When a chemical equation in balanced, it will have a set of whole number coefficients that cannot be reduced to smaller whole numbers. what is the coefficient for 02 (2 is a subscript) when the following combustion reaction of a hydrocarbon is balanced? basically, just balance the following
_c7h14+_o2 to _co2+h20 (where 7, 14, 2, 2, and 2 are subscripts)

Answer : The balanced chemical reaction will be,

Explanation :

Balanced chemical reaction : It is defined as the reaction in which the number of atoms of individual elements present on reactant side must be equal to the product side.

Combustion reaction : It is a chemical reaction in which a hydrocarbon reaction with the oxygen to give product as carbon dioxide and water.

The unbalanced chemical reaction is,

This reaction is an unbalanced chemical reaction because in this reaction number of carbon , hydrogen and oxygen atoms are not balanced.

In order to balance the chemical equation, the coefficient '2' put before the , the coefficient '21' put before the , the coefficient '14' put before the and the coefficient '14' put before the then we get the balanced chemical equation.

The balanced chemical reaction will be,

the correct option is c.

isotopes refers to different forms of the same element with the same atomic number but different mass number as a result of difference in their neutron numbers. isotopes have different relative atomic mass but share the same chemical properties.

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol