Answer:
Diphosphorus pentoxide
Carbon dichloride
BCl3
N2H4
Explanation:
These are all covalent compounds. To name covalent compounds, you add prefixes to the beginning of their names depending on what the subscript is of each element. The prefixes are:
1: Mono
2: Di
3: Tri
4: Tetra
5: Penta
6: Hexa
7: Hepta
8: Octa
9: Nona
10: Deca
For example, since the first one is Phopsphorus with a 2 next to it, you add the prefix Di to it.
If the first element in the compound only has one, meaning no number next to it, you do not say mono. This is why we just say "Carbon" for the second one instead of "Monocarbon."
Finally, you always have to end the second element in the compound with "ide." So, "chlorine" becomes "chloride," "oxygen" becomes "oxide," and so on.
Answer: 12.0 milliliters of 6.50 M HCl ( aq ) are required to react with 2.55 g Zn.
Explanation:
moles =
moles of zinc =
The balanced chemical equation is :

According to stoichiometry:
1 mole of zinc reacts with = 2 moles of HCl
Thus 0.0390 moles of zinc reacts with =
moles of HCl
To calculate the volume for given molarity, we use the equation:
.....(1)
Molarity of
solution = 6.50 M
Volume of solution = ?
Putting values in equation 1, we get:


Thus 12.0 ml of 6.50 M HCl ( aq ) are required to react with 2.55 g Zn
Answer would be A)the atom
Answer:
photoelectric effect, phenomenon in which electrically charged particles are released from or within a material when it absorbs electromagnetic radiation. The effect is often defined as the ejection of electrons from a metal plate when light falls on it.
Answer:
0.7g of HCl
Explanation:
First, let us write a balanced equation for the reaction between HCl and Al(OH)3.
This is illustrated below:
Al(OH)3 + 3HCl —> AlCl3 + 3H2O
Next, let us obtain the masses of Al(OH)3 and HCl that reacted together according to the equation. This can be achieved as shown below:
Molar Mass of Al(OH)3 = 27 + 3(16+1)
= 27 + 3(17) = 27 + 51 = 78g/mol.
Molar Mass of HCl = 1 + 35.5 = 36.5g/mol
Mass of HCl from the balanced equation = 3 x 36.5 = 109.5g
Now we can obtain the mass of HCl that would react with 0.5g of Al(OH)3. This can be achieved as follow:
Al(OH)3 + 3HCl —> AlCl3 + 3H2O
From the equation above,
78g of Al(OH)3 reacted with 109.5g of HCl.
Therefore, 0.5g of Al(OH)3 will react with = (0.5 x 109.5)/78 = 0.7g of HCl