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Ostrovityanka [42]
3 years ago
9

General Information. A brief introduction about the caffeine.​

Chemistry
2 answers:
tia_tia [17]3 years ago
7 0
Caffeine, a central nervous system stimulant, is arguably the most frequently ingested pharmacologically active substance in the world. Occurring naturally in more than 60 plants, including coffee beans, tea leaves, cola nuts, and cocoa pods, caffeine has been part of innumerable cultures for centuries.
viva [34]3 years ago
3 0

Answer: A central nervous system stimulant

Explanation: arguably the most frequently ingested pharmacologically active substance in the world.

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Match the following reactions with the reaction type. You may use each type more than once.
vaieri [72.5K]

Answer:

1) Decomposition

2)Double-replacement reaction

3)synthesis

4)Double-replacement reaction

5)single-replacement reaction

6)Double-replacement reaction

Explanation:

Step 1: Data given

Synthesis: This is a reaction where two or more reactants will combine to form a new, single product. This is  occurs when two or more reactants combine to form a single product.  This can be shown with the following equation.

A + B → C

A and B are the reactants to form a new product C

Decomposition: This is a reaction where 1 (more complex) compound will be broken down into 2 or more (more simple)  products.

This can be shown by the following equation:

AB → A + B

A single replacement: This is a reaction where one element wil be replaced by another element in the same compound. This can be shown by the following equation:

A + BC → B + AC

Here is the elemnt B in the compound BC, replaced by the element A, to form a new compound AC

A double-replacement: This is a reaction where the positive and negative ions of two ionic compounds will be exchanged and 2 new compounds willbe formed. This can be shown by the following equation:

AB+CD→AD+BC

A combustion reaction requires oxygen gas (O2) to produce the  products CO2 and H2O

Step 2:

1) 2HgO(s) → 2Hg(l) + O2(g)

⇒ Decomposition

2) KCl(aq) + AgNO3(aq) → AgCl(s) + KNO3(aq)

⇒  Double-replacement reaction

3) 2Na(s) + H2(g) → 2NaH(s)

⇒ synthesis

4) Mg(OH)2(aq) + 2HNO2(aq) → Mg(NO2)2(aq) + 2 H2O(l)

⇒  Double-replacement reaction

5) Ca(s) + 2AgNO3(aq) → Ca(NO3)2(aq) + Ag(s)

⇒  Single-replacement reaction

6) Al2O3(s) + H2SO4(aq) → Al2(SO4)3(aq) + H2O(l)

⇒  Double-replacement reaction

8 0
3 years ago
For a particular isomer of C 8 H 18 , C8H18, the combustion reaction produces 5104.1 kJ 5104.1 kJ of heat per mole of C 8 H 18 (
vlada-n [284]

Answer:

The standard enthalpy of formation of this isomer of octane is -220.1 kJ/mol

Explanation:

Step 1: Data given

The combustion reaction of octane produces 5104.1 kJ per mol octane

Step 2: The balanced equation

C8H18(g) + 12.5 O2 ⟶ 8CO2 (g) + 9 H2O (g)  ∆H°rxn = -5104.1 kJ/mol

Step 3:

∆H°rxn = ∆H°f of products minus the ∆H° of reactants

∆H°rxn = ∆H°f products - [∆H°f reactants]

-5104.1 kJ/mol = (8*∆H°fCO2 + 9*∆H°fH20) - (∆H°fC8H18 + 12.5∆H°fO2)

∆H°f C8H18 = ∆H°f 8CO2 + ∆H°f 9H2O+ 5104.1 kJ/mol

∆H°f C8H18 = 8 * (-393.5 kJ)/mol + 9 * (-241.8 kJ/mol)] + 5104.1 kJ /mol

∆H°f C8H18 = -220.1 kJ/mol

The standard enthalpy of formation of this isomer of octane is -220.1 kJ/mol

7 0
3 years ago
What is in with those numbers? H2O?
vlabodo [156]

Answer:

hydrogen atom and oxygen atom

8 0
3 years ago
Calculate the solubility (in M units) of ammonia gas in water at 298 K and a partial pressure of 7.00 bar . The Henry’s law cons
Juliette [100K]

Explanation:

just for friendship kvp-hxns-jst

6 0
3 years ago
Calculate the fraction of acetic acid that is in the dissociated form in his solution. Express your answer as a percentage. You
Sliva [168]

Answer:

0.11%

Explanation:

Without mincing words, let us dive straight into the solution to the question/problem. The first step to solve this question is to write out the chemical reaction, that is the reaction showing the dissociation of acetic acid.

CH3COOH <=======================================> CH3COO⁻ + H⁺

Initially, the amount present in the acetic acid which is = 12M, the concentration for CH3COO⁻  and H⁺ is 0 respectively.

At equilibrium, the amount present in the acetic acid which is = 12 - x, the concentration for CH3COO⁻ = x  and H⁺ = x respectively. Note that the ka for acetic acid = 1.8 × 10⁻⁵.

1.8 × 10⁻⁵ = x²/ 14 - x. Therefore, x = 0.0158 M.

The next thing to do is to calculate for the percentage of dissociation, this can be done as given below:

percentage of dissociation = x/14 × 100. Recall that the value that we got for x = 0.0158 M. Hence, the percentage of dissociation = 0.0158 M/ 14m × 100 = 0.11%

7 0
3 years ago
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