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snow_tiger [21]
3 years ago
6

Given that nitrogen forms three bonds with hydrogen to make

="NH_{3}" alt="NH_{3}" align="absmiddle" class="latex-formula">, how many hydrogen atoms do you think will bond with an atom of phosphorus, which is in the same family as nitrogen? Explain your reasoning.
Chemistry
1 answer:
lbvjy [14]3 years ago
3 0

Answer:

Three hydrogen atoms to form PH₃.

Explanation:

Hello!

In this case, since the elements belonging to the nitrogen family (N, P, As, Sb and Bi) show five valence electrons, because there are five electrons at their outer shell, it is clear that if phosphorous bonds with hydrogen, it is going to require the same amount of oxygen atoms (3) because elements having five valence electrons need 3 bonds in order to attain the octet (5+3=8).

Therefore the compound would be:

PH_3

Which is phosphine.

Best regards!

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D= Sx2
lbvjy [14]

Answer:

2160 times 4 would be the answer

Explanation:

Took it

7 0
2 years ago
Which of the following is a solution? <br> Steel <br> Milk <br> Plastic <br> Water
Anit [1.1K]
Milk. Everything else is a compound/alloy.
4 0
4 years ago
(a) (1)
Elis [28]

Explanation:

The ionization energy of an atom is the amount of energy that is required to remove an electron from a mole of atoms in the gas phase:

M(g)  ®  M+(g)  +  e-

It is possible to remove more electrons from most elements, so this quantity is more precisely known as the first ionization energy, the energy to go from neutral atoms to cations with a 1+ charge.  The second ionization energy is the energy that is required to remove a second electron, to form 2+ cations from 1+ cations:

M+(g)  ®  M2+(g)  +  e-

The third ionization energy is the energy required to form 3+ cations:

M2+(g)  ®  M3+(g)  +  e-

and so on.  Ionization energies are always positive numbers, because energy must be supplied (an endothermic energy change) to separate electrons from atoms.  The second ionization energy is always larger than the first ionization energy, because it requires even more energy to remove an electron from a cation than it is from a neutral atom.

The first ionization energy varies in a predictable way across the periodic table.  The ionization energy decreases from top to bottom in groups, and increases from left to right across a period.  Thus, helium has the largest first ionization energy, while francium has one of the lowest.

From top to bottom in a group, orbitals corresponding to higher values of the principal quantum number (n) are being added, which are on average further away from the nucleus.  Since the outermost electrons are further away, they are less strongly attracted by the nucleus, and are easier to remove, corresponding to a lower value for the first ionization energy.From left to right across a period, more protons are being added to the nucleus, but the number of electrons in the inner, lower-energy shells remains the same.  The valence electrons feel a higher effective nuclear charge — the sum of the charges on the protons in the nucleus and the charges on the inner, core electrons.  The valence electrons are therefore held more tightly, the atom decreases in size (see atomic radius), and it becomes increasingly difficult to remove them, corresponding to a higher value for the first ionization energy.

 

The following charts illustrate the general trends in the first ionization energy:

Dunno kung tama beng pero trysorry kung mali

8 0
3 years ago
A 50-mL beaker only has a scale that measures 10, 20, 30, and 40 mL. What is the uncertainty associated with the 50 mL beaker.
Juliette [100K]
If the uncertainty of a certain measurement instrument is not given, then it is assumed to be equal to half of the least count of that instrument. In this case, the least count is 10 ml, so half of this is 5 ml. Therefore, the graduated cylinder has an uncertainty of +/- 5 ml
8 0
3 years ago
44.8% of a 250. mL acid solution is used in an experiment, what volume of acid (in mL) was used?
postnew [5]

Answer:

112 mL  

Explanation:

The formula for percent by volume is

\text{Percent by volume} = \dfrac{\text{Volume of solute}}{\text{Volume of solution}}\times 100 \, \%

If you have 250 mL of a solution that is 44.8 % v/v,

\begin{array}{rcl}44.8\, \% & = & \dfrac{\text{Volume of solute}}{\text{250 mL}}\times 100 \, \%\\\\44.8 \times \text{ 250 mL} & = & \text{Volume of solute} \times 100\\\text{Volume of solute} & = & \dfrac{44.8 \times 250\text{ mL}}{100}\\\\ & = & \textbf{112 mL}\\\end{array}

7 0
3 years ago
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