 , 08.08.2019 03:10 Martinemaberry1944

# complete and balance the following equations. for each, enter the molecular, lonic, and net lonic equations. all the products are soluble in water.) include states in your answer enter the balanced molecular equation: hc2h302(aq) + mg(oh)2(5) ► enter the balanced ionic equation: hc2h302(aq) + mg(oh)2(5) - 6CO2 + 6H2O YOU JUST ADD THEM TOGETHER because a complete equation must have all the elements present on the left hand side of the arrow on the right hand side of the arrow as well, and to be balanced there must be the same number of each type of atom, so both A and C are full equations, but A is not balanced because although there are two oxygen atoms on each side of the arrow, there is only one hydrogen atom on the left and two on the right, thus the right hand side is 'heavier' than the left and the equation is unbalanced.

Explanation: 1. S + Sr(OH) ⇒ SrSO4 + 2 H2O is the balanced reaction.

2. 0.034 liters of KI will be required  completely react with 2.43 g of Cu(NO3)2.

3. 0.55 liters is the volume in L of a 0.724 M Nal solution contains0.405 mol of NaI.

4. 33.3 ml  many mL of 0.300 M NaF would be required to make a 0.0400 M solution of NaF when diluted to 250.0 mL with water

Explanation:

Balance chemical reaction of neutralization:

1. S + Sr(OH) ⇒ SrSO4 + 2 H2O

2. Data given:

balance chemical reaction:

2Cu(NO₃)₂(aq) + 4KI(aq) → 2CuI(aq) + I₂(s) + 4KNO₃(aq)

molarity of KI = 0.209 M

mass of Cu(NO₃)₂ = 2.43 grams

number of moles of Cu(NO₃)₂ will be calculated as:

number of moles = atomic mass of Cu(NO₃)₂ = 187.56 grams/mole

putting the values in the equation,

number of moles= = 0.0129 moles

2 moles of Cu(NO₃)₂ will react with 4 moles of KI

0.0129 moles will react with x moles of KI = x = 0.0258

atomic mass of KI = 166.00 grams/mole

mass = 166.00 x 0.0258

= 4.28 grams or ml is the final volume of KI

so molarity = so molarity of KI is 0.0258 M, volume is 1 litre.

Using the formula

Minitial x Vinitial = M final Vfinal

V initial = = = 34.67 ml 0.034 liters of KI will be required.

3) Data given:

molarity of NaI = 0.724

number of moles of NaI =?

Volume in litres =?

formula used:

molarity = volume in litres = = 0.55 liters is the volume

4) Data given:

Initial molarity = 0.3 M

initial volume = ?

final molarity = 0.04 M

final volume diluted by = 250 ml

formula used:

M initial X Vinitial = Mfinal X V final

putting the values in the equation:

Vinitial = = 33.3 ml of 0.3 M solution will be required. Code is completed below

Explanation:

Source Code in Java:

class Parenthesis

{

boolean hasBalancedParentheses(String eq) //method to check and return if a set of parenthesis is balanced or not

{

int count=0; //to store count of current unclosed opening brackets

for(int i=0;i<eq.length();i++)

{

if(eq.charAt(i)=='(')

count++;

else if(eq.charAt(i)==')')

count--;

if(count<0) //if count falls below zero, there were more closing brackets than opening brackets till this point

return false;

}

if(count>0) //if count is still more than zero, there were less closing brackets than opening brackets

return false;

return true; //if program reaches this point, it has passed all the tests

}

public static void main(String args[])

{

//testing the function

Parenthesis ob=new Parenthesis();

System.out.println(ob.hasBalancedParentheses("()(()())((())())"));

System.out.println(ob.hasBalancedParentheses(")((()())(()))())"));

}

} Cu + HNO3 > Cu (NO3) 2 + NO2 + H2O

Explanation:

This equation is the equation for dissolving copper in nitric acid, this reaction is a REDOX reaction, that is, an oxide-reduction reaction. Explanation:

17. it goes from solid copper to aqueous copper:

Cu(s) --> Cu₂(aq) + 2e⁻

18. complete ionic:

Cu(s) --> Cu₂(aq) + 2e⁻

19. net ionic, must include only reacting species, so

Cu(s) --> Cu₂(aq) + 2e⁻

20. this type of reaction is dissolution reaction(redox reaction)

copper reduced from Cu²⁺ to Cu. NO2- + 5H2O + 2Al + OH- → NH3 + 2Al(OH)4-

There is 1 hydroxide ion, on the reactant side

Explanation:

NO2−(aq) + Al(s) → NH3(aq) + Al(OH)4−(aq)

Step 1: The half reactions

NO2- (aq) → NH3(g)

Al(s) → Al(OH)4-

Step 2: Balancing electrons

NO2- → NH3

On the left side N has an oxidation number of +3 and on the right side -3.

NO2- +6e-→ NH3

Al(s) → Al(OH)4-

On the left side, Al has an oxidation number of 0 and on the right side +3.

Al(s) → Al(OH)4- +3e-

To have the same amount of electrons transfered, we have to multiply the second reaction by 2

NO2- +6e-→ NH3

2(Al(s) → Al(OH)4- +6e-)

Step 3: Balance with OH/H2O

NO2- +6e +5H2O → NH3 +7OH-

2Al +8OH- → 2Al(OH)4- + 6e-

Step 4: The netto reaction

NO2- + 5H2O + 2Al + 8OH-  → NH3 +7OH- + 2Al(OH)4-

NO2- + 5H2O + 2Al + OH- → NH3 + 2Al(OH)4-

There is 1 hydroxide ion, on the reactant side Theoretical yield: 36,3g

Percent yield: 71,9%

Explanation:

The reaction of tert-butyl alcohol with hydrobromic acid to produce tert-butyl bromide is a 1:1 reaction.

Moles of tert-butyl alcohol in 25,0 mL are:

25,0mL × (0,786g / mL) × (1mol / 74,12g) = 0,265 moles of tert-butyl alcohol

Moles of hydrobromic acid are:

60,0mL× 47,0% × (1,49g / mL) × (1mol / 80,91g) = 0,519 moles of HBr

As moles of tert-butyl alcohol are less than moles of HBr, limiting reactant is tert-butyl alcohol.

Theoretical yield of tert-butyl bromide in grams is:

0,265 moles × (137,02g / 1mol) = 36,3g

Percent yield:

26,1g / 36,3g × 100 = 71,9%

The reactions are in the image.

There are many ways to produce alkyl halides from primary and secondary alcohols. The reaction in this case is the Appel reaction.

The reaction of tertiary alkyl halides may undergo a hydrolysis reaction -under proper conditions- to form an alcohol and a hydrogen halide.

I hope it helps!   There is one 1 hydroxide ion in the balanced equation.

Explanation:

Oxidation reaction is defined as the chemical reaction in which an atom looses its electrons. The oxidation number of the atom gets increased during this reaction. Reduction reaction is defined as the chemical reaction in which an atom gains electrons. The oxidation number of the atom gets reduced during this reaction. For the given chemical reaction: The half cell reactions for the above reaction follows:

Reduction half reaction: Oxidation half reaction: Multiplying the Oxidation half reaction by 2 and we get that:- Adding the half reactions, we get that:- There is one 1 hydroxide ion in the balanced equation. A balanced chemical equation  ### Other questions on the subject: Chemistry Chemistry, 21.06.2019 18:20, datboyjulio21
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