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Gekata [30.6K]
3 years ago
15

How can you find the charge on po4​

Chemistry
2 answers:
Anvisha [2.4K]3 years ago
7 0

Answer:

Answer:

Explanation:

I hope it's helpful!

Kamila [148]3 years ago
4 0
In all of these the P has a formal charge of 0 and one oxygen is also 0.
...
PO43- Step 8.
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A single covalent bond involves two atoms sharing one pair of electrons.<br><br> True<br> False
adelina 88 [10]
True

Explanation: Covalent bonds occur when electrons are shared between two atoms. A single covalent bond is when only one pair of electrons is shared between atoms.

4 0
3 years ago
Read 2 more answers
1. An atom of an elephant has atomic mass of 39 and has number of neutrons as 20 calculate protons number.
r-ruslan [8.4K]

Answer:

See Explanation

Explanation:

1) Let us recall that;

Mass number = number of protons + number of neutrons

Where;

Number of protons = ?

Number of neutrons =20

Mass number =39

Number of protons = mass number - number of neutrons

Mass of protons =39 - 20= 19 protons

2) Atomic number = 13

Number of neutrons =14

Then the mass number =13 +14 =27

The symbol of the element is 27X13

The number of protons= the number of electrons =13

The atomic number is the same as the proton number =13

5 0
3 years ago
Galaxies that are moving away from Earth are
sweet [91]

Answer: (D) red shifted

6 0
3 years ago
Scientists who always report their observations and results truthfully possess the attitude of
Firdavs [7]
The scientific method use, and thus going "for, the idea of following the ideal of scientific evidence.
4 0
3 years ago
Copper crystallizes in a face centered cubic lattice. if the edge of the unit cell is 351 pm what is the radius of the copper at
Troyanec [42]
Density of unit cell is mathematically expressed as
D = \frac{\text{Z  X Atomic Weight}}{\text{Avagadro's Number X a^3}}

where, Z = number of atoms/unit cell = 4 (For FCC structure)
Atomic weight of Cu = 63.5 g
a = edge length = 351 pm = 351 X 10^-10 cm
Avagadro's number = 6.023 X 10^23

∴ Density of unit cell = \frac{4X63.5}{6.023X10^2^3X(351X10^-^1^0)^3}
                                 = 9.752 g/cm3

Now, for FCC structure a = √8 r
where r = radius of Cu
∴ r = a/√8 = (351 X 10^-10)/√8 = 1.24 X 10^-8 cm = 124 pm
8 0
3 years ago
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