Ineed balancing these equations,

1. __ na + __ clz = __ nael
2. __ na + __ h2o = __ naoh + __ hz
3. __ co + __ no = __ coz + __ nz
4. __ fezo3 + __ co = __ fe + __ a²
5. __ c + __ fezo3 = __ co + __ fe

you are supposed to fill in the blanks, u : )

1. 2 Na + cl2 = 2NaCl

2. 2 Na + 2 H2O = 2 NaOH + H2

3. Co + no = co2 + N2

4. Fe2O3+ 3 co = 2 Fe + 3 Co2

5. C + Fe2o3 = co + Fe2

Explanation:

Btw did u copy them wrongly cuz one look and i could tell that u werent even writing it properly

the number of valence electrons for each molecule or ion is shown beneath the structure. remember hydrogen will not have more than two electrons. this hydrogen is part of a covalent bond (sharing two electrons). a double bond here would cause hydrogen to share four electrons with phosphorus.

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