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Delvig [45]
3 years ago
8

Explain the roles of products, reactants, and limiting reactant in chemical reaction.

Physics
1 answer:
andreev551 [17]3 years ago
4 0
The <span>roles of products, reactants, and limiting reactant are significant in order for a chemical reaction to take place. The reactants are the components that are interacted in such a way that it would eventually produce a product, while on the other hand, a limiting reactant is used to regulate the products intended to produce.</span>
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What factor affects the polarization of light by scattering?
MatroZZZ [7]

Answer:

The correct answer would be A.

Explanation:

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2 years ago
Click on the boxes below to indicate the number of electrons or protons in each labeled position for a zinc atom (z=30, a=65). P
g100num [7]

Answer:

Name: Zinc

Symbol: Zn

Atomic Number: 30

Atomic Mass: 65.39 amu

Melting Point: 419.58 °C (692.73 K, 787.24396 °F)

Boiling Point: 907.0 °C (1180.15 K, 1664.6 °F)

Number of Protons/Electrons: 30

Number of Neutrons: 35

Classification: Transition metal

Crystal Structure: Hexagonal

Density at 293 K: 7.133 g/cm3

Color: bluish

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7 0
3 years ago
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Tom [10]

Answer:

4th answer

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6 0
2 years ago
A 23.5 g piece of aluminum metal is initially at 100.0°C. It is dropped into a coffee cup-calorimeter containing 130.0 g of wate
vivado [14]

Answer: The molar heat capacity of aluminum is 25.3J/mol^0C

Explanation:

heat_{absorbed}=heat_{released}

As we know that,  

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]         .................(1)

where,

q = heat absorbed or released

m_1 = mass of water = 130.0 g

m_2 = mass of aluminiunm = 23.5 g

T_{final} = final temperature = 26.0^oC=(273+26)K=299K

T_1 = temperature of water = 23^oC=(273+23)K=296K

T_2 = temperature of aluminium = 100^oC=273+100=373K

c_1 = specific heat of water= 4.184J/g^0C

c_2 = specific heat of aluminium= ?

Now put all the given values in equation (1), we get

130.0\times 4.184\times (299-296)=-[23.5\times c_2\times (299-373)]

c_2=0.938J/g^0C

Molar mass of Aluminium = 27 g/mol

Thus molar heat capacity =0.938J/g^0C\times 27g/mol=25.3J/mol^0C

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