Question 1 :
V1/T1 = V2/T2
3.0L/273K = V2/373K
To get the value of Z, cross multiply
3.0L x 373K = 273K x V2
1119 = 273V2
Divide both sides by 273
1119/273 = 273V2/273
4.10L = V2
The new volume is 4.10 liters
Question 2 :
P1/T1 = P2 /T2
P1 = 880 kPA= 880 *10^3 Pa
T1 = 250 K
T2 = 303 K
P2 =?
Substituting for P2
P2 = P1 T2/ T1
P2 = 880 kPa * 303 / 250
P2 = 266,640 kPa/ 250
P2 = 1066.56 kPa.
The new pressure of the gas is 1066.56 kPa
Question 3 :
Given that:
Volume of gas V = 4.80L
(since 1 liter = 1dm3
4.80L = 4.80dm3)
Temperature T = 62°C
Convert Celsius to Kelvin
(62°C + 273 = 335K)
Pressure P = 2.9 atm
Number of moles of gas N = ?
Apply ideal gas equation
pV = nRT
2.9atm x 4.8dm3 = n x (0.0082 atm dm3 K-1 mol-1 x 335K)
13.92 atm dm3 = nx 2.747 atm dm3 mol-1
n = 13.92/2.747
n = 5.08 moles
There are 5.08 moles of gas contained in the sample
Question 4 :
Volume of gas V = 3.47L
(since 1 liter = 1dm3
3.47L = 3.47dm3)
Temperature T = 85.0°C
Convert Celsius to Kelvin
(85.0°C + 273 = 358K)
Pressure P = ?
Number of moles of gas N = 0.100 mole
Apply ideal gas equation
pV = nRT
p x 3.47dm3 = 0.10 x (0.0082 atm dm3 K-1 mol-1 x 358K)
p x 3.47dm3 = 0.29 atm dm3
p = (0.29 atm dm3 / 3.47 dm3)
p = 0.085 atm
If 1 atm = 760 mm Hg
0.085atm = 0.085 x 760
= 64.6 mm Hg
The pressure of the gas is 64.6 mm hg
Body elements and trace fossils I believe
Answer: The correct answer is Option 1.
Explanation:
Control group is the group used in the experiment by the researchers in which no change in the variable is done. It is then set as a benchmark for the groups which are being tested.
We are given 4 groups which are being experimented for the flotation of an egg in water. As in Cup 1, there is no addition of salt and hence there is no change in the variables. So, this is set as a benchmark fro the cups which are further used in the experiment conducted.
Hence, the correct answer is Option 1.
Amino Acids undergo a kind of acid base reaction within a molecule. The
Amino group acts as a base and accepts proton from
Carboxylic functional group resulting in formation of
Ammonium with positive charge and a negative charge on
Carboxylate. Such molecules containing both positive and negative charges are called as
Zwittor Ions.
Phenylalanine both in Un-ionized and Ionized form is shown below,
Answer:
Percent yield of PI3 = 95.4%
Explanation:
This is the reaction:
2P (s) + 3I2 (g) > 2PI3 (g)
Let's determine the moles of iodine that has reacted.
58.6 g / 253.8 g/mol = 0.231 mol
Ratio is 3:2. Let's make a rule of three to state the moles produced at 100 % yield reaction.
3 moles of I2 react to make 2 moles of PI3
0.231 moles of I2 would make (0.231 .2) / 3 = 0.154 moles of PI3
As we have produced 0.147 moles let's determine the percent yield.
(Yield produced / Theoretical yield) . 100 > (0.147 / 0.154) . 100 = 95.4%
