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Tom [10]
3 years ago
14

Ozone(O3) number of atoms

Chemistry
1 answer:
spayn [35]3 years ago
8 0
As the O3 suggests, the Ozone molecule is made up of 3 Oxygen atoms.
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What happens when atoms bonds?​
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Atoms form chemical bonds to make their outer electron shells more stable. ... An ionic bond, where one atom essentially donates an electron to another, forms when one atom becomes stable by losing its outer electrons and the other atoms become stable (usually by filling its valence shell) by gaining the electrons.
8 0
4 years ago
Which rule for assigning oxidation numbers is correct?
AURORKA [14]

The incorrect rule for assigning oxidation numbers is Hydrogen is usually –1.

Hydrogen is usually +1

<h3>What is oxidation number?</h3>

Oxidation numbers can be defined as that number which is assigned to an element in chemical reaction which represents the number of electrons lost or gained.

So therefore, the incorrect rule for assigning oxidation numbers is Hydrogen is usually –1.

Learn more about oxidation numbers:

brainly.com/question/27239694

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7 0
2 years ago
Can some one tell me the best position in sex
masya89 [10]

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8 0
3 years ago
An unknown metal has a mass of 86.8 g. When 5040 J of heat are added to the sample, the sample temperature changes by 64.7 ∘ C .
grandymaker [24]

Answer: The specific heat of the unknown metal is 0.897J/g^0C

Explanation:

The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.

Q=m\times c\times \Delta T

Q = Heat absorbed=5040 Joules

m= mass of substance = 86.8 g

c = specific heat capacity = ?

Initial temperature of the water = T_i

Final temperature of the water = T_f

Change in temperature ,\Delta T=T_f-T_i=(64.7)^0C

Putting in the values, we get:

5040=86.8\times c\times 64.7^0C

c=0.897J/g^0C

The specific heat of the unknown metal is 0.897J/g^0C

4 0
3 years ago
Carbon-14 has a half-life of 5,700 years.
Maksim231197 [3]

Answer:

3.106\ \text{g}

Explanation:

t_{1/2} = Half-life of carbon = 5700 years

t = Time at which the remaining mass is to be found = 10400 years

m_0 = Initial mass of carbon = 11 g

Decay constant is given by

\lambda=\dfrac{\ln2}{t_{1/2}}

Amount of mass remaining is given by

m=m_0e^{-\lambda t}\\\Rightarrow m=m_0e^{-\dfrac{\ln2}{t_{1/2}} t}\\\Rightarrow m=11e^{-\dfrac{\ln 2}{5700}\times 10400}\\\Rightarrow m=3.106\ \text{g}

The amount of the substance that remains after 10400 years is 3.106\ \text{g}.

5 0
3 years ago
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