Answer:
1.17 grams
Explanation:
Let's consider the balanced equation for the combustion of ethylene.
C₂H₄(g) + 3 O₂(g) → 2 CO₂(g) + 2 H₂O(l)
We can establish the following relations:
- 1411 kJ are released (-1411 kJ) when 1 mole of C₂H₄ burns.
- The molar mass of C₂H₄ is 28.05 g/mol.
The grams of C₂H₄ burned to give 59.0 kJ of heat (q = -59.0 kJ) is:
Using ideal gas equation,
P\times V=n\times R\times T
Here,
P denotes pressure
V denotes volume
n denotes number of moles of gas
R denotes gas constant
T denotes temperature
The values at STP will be:
P=100 kPa
T=293 K
R=8.314472 L kPa K⁻¹ mol⁻¹
Number of moles of gas=3.43 mole
Putting all the values in the above equation,
V=83.55 L
So the volume will be 83.55 L.
83.55 L of radon gas would be in 3.43 moles at room temperature and pressure (293 K and 100 kPa).
Answer:
5' RG GWCCY 3'
3' YCCWG GR 5'
Explanation:
The enzyme PpuMI is a restriction endonuclease enzyme, it has a specific recognition site where it cut the DNA. The source of the enzyme is from an E. coli strain that carries the PpuMI gene from Pseudomonas putida (R. Morgan).
The enzyme PpuMI recognizes specific sequence with palindrome arrangement. It target the sequence 5' RGGWCCY 3'
target Sequence: 5' RGGWCCY 3'
3' YCCWGGR 5'
The enzyme cleavage point is at:
5' RG^GWCCY 3'
3' YCCWG^GR 5'
The product of the cleavage will give a sticky end Cleavage:
5' RG GWCCY 3'
3' YCCWG GR 5'
Note: R stands for purines (adenine and guanine). Y stands for pyrimidines (cytosine, thymine, and uracil). And W represents adenine or thymine.
<h3>
Answer:</h3>
112.08 mL
<h3>
Explanation:</h3>
From the question we are given;
- Initial volume, V1 = 100.0 mL
- Initial temperature, T1 = 225°C, but K = °C + 273.15
thus, T1 = 498.15 K
- Initial pressure, P1 = 1.80 atm
- Final temperature , T2 = -25°C
= 248.15 K
- Final pressure, P2 = 0.80 atm
We are required to calculate the new volume of the gases;
- According to the combined gas law equation;
Rearranging the formula;
Therefore;
Therefore, the new volume of the gas is 112.08 mL
The first step is to calculate the molarity of each compound:
final volume of solution = 157 + 139 = 296 mL
molarity of <span>nac2h3o2 = (157 x 0.35) / 296 = 0.1856 molar
molarity of </span><span>hc2h3o2 = (139 x 0.46) / 296 = 0.216 molar
Then, we calculate the pH as follows:
pKa of acetic acid = -log(</span><span>1.75 × 10^-5) = 4.7569
pH = pKa + </span><span> log ([salt] / [acid])
= </span>4.7569 + log(0.1856 / 0.216)
= 4.691
