Answer:
Alkali metals are highly reactive elements that appear to be silver and they are found in group 1 of the periodic table. It consists of lithium (Li), sodium (Na), potassium (K), rubidium (Rb), cesium (Cs), and francium (Fr). As you go further down the group, the more reactive they are. Those elements all react to water and air, so they must be kept in oil to preserve their state.
Listed below are the inter-molecular bonding that exists in the compound <span>ch3ch2ch2ch2ch2ch2oh.
1. Hydrogen Bonding- The type of bonding that exist between O-H in the compound.
2. Dipole-Dipole Bonding- The type of Bonding that is created when the electronegative draws more electron to its self. Exists between C-O
3. London Dispersion- Created between C-H bonding.</span>
Answer:
The shape of the BF3 molecule is best described as trigonal planar.
Explanation:
The Lewis Structure for BF3 is like this:
_ _
| F | | F |
\ /
B
|
| F |
---
It forms three angles of 120° each. The bonds are in the same planar that's why it is trigonal planar and they are exactly the same.
Boron and Fluorine have 3 covalent bonds, produced by electronic promotion that enables the 2py and 2pz orbitals, leaving an electron to pair in the 2px. So boron will have 3 possible electrons to pair in 2s1, 2px and 2py, remember that electronic configuration for B is 1s2, 2s2, 2p1
By hybridization between the orbitals 2s2 and 2p1, the electrons of F, can joined to make the covalent bond. The new B configuration is 1s2, 2s1, 2px1, 2py1 (these last three, hybrid orbitals)
Answer:
A sample of a gas (5.0 mol) at 1.0 atm is expanded at constant temperature from 10 L to 15 L. The final pressure is 0.67 atm.
Step by Step Explanation?
Boyle's law states that in constant temperature the variation volume of gas is inversely proportional to the applied pressure.
The formula is,
P₁ x V₁ = P₂ × V₂
Where,
P₁ is initial pressure = 1 atm
P2 is final pressure = ? (Not Known)
V₁ is initial volume = 10 L
V₂ is final volume = 15 L
Now put the values in the formula,
\begin{gathered}\rm 1\times 10 = P_2\times 15\\\\\rm P_2 = \frac{10}{15\\} \\\\\rm P_2 = 0.67\end{gathered]
Therefore, the answer is 0.67 atm.
Answer is: <span>
time of reaction, in minutes (data in attachment).
</span>Glucose 6-phosphatase, which is found in mammalian liver cells, is a key enzyme in control of blood glucose levels. The enzyme catalyzes the breakdown of glucose 6-phosphate into glucose and inorganic phosphate.<span> These products are transported out of liver cells into the blood, increasing blood glucose levels. In this exercise, you will graph data from a time-course experiment that measured </span>concentration in the buffer outside isolated liver cells, thus indirectly measuring glucose 6-phosphatase activity inside the cells.
