Answer:
29.98kg
Explanation:
12.0 gallons * (3.78541178 liters/gallon) * (1000 mL/liter) * (0.66 g/mL) * (1 kg/1000 g) = 29.98 kg
This is an incomplete question, the given sketch is shown below.
Answer : The name of given unit cell is, FCC (face-centered cubic unit cell)
Explanation :
Unit cell : It is defined as the smallest 3-dimensional portion of a complete space lattice which when repeated over the and again in different directions produces the complete space lattice.
There are three types of unit cell.
- SCC (simple-centered cubic unit cell)
- BCC (body-centered cubic unit cell)
- FCC (face-centered cubic unit cell)
In SCC, the atoms are arranged at the corners.
The number of atoms of unit cell = Z = 1
In BCC, the atoms are arranged at the corners and the body center.
The number of atoms of unit cell = Z = 2
The given unit cell is, FCC because the atoms are arranged at the corners and the center of the 6 faces.
The number of atoms of unit cell = Z = 4
Thus, the name of given unit cell is, FCC (face-centered cubic unit cell)
Answer:
Option A
Explanation:
Number of millimoles of Na3PO4 = 1 × 100 = 100
Number of millimoles of AgNO3 = 1 × 100 = 100
When 1 mole of Na3PO4 is dissociated we get 3 moles of sodium ions and 1 mole of phosphate ion
When 1 mole of AgNO3 is dissociated, we get 1 mole of Ag+ and 1 mole of NO3-
As Ag+ concentration is negligible, the dissociated Ag+ ion must have form the precipitate with phosphate ion and as number of moles of Ag+ and phosphate ion are same, therefore the concentration of phosphate ion must be negligible
Here as 100 millimoles of Na3PO4 is there, we get 300 millimoles of Na+ and 100 millimoles of PO43-
And as 100 millimoles of AgNO3 is there, we get 100 millimoles of Ag+ and 100 millimoles of NO3-
∴ Increasing order of concentration will be PO43- < NO3- < Na+
The rate constant, k, for the decomposition reaction : k = 0.0124 / days
<h3>Further explanation</h3>
Given
The half-life of 56 days
Required
The rate constant, k
Solution
For first-order, rate law : ln[A]=−kt+ln[A]o
The half-life : the time required to reduce to half of its initial value.
The half life :
t1/2 = (ln 2) / k
k = (ln 2) / t1/2
k = 0.693 / 56 days
k = 0.0124 / days
The testing of the acidities of the compounds to determine the combination that will keep the drug intact before it reaches the intestines is a part of EXPERIMENTATION process. The experimentation is done in order to prove or answer different queries that were raised during the hypothesis making.
