Use the following chemical equation to answer the question: 2 K + MgBr2 ⟶ 2 KBr + Mg

Which elements or compounds are the reactants?
1. K and MgBr2
2. MgBr2 and Mg
3. KBr and Mg
4. K and KBr

Option 1. K and MgBr2

Explanation:

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

2K + MgBr2 —> 2KBr + Mg

In a chemical reaction, the reactants are located on the left side of the equation while the products are located on the right side.

The reactants are separated from the products by an arrow sign.

Considering the question given above,

Element /compound >>> location

K, MgBr2 >>>>>>>>>>>>> Left

KBr, Mg >>>>>>>>>>>>>> Right

From the illustration above, the reactants (i.e element/compound on the right side) are K and MgBr2

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