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3 years ago

3 point different between reactant and product

Chemistry

2 answers:

nevsk [136]3 years ago

5 0

Answer:

reactants

The substance(s) to the left of the arrow in a chemical equation are called reactants. A reactant is a substance that is present at the start of a chemical reaction

Explanation:

product_The substance(s) to the right of the arrow are called products . A product is a substance that is present at the end of a chemical reaction. In the equation above, the zinc and sulfur are the reactants that chemically combine to form the zinc sulfide product.

There is a standard way of writing chemical equations. The reactants are all written on the left-hand side of the equation, with the products on the right-hand side. An arrow points from the reactants to the products to indicate the direction of the reaction:

reactants → products

Lelechka [254]3 years ago

3 0

Reactants

1. The substances used as starting materials and which react with one another are reactants.

2. Example: In this reaction Mg and O2 are reactants.

Products

1. The substances which are formed as a result of reaction are products.

2. Example: In this reaction MgO is a product.

You might be interested in

What happens if more product is added to a system at equalibrium

telo118 [61]

Answer:

When additional product is added, the equilibrium shifts to reactants to reduce the stress. If reactant or product is removed, the equilibrium shifts to make more reactant or product, respectively, to make up for the loss.

5 0

3 years ago

A chemist prepares a solution of aluminum sulfate Al2SO43 by weighing out 101.g of aluminum sulfate into a 200.mL volumetric fla

qaws [65]

Answer:

50.5 g/dL

Explanation:

From the question given above, the following data were obtained:

Mass of Al₂(SO₄)₃ = 101 g

Volume (V) = 200 mL

Concentration of solution (g/dL) =?

Next, we shall convert 200 mL to decilitre (dL).

This is illustrated below:

1 mL = 0.01 dL

Therefore,

200 mL = 200 mL / 1 mL × 0.01 dL

200 mL = 2 dL

Therefore, 200 mL is equivalent to 2 dL.

Finally, we shall determine the concentration of the solution in g/dL as follow:

Mass of Al₂(SO₄)₃ = 101 g

Volume (V) = 2 dL

Concentration of solution (g/dL) =?

Concentration of solution (g/dL) = mass /volume

Concentration of solution (g/dL) = 101/2

= 50.5 g/dL

Therefore, the concentration of the Al₂(SO₄)₃ solution is 50.5 g/dL

8 0

4 years ago

A person's heartbeat is 66 beats per minute. If his/her heart beats 3.1e9 times in a lifetime, how long (in whole years) does th

BartSMP [9]

88(60) 24(365) that should be the answer after you figure it out.

5 0

4 years ago

Determine the approximate amount of potassium hydrogen phthalate, KHP, that you will need to neutralize 6.00 ml of 0.100 M NaOH.

Sveta_85 [38]

Answer:

potassium hydrogen phthalate KHP MOLAR MASS = 204.233 glmol

to get 1000 ml

Molar concentration = Mass concentration/Molar Mass

mass concentration = molar concentration x molar mass

mass concentration=0.1 M,

molar mass= 204.233 g/mol

so to get 1L

mass conc = 204.233 x 0.1

= 20.4233g for 1L or 1000 ml

to get 6.00 ml

if 20.4233g is for 1000ml

then to 6.00 ml

= 20.4233 x 6 / 1000

= 0.123g for 6.00 ml

according to the equation below

NaOH(aq) + KHC8H4O4(aq) --> KNaC8H4O4(aq) + H2O(l)

number of moles of NaOH is equal to that of KHP

so the same amount will be needed too, which is

= 0.123g

6 0

3 years ago

Pentaborane−9 (B5H9) is a colorless, highly reactive liquid that will burst into flames when exposed to oxygen. The reaction is

mote1985 [20]

Answer : The heat released per gram of the compound reacted with oxygen is 71.915 kJ/g

Explanation :

Enthalpy change : It is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as $\Delta H^o$

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)]$

The equilibrium reaction follows:

$2B_5H_9(l)+12O_2(g)\rightleftharpoons 5B_2O_3(s)+9H_2O(l)$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(n_{(B_2O_3)}\times \Delta H^o_f_{(B_2O_3)})+(n_{(H_2O)}\times \Delta H^o_f_{(H_2O)})]-[(n_{(B_5H_9)}\times \Delta H^o_f_{(B_5H_9)})+(n_{(O_2)}\times \Delta H^o_f_{(O_2)})]$

We are given:

$\Delta H^o_f_{(B_5H_9(l))}=73.2kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(B_2O_3(s))}=-1271.94kJ/mol\\\Delta H^o_f_{(H_2O(l))}=-285.83kJ/mol$