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leonid [27]
3 years ago
10

Classify the four reactions that your group analyzed today as being either endothermic or exothermic

Chemistry
1 answer:
ArbitrLikvidat [17]3 years ago
4 0
Chemical reactions are basically divided into two major classes depending on whether the reaction lose energy or gain energy from the environment during the course of the reaction. The two classes of reaction are exothermic and endothermic reaction.
An exothermic reaction is a type of reaction in which the reaction system lose energy to the environment and thus, the energy content of the reactants is more than that of the product formed. Because of this, the enthapyl change of an exothermic reaction is always negative. 
An endothermic reaction is a type of reaction in which the reaction system absorb energy from the environment. Thus, the energy contents of the products is always higher than that of the reactants and the enthapyl change of the reaction is always positive. During the course of the reaction, the reaction container is usually cold to the touch because energy is been absorbed from the environment.
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using the equaation 2h2+o2-->2h2o if 10.0g of hydrogen are used in the presence of excess oxygen how many grams of water will
astra-53 [7]

Answer:

90g of H2O

Explanation:

2H2 + O2 —> 2H2O

First, we calculate the molar masses of H2 And H20.

Molar Mass of H2 = 2g/mol

Mass conc of H2 from the balanced equation = 2 x 2 = 4g

Molar Mass of H2O = 2 + 16 = 18g/mol

Mass conc of H2O from the balanced equation = 2x18 = 36g

From the equation,

4g of H2 produced 36g of H2O

Therefore, 10g of H2 will be produce = (10x36)/4 = 90g of H2O

7 0
3 years ago
What does variable mean in a science project.
Paul [167]
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6 0
3 years ago
Calculate the solubility of mn(oh)2 in grams per liter when buffered at ph=7.0. assume that buffer capacity is not exhausted
Vanyuwa [196]
When PH + POH = 14 
∴ POH = 14 -7 = 7

when POH = -㏒[OH-]

          7    = -㏒ [OH-]
∴[OH-] = 10^-7

by using ICE table:

           Mn(OH)2(s) ⇄  Mn2+ (aq)  + 2OH-(aq)
initial                              0                     10^-7
change                           +X                      +2X
Equ                                 X                  (10^-7 + 2X)

when Ksp = [Mn2+][OH-]^2

when Ksp of Mn(OH)2 = 4.6 x 10^-14

by substitution:

4.6 x 10^-14 = X*(10^-7+2X)^2  by solving this equation for X

∴ X =2.3 x 10-5 M

∴ The solubility of Mn(OH)2 in grams per liter (when the molar mass of Mn(OH)2 = 88.953 g/mol
= 2.3 x10^-5 moles/L * 88.953 g/mol

= 0.002 g/ L
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3 years ago
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Dominik [7]
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