1.1214 mL will a 0.205-mole sample of He occupy at 3.00 atm and 200 K.
<h3>What is an ideal gas equation?</h3>
The ideal gas law (PV = nRT) relates the macroscopic properties of ideal gases. An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).
Using equation PV=nRT, where n is the moles and R is the gas constant. Then divide the given mass by the number of moles to get molar mass.
Given data:
P= 3.00 atm
V= ?
n=0.205 mole
R= 
T=200 K
Putting value in the given equation:
V= 1.1214 mL
Learn more about the ideal gas here:
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Answer:
The enthalpy change during the reaction is -199. kJ/mol.
Explanation:
Mass of solution = m
Volume of solution = 100.0 mL
Density of solution = d = 1.00 g/mL
First we have to calculate the heat gained by the solution in coffee-cup calorimeter.
where,
m = mass of solution = 100 g
q = heat gained = ?
c = specific heat = 
= final temperature = 
= initial temperature = 
Now put all the given values in the above formula, we get:
Now we have to calculate the enthalpy change during the reaction.
where,
= enthalpy change = ?
q = heat gained = 2.242 kJ
n = number of moles fructose = 
Therefore, the enthalpy change during the reaction is -199. kJ/mol.
Answer: pOH = 4.68
Explanation:
pOH = 14 - pH
pH = - Log [H+]
= - Log [4.8 x 10^-10]
= -(-9.32)
pH =+9.32
Therefore, pOH= 14 - 9.32
= 4.68
Answer:
a) 6312.12 lbm
b) D = 11.843 inch
Explanation:
GIVEN DATA:
Height of person = 1.89 m = 74.41 inch
mass of person m = 89 kg = 196.211 lbm
a) person earth weight =mg
where g is acceleration due to gravity = 32.17 ft/sec^2
b) waist to height ratio = 0.5
circumference of waist =
=37.21 inch
we know that circumference is given as 
D = 11.843 inch
In this compound (Phosgene) the central atom (carbon is Sp² Hybridized).
Sp, Sp² and Sp³ can be calculated very simply by doing three steps,
Step 1:
Assume triple bond and double bond as one bond and assign s or p to it. In this example carbon double bond oxygen is considered once and let suppose it is s. Now we are having our s.
Step 2:
Count lone pair of electron, each lone pair counts for s and p. In this case there is no lone pair of electron on carbon, so not included.
Step 3:
Count single bonds for s and p. As we have already assigned s to the double bond, now one p for one single bond, and other p for the other single bond.
Result:
So, we counted 1 s for double bond, 1 p for one single and other p for second single bond. As a whole we got,
Sp²
Practice:
You can practice for hybridization of Oxygen in this molecule. Oxygen has 2 lone pair of electrons. (Hint: Sp² Hybridization)
