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gregori [183]
2 years ago
8

Give the meaning of each of the following symbols when used in a chemical equation.

Chemistry
1 answer:
nlexa [21]2 years ago
6 0

Reaction arrows are used to describe the state or progress of a reaction. 2.1 The Chemical Reaction Arrow. The chemical reaction arrow is one straight arrow pointing from reactant(s) to product(s) and by-products, sometimes along with side products. A → B. It is the most widely used arrow.

Boron is a chemical element with the symbol B and atomic number 5. Produced entirely by cosmic ray spallation and supernovae and not by stellar nucleosynthesis, it is a low-abundance element in the Solar System and in the Earth's crust

The C5 (C5) fraction is a co-product of naphtha cracking and is used as a raw material for synthetic rubber and petroleum resins.

Deuterium

Deuterium is frequently represented by the chemical symbol D. Since it is an isotope of hydrogen with mass number 2, it is also represented by 2. H. .

Unimolecular Elimination (E1) is a reaction in which the removal of an HX substituent results in the formation of a double bond. It is similar to a unimolecular nucleophilic substitution reaction (SN1) in various ways. One being the formation of a carbocation intermediate.

Aqueous (aq.): In the presence of water, often meaning water is the solvent. Aqueous NaCl. Anhydrous NaCl.

Hydrogen is the chemical element with the symbol H and atomic number 1. ... Since hydrogen readily forms covalent compounds with most nonmetallic elements, most of the hydrogen on Earth exists in molecular forms such as water or organic compounds.

Catalyst, in chemistry, any substance that increases the rate of a reaction without itself being consumed. Enzymes are naturally occurring catalysts responsible for many essential biochemical reactions.

Kp is the equilibrium constant calculated from the partial pressures of a reaction equation. It is used to express the relationship between product pressures and reactant pressures. It is a unitless number, although it relates the pressures.

Hope this helps a bit?

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Answer: it’s number 2 add a catalyst

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4 0
3 years ago
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A 225-g sample of aluminum was heated to 125.5 oc, then placed into 500.0 g water at 22.5 oc. (the specific heat of aluminum is
Sidana [21]
when  heat gained = heat lost 

when AL is lost heat and water gain heat

∴ (M*C*ΔT)AL = (M*C*ΔT) water

when M(Al) is the mass of Al= 225g 

C(Al) is the specific heat of Al = 0.9 

ΔT(Al) = (125.5 - Tf) 

and Mw is mass of water  = 500g

Cw is the specific heat of water = 4.81 

ΔT = (Tf - 22.5) 

so by substitution:

∴225* 0.9 * ( 125.5 - Tf) = 500 * 4.81 * (Tf-22.5)

∴Tf = 30.5 °C
7 0
3 years ago
Read 2 more answers
10. Which is an example of mechanical energy?*
gtnhenbr [62]

Answer: A golfer hitting a golf ball.

Explanation:

The atomic particles move more in this option than the others.

6 0
3 years ago
A 1.00 g sample of octane (C8H18) is burned in a bomb calorimeter with a heat capacity of 837J∘C that holds 1200. g of water at
lubasha [3.4K]

Answer:

The heat of combustion for 1.00 mol of octane is  -5485.7 kJ/mol

Explanation:

<u>Step 1:</u> Data given

Mass of octane = 1.00 grams

Heat capacity of calorimeter = 837 J/°C

Mass of water = 1200 grams

Temperature of water = 25.0°C

Final temperature : 33.2 °C

<u> Step 2:</u> Calculate heat absorbed by the calorimeter

q = c*ΔT

⇒ with c = the heat capacity of the calorimeter = 837 J/°C

⇒ with ΔT = The change of temperature = T2 - T1 = 33.2 - 25.0 : 8.2 °C

q = 837 * 8.2 = 6863.4 J

<u>Step 3:</u> Calculate heat absorbed by the water

q = m*c*ΔT

⇒ m = the mass of the water = 1200 grams

⇒ c = the specific heat of water = 4.184 J/g°C

⇒ ΔT = The change in temperature = T2 - T1 = 33.2 - 25  = 8.2 °C

q = 1200 * 4.184 * 8.2 =  41170.56 J

<u>Step 4</u>: Calculate the total heat

qcalorimeter + qwater = 6863.4 + 41170. 56 = 48033.96 J  = 48 kJ

Since this is an exothermic reaction, there is heat released. q is positive but ΔH is negative.

<u>Step 5</u>: Calculate moles of octane

Moles octane = 1.00 gram / 114.23 g/mol

Moles octane = 0.00875 moles

<u>Step 6:</u> Calculate heat combustion for 1.00 mol of octane

ΔH = -48 kJ / 0.00875 moles

ΔH = -5485.7 kJ/mol

The heat of combustion for 1.00 mol of octane is  -5485.7 kJ/mol

8 0
3 years ago
In the following reaction, what element is gaining mass?
vagabundo [1.1K]

Mg gained mass because it went from being a single element (on the reactant side) to being a molecule (on the product side).

8 0
3 years ago
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