The oxidation number of oxygen is -2, and that of chlorine is -1. Use this information to match the names of the compounds to their chemical formulas. manganese(II) chloride manganese(IV) chloride manganese(II) oxide manganese(IV) oxide MnCl2 arrowRight MnO2 arrowRight

Answer : The correct formula of manganese(II) chloride and manganese(II) oxide is and respectively.

Explanation :

As we know that the chemical formula is formed by the criss-cross method.

In criss-cross method, the oxidation state of the ions gets exchanged and they form the subscripts of the other ions.

For formation of manganese(II) chloride, manganese will form having an oxidation state of +2 called as cation and anion is chloride with oxidation state of -1.Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral  .

For formation of manganese(II) oxide, manganese will form having an oxidation state of +2 called as cation and anion is oxide with oxidation state of -2.Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral  .

Hence, the correct formula of manganese(II) chloride and manganese(II) oxide is and respectively.

The oxidation number of oxygen is -2, and that of chlorine is -1

The top guy is wrong↑

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the answer for sodium is 0.186

the answer for rubidium is 0.251

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