Answer:
Br
|
Br-P-Br
|
Br
Explanation:
To calculate the valance electrons, look at the periodic table to find the valance electrons for each atom and add them together. P is in column 5A, so it has 5, Br is in column 7A, so it has 7 (multiply by 4 since there are 4 Br atoms to give 28) and there is a 1- charge, so add one more electron. 5+28+1=34, so there are 34 electrons to place. P would be the central atom, so place it in the middle. Place each Br around the P (as shown above) with a a single line connecting it. Each line represents 2 electrons, so 8 total have been place, leaving 26 remaining. Place 6 electrons around each Br (2 on each of the unbonded sides), which leaves 2 electrons remaining. The remaining pair of unbound electrons will be attached to the P between any two Br atoms. Phosphorus doesn't have to follow the octet rule, so it actually ends up with 10 valance electrons.
I would say D, because you need to start with nothing to measure the different sizes as they start to grow. hope this helps!
B.) Helium have <span>two protons and two electrons
Hope this helps!</span>
The mass of hydrazine (N₂H₄) required to produce 96 g of water (H₂O) is 85.4 g (Option C)
<h3>Balanced equation </h3>
N₂H₄ + O₂ —> N₂ + 2H₂O
Molar mass of N₂H₄ = (2×14) + (4×1) = 32 g/mol
Mass of N₂H₄ from the balanced equation = 1 × 32 = 32 g
Molar mass of H₂O = (2×1) + 16 = 18 g/mol
Mass of H₂O from the balanced equation = 2 × 18 = 36 g
SUMMARY
From the balanced equation above,
36 g of H₂O were produced by 32 g of N₂H₄
<h3>How to determine the mass of N₂H₄</h3>
From the balanced equation above,
36 g of H₂O were produced by 32 g of N₂H₄
Therefore,
96 g of H₂O will be produced by = (96 × 32) / 36 = 85.4 g of N₂H₄
Thus, 85.4 g of N₂H₄ is needed for the reaction
Learn more about stoichiometry:
brainly.com/question/14735801
Answer:
.
Explanation:
Electrons are conserved in a chemical equation.
The superscript of 
indicates that each of these ions carries a charge of 
. That corresponds to the shortage of one electron for each 
ion.
Similarly, the superscript 
on each 
ion indicates a shortage of three electrons per such ion.
Assume that the coefficient of (among the reactants) is 
, and that the coefficient of (among the reactants) is 
.
.
There would thus be silver (
) atoms and aluminum (
) atoms on either side of the equation. Hence, the coefficient for 
and 
would be 
and 
, respectively.
.
The ions on the left-hand side of the equation would correspond to the shortage of electrons. On the other hand, the 
ions on the right-hand side of this equation would correspond to the shortage of 
electrons.
Just like atoms, electrons are also conserved in a chemical reaction. Therefore, if the left-hand side has a shortage of electrons, the right-hand side should also be electrons short of being neutral. On the other hand, it is already shown that the right-hand side would have a shortage of electrons. These two expressions should have the same value. Therefore, 
.
The smallest integer and that could satisfy this relation are 
and 
. The equation becomes:
.
