substance formed when two or more chemical elements are chemically bonded together." There are two main types of chemical bonds: covalent and ionic.

Explanation:

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3 years ago

A man ran 60.5 km, how far is this in yards?

pickupchik [31]

Answer:

66163.604

Explanation:

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3 years ago

a) Calculatethe molality, m, of an aqueous solution of 1.22 M sucrose, C12H22O11. The density of the solution is 1.12 g/mL.b) Wh

Contact [7]

Answer:

a) 1,74 molal

b) 37,2 %

c) 0,03

Explanation:

We are going to define sucrose as solute, water as solvent and the mix of both, the solution.

Let´s start with the data:

$Molarity = M = \frac{1,22 mol solute}{lts solution}$

We can assume as a calculus base, 1 liter of solution. So, in 1 liter of solution we have 1,22 moles of solute:

$1 lts solution * \frac{1,22 moles solute}{lts solution}=1,22 moles solute$

Knowing that the molality (m) is defined as mol of solute/kgs solvent, we have to calculate the mass of solvent on the solution. Remember our calculus base (1 lts of solution). In 1 lts of solution we have 1120 grams of solution.

$1 lts solution * \frac{1,12 grs solution}{mL solution}*\frac{1000 mL solution}{1 lts solution} = 1120 grs of solution$

With the molecular weight of solute (Sum of: for carbon = 12*12=144; for hydrogen = 1*22=22 and for oxygen = 16*11=176. Final result = 342 grs per mol), we can obtain the mass of solute:

$1,22 mol solute*\frac{342 grs solute}{1 mol solute} = 417,24 grs solute$

Now, the mass of solvent is: mass solvent = mass of solution - mass of solute. So, we have: 1120 - 417,24 = 702,76 grs of solvent = 0,70276 Kgs of solvent

$molality = m = \frac{1,22 mol solute}{0,70276 kgs solvent}= 1,74 molal$

For b) question we have that the mass percent of solute is hte ratio between the mass of solute and the mass of solution. So,

$%(w/w) = \frac{417,24 grs solute}{1120 grs solution} = 37,2%$

For c) question we have that the mole fraction of solute is the ratio between moles of solute and moles of solution. Let's calculate the moles of solution as follows: Moles solution = moles solute + moles solvent. First we have that the moles of solvent are (remember that the molecular weight of water for this calculus is 18 grs per mol):

$702,76 grs solvent*\frac{1 mol solvent}{18 grs solvent} = 39,04 moles solvent$