To find the midpoint M of a line segment with endpoints Q(0,5) and R(2,1), you can use the midpoint formula:
Midpoint M = ((x₁ + x₂) / 2, (y₁ + y₂) / 2)
In this formula:
- (x₁, y₁) are the coordinates of the first endpoint (Q in this case).
- (x₂, y₂) are the coordinates of the second endpoint (R in this case).
Plug in the values:
M = ((0 + 2) / 2, (5 + 1) / 2)
M = (2 / 2, 6 / 2)
M = (1, 3)
So, the midpoint M of the line segment with endpoints Q(0,5) and R(2,1) is (1, 3).
1. Nucleus
2. Neutrons
3. Electrons
4. Protons
Molecular compounds often have lower melting and boiling points. Ionic compounds cannot conduct electricity when they are in a solid form.
Any substance made up of similar molecules with atoms from two or more different chemical elements is referred to as a chemical compound. Atoms from more over 100 specific chemical elements, both in their pure forms and when mixed to form chemical compounds, make up all matter in the universe.
For instance, the atoms that make up carbon differ from those which make it up iron, which differ from the ones that make up gold. Molecular compounds often have lower melting and boiling points. Ionic compounds cannot conduct electricity when they are in a solid form.
Therefore, molecular compounds often have lower melting and boiling points. Ionic compounds cannot conduct electricity when they are in a solid form.
To know more about molecular compounds, here:
#SPJ2
manslaughter
murder
assult
Answer:
assault
Explanassaultation:
b. the number of neutrons.
c. the number of protons and neutrons.
d. simply what element the atom belongs to.
Answer:
C. the number of protons and neutrons
how many moles of O2 is required?
1. 27.79
2. 7.63
3. 8.4
4. 25.48
5. 12.635
6. 21.035
7. 23.8
8. 19.04
9. 22.715
10. 26.775
Answer in units of mol.
Answer:
22.715 moles of oxygen are used
Explanation:
Given data:
Number of moles of ethane = 6.49 mol
Number of moles of O₂ required = ?
Solution:
Chemical equation:
2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O
Now we will compare the moles of oxygen with ethane.
C₂H₆ : O₂
2 : 7
6.49 : 7/2×6.49 = 22.715 mol
Thus, 22.715 moles of oxygen are used.
b. 2H2CO3
c. CO2 2H2O