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gulaghasi [49]
3 years ago
8

A(n) _____ bond joins these two oxygen atoms. Two oxygen atoms with two electron shells. The atoms share four electrons located

in their outer electron shells. Each atom also has one electron pair in the inner electron shell and two electron pairs in the outer electron shell.
Chemistry
1 answer:
Lisa [10]3 years ago
6 0

Answer:

double covalent bond  

Explanation:

Double covalent bond -

It is the type of interaction observed between two species , which share the electrons in order to attain stability , is referred to as covalent bond.

The shared electrons are referred to as the bonding pairs or the shared pairs .

Stability and completion of the octet is the driving force for the formation covalent bond.

The molecules of the organic compound usually shows this type bonding .

A double covalent bond is the one which have four shared pair of electrons , i.e. two covalent bonds.

Hence , from the question,

Oxygen is capable to show double covalent bond .

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Aleksandr-060686 [28]
One mole of a substance contains 6.02×10∧23 particles,
1 mole of a aluminium contains 27 g
35 g of aluminium contains 35/27 =1.296 moles
 Thus, the number of particles will be 1.296 × 6.02 ×10∧23
      = 7.804 × 10∧23 particles,
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3 years ago
A 19.0 g piece of metal is heated to 99.0 °C and then placed in 150 mL of water at 21°C. The temperature of the water rose to 23
Anna35 [415]

Answer:

0.8696\ \text{J/g}^{\circ}\text{C}

Explanation:

m_m = Mass of metal = 19 g

c_m = Specific heat of the metal

\Delta T_m = Temperature difference of the metal = 99-23=76^{\circ}\text{C}

V = Volume of water = 150 mL = 150\ \text{cm}^3

\rho = Density of water = 1\ \text{g/cm}^3

c_w = Specific heat of the water = 4.186 J/g°C

\Delta T_w = Temperature difference of the water = 23-21=2^{\circ}\text{C}

Mass of water

m_w= \rho V\\\Rightarrow m_w=1\times 150\\\Rightarrow m_w=150\ \text{g}

Heat lost will be equal to the heat gained so we get

m_mc_m\Delta T_m=m_wc_w\Delta T_w\\\Rightarrow c_m=\dfrac{m_wc_w\Delta T_w}{m_m\Delta T_m}\\\Rightarrow c_m=\dfrac{150\times 4.186\times 2}{19\times 76}\\\Rightarrow c_m=0.8696\ \text{J/g}^{\circ}\text{C}

The specific heat of the metal is 0.8696\ \text{J/g}^{\circ}\text{C}.

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Explanation:

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