Hi anyone there who can talk to. me ans this and talk to me in comments

Write the complete balanced equation for the neutralization reaction that occurs when aqueous hydroiodic acid, HI, and sodium hy

Serhud [2]

Answer:

H+ ( aq ) + HCO3- ( aq ) ------> H2O( l ) + CO2 ( g )

Explanation:

The complete reaction when hydroiodic acid and sodium hydrogen carbonate combine, would be as follows -

HI + NaHCO3 ----> NaI + H2O + CO2

net reaction

H2CO3 is highly unstable, and thus decomposes into the water and carbon dioxide you see present as the reactants. If you didn't know already, H2CO3 is also reffered to as carbonic acid. The rest of the elements present on the reactant side are Iodine and Sodium, which is why they are present on the product side as NaI.

Let me include the " physical states " in this reaction as well -

HI ( aq ) + NaHCO3 ( aq ) ----> NaI ( aq ) + H2O ( l ) + CO2 ( g )

Now the complete ionic equation would simply be each compound present as ions in an aqueous solution, so there is no need for an explanation on this step -

H+ ( aq ) + I- ( aq ) + Na+ ( aq ) + HCO3- ( aq ) -------> Na+ ( aq ) + I- ( aq ) + H2O( l ) + CO2 ( g )

The spectator ions in this reaction are I- and Na+, so canceling them out, you would receive the following net ionic equation -

H+ ( aq ) + HCO3- ( aq ) ------> H2O( l ) + CO2 ( g )

Hope that helps!

4 0

3 years ago

Using the equations 2 Sr(s) + O₂ (g) → 2 SrO (s) ∆H° = -1184 kJ/mol SrO (s) + CO₂ (g) → SrCO₃ (s) ∆H° = -234 kJ/mol CO₂ (g) → C(

kkurt [141]

Answer: The $\Delta H^o_{rxn}$ for the reaction is 72 kJ.

Explanation:

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The given chemical reaction follows:

$2SrCO_3(s)\rightarrow 2Sr(s)+2C(s)+3O_2(g)$ $\Delta H^o_{rxn}=?$

The intermediate balanced chemical reaction are:

(1) $2Sr(s)+O_2(g)\rightarrow 2SrO(s)$ $\Delta H_1=-1184kJ$

(2) $SrO(s)+CO_2(g)\rightarrow SrCO_3(s)$ $\Delta H_2=-234kJ$ ( × 2)

(3) $CO_2(g)\rightarrow C(s)+O_2(g)$ $\Delta H_3=394kJ$ ( × 2)

The expression for enthalpy of the reaction follows:

$\Delta H^o_{rxn}=[1\times (\Delta H_1)]+[2\times (-\Delta H_2)]+[2\times (\Delta H_3)]$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(1\times (-1184))+(2\times -(-234))+(2\times (394))]=72kJ$

Hence, the $\Delta H^o_{rxn}$ for the reaction is 72 kJ.

4 0

3 years ago

Which statement best explains the difference between polar and nonpolar covalent bonds?(1 point)

miv72 [106K]

Answer:

Answer is 4

Explanation:

Polar covalent bonding is a type of chemical bond where a pair of electrons is unequally shared between two atoms. ... If the electronegativity of two atoms is basically the same, a nonpolar covalent bond will form, and if the electronegativity is slightly different, a polar covalent bond will form.

6 0

3 years ago

Read 2 more answers

Could someone help me with this?

Varvara68 [4.7K]

Solving part-1 only

#1

KMnO_4

Transition metal is Manganese (Mn)

#2

Actually it's the oxidation number of Mn

Let's find how?

$\\ \tt\Rrightarrow x+1+4(-2)=0$

$\\ \tt\Rrightarrow x+1-8=0$

$\\ \tt\Rrightarrow x-7=0$

$\\ \tt\Rrightarrow x=+7$

x is the oxidation number

#3

Purple as per the color of potassium permanganate

#4

$\boxed{\begin{array}{c|c|c}\boxed{\bf Tube} &\boxed{\bf Charge} &\boxed{\bf No\:of\; electrons\: loss}\\ \sf 2 &\sf +6 &\sf 6e^-\\ \sf 3& \sf +2 &\sf 2e- \\ \sf 4 &\sf 4 &\sf 4e^-\end{array}}$

7 0

2 years ago

The particle picture is best described as...

mixas84 [53]

Answer:

Both are pure substances and a compound

8 0

3 years ago

Hi anyone there who can talk to. me ans this and talk to me in comments​​

Hi anyone there who can talk to. me ans this and talk to me in comments