Answer:
Option (3).
Explanation:
Transfer RNA or t RNA's is one of the type of the RNA molecule necessary for the synthesis of protein translation. This RNA works as a adaptor molecule and contains 70 to 90 nucleotide in length.
Transfer RNA contains an anti codon loop. This anti codon loop has the ability to recognize an mRNA codon and result in the formation of protein product.
Thus, the correct answer is option (3).
We first need to find the number of moles of gas in the container
PV = nRT
where;
P - pressure - 2.87 atm x 101 325 Pa/atm = 290 802.75 Pa
V - volume - 5.29 x 10⁻³ m³
n - number of moles
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature - 230 K
substituting these values in the equation
290 802.75 Pa x 5.29 x 10⁻³ m³ = n x 8.314 Jmol⁻¹K⁻¹ x 230 K
n = 0.804 mol
the molar mass = mass present / number of moles
molar mass of gas = 56.75 g / 0.804 mol
therefore molar mass is 70.6 g/mol
Answer: The correct answer is A. Hydrogen and chlorine need to be balanced. There's an equal amount of magnesium on each side.
<span>361.4 pm is the length of the edge of the unit cell.
First, let's calculate the average volume each atom is taking. Start with calculating how many moles of copper we have in a cubic centimeter by looking up the atomic weight.
Atomic weight copper = 63.546
Now divide the mass by the atomic weight, getting
8.94 g / 63.546 g/mol = 0.140685488 mol
And multiply by Avogadro's number to get the number of atoms:
0.140685488 * 6.022140857x10^23 = 8.472278233x10^22
Now examine the face-centered cubic unit cell to see how many atoms worth of space it consumes. There is 1 atom at each of the 8 corners and each of those atoms is shared between 8 unit cells for for a space consumption of 8/8 = 1 atom. And there are 6 faces, each with an atom in the center, each of which is shared between 2 unit cells for a space consumption of 6/2 = 3 atoms. So each unit cell consumes as much space as 4 atoms. Let's divide the number of atoms in that cubic centimeter by 4 to determine the number of unit cells in that volume.
8.472278233x10^22 / 4 = 2.118069558x10^22
Now calculate the volume each unit cell occupies.
1 cm^3 / 2.118069558x10^22 = 4.721280262x10^-23 cm^3
Let's get the cube root to get the length of an edge.
(4.721280262x10^-23 cm^3)^(1/3) = 3.61426x10^-08 cm
Now let's convert from cm to pm.
3.61426x10^-08 cm / 100 cm/m * 1x10^12 pm/m = 361.4 pm
Doing an independent search for the Crystallographic Features of Copper, I see that the Lattice Parameter for copper at at 293 K is 3.6147 x 10^-10 m which is in very close agreement with the calculated amount above. And since metals expand and contract with heat and cold, I assume the slight difference in values is due to the density figure given being determined at a temperature lower than 293 K.</span>
