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3 years ago

How many liters are in 5.98 moles of nitrogen gas at STP?

Chemistry

1 answer:

Shtirlitz [24]3 years ago

8 0

Answer:

The correct answer is c) 134L

Explanation:

We use the formula PV =nRT. The normal conditions of temperature and pressure are 273K and 1 atm, we use the gas constant = 0, 082 l atm / K mol.

1 atm x V = 5, 98 mol x 0, 082 l atm / K mol x 273 K

V = 5, 98 mol x 0, 082 l atm / K mol x 273 K / 1 atm

V = 133, 86828 l

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How many decigrams are in 4.6 decagrams?

11111nata11111 [884]

Answer:

$\large \boxed{\text{460 dg}}$

Explanation:

1. Convert decagrams to grams

$\text{Mass} = \text{4.6 dag} \times\dfrac{\text{10 g}}{\text{1 dag}} = \text{46 g}$

2. Convert grams to decigrams

\ $\text{Mass} = \text{46 g} \times\dfrac{\text{10 g}}{\text{1 dg}} = \text{460 dg}\\\\\large \boxed{\textbf{4.6 dag = 460 dg}}$

3 0

3 years ago

Read 2 more answers

All atoms of the same elements have the same what?

Flura [38]

Answer:

The same number of proteins in their nucleus.

4 0

2 years ago

Read 2 more answers

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 (<em>Sum of: for carbon = 12*12=144; for hydrogen = 1*22=22 and for oxygen = 16*11=176. Final result = 342 grs per mol</em>), 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: <em>Moles solution = moles solute + moles solvent.</em> First we have that the moles of solvent are (remember that the molecular weight of water for this calculus is 18 grs per mol):

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