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gizmo_the_mogwai [7]
3 years ago
8

How many moles of gas are contained in a rigid, sealed, 5 L container, at 1 atm of pressure and 22 oc?

Chemistry
1 answer:
Ugo [173]3 years ago
7 0
The ideal gas law is: PV=nRT
Pressure
Volume
n= moles
R= gas constant
Temperature in Kelvin
(Degrees in celsius +273)
n= PV/RT
(1.00atm)(5.00L)/(.08026)(295K)= .207mol of gas
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1. How many ATOMS of boron are present in 2.20 moles of boron trifluoride ? atoms of boron.
maks197457 [2]

Answer:

1. How many ATOMS of boron are present in 2.20 moles of boron trifluoride? atoms of boron.

2. How many MOLES of fluorine are present in  of boron trifluoride? moles of fluorine.​

Explanation:

The molecular formula of boron trifluoride is BF_3.

So, one mole of boron trifluoride has one mole of boron atoms.

1. The number of boron atoms in 2.20 moles of boron trifluoride is 2.20 moles.

The number of atoms in 2.20 moles of boron is:

One mole of boron has ---- 6.023x10^2^3 atoms.

Then, 2.20 moles of boron has

-=2.20 mol. x 6.023 x 10^2^3 atoms /1 mol\\=13.25x10^2^3 atoms

2. Calculate the number of moles of BF3 in 5.35*1022 molecules.

(5.35x10^2^2 molecules/6.023x10^2^3)x 1mol\\=0.0888mol

One mole of boron trifluoride has three moles of fluorine atoms.

Hence, 0.0888moles of BF3 has 3x0.0888mol of fluorine atoms.

=0.266mol of fluorine atoms.

5 0
2 years ago
7. What must you do before you make a hypothesis?
almond37 [142]

Answer:

Ask a question?

Explanation:

look at the scientific method uwu

6 0
3 years ago
Harriet has an itch in her throat, and she begins to cough. she reaches for a cup of water to drink. which action is carried out
Nikitich [7]
when she reaches for the cup that is somatic nervous system. when she begins to cough
8 0
2 years ago
Aristotle provided a system to (blank) the elements.
Lena [83]

Explanation:

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8 0
2 years ago
General Chemistry fourth edition by McQuarrie, Rock, and Gallogly. University Science Books presented by Macmillan Learning.
Helen [10]

Answer:

3.07 Cal/g

Explanation:

Step 1: Calculate the heat absorbed by the calorimeter

We will use the following expression.

Q = C × ΔT

where,

  • Q: heat absorbed
  • C: heat capacity of the calorimeter (37.60 kJ/K = 37.60 kJ/°C)
  • ΔT: temperature change (2.29 °C)

Q = 37.60 kJ/°C × 2.29 °C = 86.1 kJ

According to the law of conservation of energy, the heat released by the candy has the same magnitude as the heat absorbed by the calorimeter.

Step 2: Convert 86.1 kJ to Cal

We will use the conversion factor 1 Cal = 4.186 kJ.

86.1 kJ × 1 Cal/4.186 kJ = 20.6 Cal

Step 3: Calculate the number of Cal per gram of candy

20.6 Cal/6.70 g = 3.07 Cal/g

3 0
3 years ago
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