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marissa [1.9K]
3 years ago
6

Atoms containing the same number of protons and different numbers of neutrons are _____. (1 point)

Chemistry
2 answers:
Scilla [17]3 years ago
4 0

Answer:

isotopes

Explanation:

The atoms of an element that have same number of protons but different number of neutrons are called as the isotopes of that element.

For example: _{1}^{1}(\textrm{H}\text{protium})\text{and}_{1}^{2}\textrm{H}(\text{deuterium}) are the isotopes of Hydrogen

Ions are the charged species of an atom that are formed by the gain or loose of electrons.

For example: \text{Na}^{+}

Neutrons are the sub-atomic particles found inside the nucleus of an atom. They carry no charge.

Compounds are the substances that are formed when two or more atoms combine in a definite proportion.

For example: NaCl

Vadim26 [7]3 years ago
3 0
I believe that it could be isotopes
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Len [333]

The hydrogen deficiency index( HDI) of strigol is = 10

<h3>How to calculate HDI:</h3>

The hydrogen deficiency index is used to measure the number of degree of unsaturation of an organic compound.

Strigol is an example of an organic compound because it contains carbons and hydrogen.

To calculate the HDI using the molecular formula given (C19H20O6) the formula for HDI is used which is:

hdi =  \frac{1}{2} (2c + 2 + n - h - x)

where C = number of carbon atoms = 19

n= number of nitrogen atoms = 0

h= number of hydrogen atoms = 20

X = number of halogen atoms = 0

Note that oxygen was not considered because it forms two bonds and has no impact.

There for HDI =

\frac{1}{2} (2 \times 19 + 2 + 0 - 20 - 0)

HDI=

\frac{1}{2} (40 - 20)

HDI =

\frac{1}{2}  \times 20

HDI = 10

Therefore, the hydrogen deficiency index of strigol is = 10

Learn more about unsaturated compounds here:

brainly.com/question/490531

7 0
3 years ago
How many moles of iron are there in 55.85g of Fe3O4
soldier1979 [14.2K]

Answer:

• Molecular mass of Iron (III) tetraoxide

\dashrightarrow \: { \tt{(56 \times 3) + (16 \times 4)}} \\  = { \tt{168 + 64}} \\  = { \tt{232\:g}}

[ molar masses: Fe → 56, O → 16 ]

\dashrightarrow \:{ \rm{232 \: g \:  = 1 \: mole}} \\ \\   \dashrightarrow \: { \rm{55.85 \: g = ( \frac{55.85}{232}) \: moles }} \\  \\ \dashrightarrow \:  { \boxed{ \tt{ = 0.24 \: moles}}}

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