Answer:
0.962 atm.
97.4 kPa.
731 torr.
14.1 psi.
97,434.6 Pa.
Explanation:
Hello.
In this case, given the available factors equaling 1 atm of pressure, each required pressure turns out:
- Atmospheres: 1 atm = 760 mmHg:
- Kilopascals:: 101.3 kPa = 760 mmHg:
- Torrs: 760 torr = 760 mmHg:
- Pounds per square inch: 14.69 psi = 760 mmHg:
- Pascals: 101300 Pa = 760 mmHg:
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Answer:
A sample of an ideal gas has a volume of 2.21 L at 279 K and 1.01 atm. Calculate the pressure when the volume is 1.23 L and the temperature is 299 K.
You need to apply the ideal gas law PV=nRT
You have the pressure, P=1.01 atm
you have the volume, V = 2.21 L
The ideal gas constant R= 0.08205 L. atm/ mole.K at 273 K
find n = PV/RT = (1.01 atm x 2.21 L / 0.08205 L.atm/ mole.K x 273 K)
n= 0.1 mole, Now find the pressure for n=0.1 mole, T= 299K and
L=1.23 L
P=nRT/V= 0.1mole x 0.08205 (L.atm/ mole.K x 299 k)/ 1.23 L
= 1.994 atm
Explanation:
Answer:
Explanation:
Hello there!
In this case, according to the given information it turns out possible for us to realize that one mole of the given compound, Mg(ClO₄)₂, has one mole of Mg, two moles of Cl and eight moles of O; thus, we proceed as follows:
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Answer:
Hexaflourine Pentaiodide?
Explanation:
f = flourine (6 = hexa)
i = iodine (5 = penta) + ide
Answer:
cinnamic acid - 150 mg
cis-stilbene - 100 μL
trans- stilbene - 100 mg
pyridinium tribromide - 200-385 mg
For this data:
moles of cinnamic acid = 0.150 g/148.16 g/mol = 0.001 mols
Theoretical mass of dibromoproduct formed = 0.001 mol x 307.97 g/mol = 0.312 g
cis-stilbene (100 ul = 0.1 ml)
moles of cis-stilbene = 0.1 ml x 1.01 g/mol/180.25 g/mol = 0.00056 mols
Theoretical mass of dibromoproduct formed = 0.00056 mol x 340.05 g/mol = 0.19 g
trans-stilbene
moles of tran-stilbene = 0.1 g/180.25 g/mol = 0.00055 mols
Theoretical mass of dibromoproduct formed = 0.00055 mol x 340.05 g/mol = 0.19 g
Explanation:
