Complete and balance the following equation: BrO3−(aq)+N2H4(aq)→Br2(l)+N2(g)

Answers

Answer 1
Answer: BrO3- + 6H+ + 6e-= Br- + 3 H2O 
N2H4 = N2 + 4 H+ + 4e- 
2 BrO3- + 3 N2H4 = 2 Br- + 3 N2 + 6 H2O 

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A toy balloon filled with air has an internal pressure of 1.25 atm and avolume of 2.50 L. If I take the balloon to the bottom of the ocean wherethe pressure is 95 atmospheres, what will the new volume of the balloonbe? How many moles of gas does the balloon hold? (Assume T 285 K)

Answers

Ideal Gas Law
 Given: P1= 1.25 atm   P2= 95 atm
            V1= 2.50 L
 
 Unknown: V2=?
 
Formula: V=nRT/P; n = RT / PV

Solution: n = (0.0821 L x atm / K x mol) (285K) / (1.25 atm) ( 2.50L)
   
                  = 23.40 / 3.125

Answer: n = 7.49 moles of gas the balloon holds

An igneous rock originally has 3 grams of uranium 238 in it. When dated the rock only contains 1.8 grams. What are the parent and daughter concentrations (in percentages) and what is its age?

Answers

Answer :

The parent and daughter concentrations (in percentages) is, 60 % and 40 % respectively.

The age of rock is 3.32* 10^9\text{ years}

Explanation :

First we have to calculate the parent and daughter concentrations (in percentages).

\text{Parent concentrations}=(1.8g)/(3g)* 100=60\%

and,

\text{Daughter concentrations}=((3-1.8)g)/(3g)* 100=40\%

As we know that, the half-life of uranium-238 = 4.5* 10^9 years

Now we have to calculate the rate constant, we use the formula :

k=(0.693)/(t_(1/2))

k=\frac{0.693}{4.5* 10^9\text{ years}}

k=1.54* 10^(-10)\text{ years}^(-1)

Now we have to calculate the time passed.

Expression for rate law for first order kinetics is given by:

t=(2.303)/(k)\log(a)/(a-x)

where,

k = rate constant  = 1.54* 10^(-10)\text{ years}^(-1)

t = time passed by the sample  = ?

a = initial amount of the reactant  = 3 g

a - x = amount left after decay process = 1.8 g

Now put all the given values in above equation, we get

t=(2.303)/(1.54* 10^(-10))\log(3)/(1.8)

t=3.32* 10^9\text{ years}

Therefore, the age of rock is 3.32* 10^9\text{ years}

How many hydrogen atoms are represented in the formula (CH3)2CH2?

Answers

There are 7 hydrogens in that formula
There are 8 hydrogens.

There are 3 within the bracket, but this is reoeated twice, giving a total of 6 for that section of the molecule.
At the end of the molecule there are two more hydrogen atoms, taking the total for the molecule to 8.
(3x2) + 2 = 8

Which compound would most likely turn litmus paper to a red color?a. CH₃CH₂CH₂OH 
b. CH₃CH(Br)CH(Br)CH₃
c. CH₃CH₂CH(Br)CH₃
d. CH₃CH₂CH₂COOH 
e. CH₃CH₂CH₂CH₃

Answers

answer D, because it's acid

How many moles of CaO are needed to react with an excess of water to form 370 grams of calcium hydroxide?

Answers

The required moles of CaO are needed to react with an excess of water to form 370 grams of calcium hydroxide is 280g.

What is the relation between mass & moles?

Relation between the mass and moles of any substance will be represented as:

n = W/M, where

  • W = given mass
  • M = molar mass

Given chemical reaction is:

CaO + H₂O → Ca(OH)₂

Moles of Ca(OH)₂ = 370g / 74g/mol = 5mol

From the stoichiometry of the reaction, 5 moles of CaO is required to produce 5 moles of Ca(OH)₂.

Mass of CaO = (5mol)(56g/mol) = 280g.

Hence required mass of CaO is 280g.

To know more about mass & moles, visit the below link:
brainly.com/question/18983376

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CaO + H20 --> Ca(OH)2 

Calcium Hydroxide has a relative molecular mass of approximately 74 grams. Therefore, with 370 grams provided, we have 370g/74g, or 5 moles. 

Therefore, due to molar ratios, 5 moles of CaO must have reacted with 5 moles of H2O to form 5 moles of Ca(OH)2. 

The answer to your question is 5 moles of CaO or 280 g of CaO 

What are the formal charges of both chlorines and iodine in icl2?

Answers

Answer:

Explanation:

 Formal charge of ICl₂⁻

Formal charge = group no - ( no of non bonding electrons +no of bonds)

In I there are 7 electrons in outermost orbit . If we add one more electrons due to - ve charge on the ion , it becomes eight . This centrally placed iodine forms two single bond with two chlorine atoms on either side.

Each of chlorine atoms also contains 7 valance electrons like iodine.

So formal charge of chlorine

= group no - ( no of non bonding electrons +no of bonds)

= 7 - ( 6 + 1 )

= 0

So formal charge of iodine

= group no - ( no of non bonding electrons +no of bonds)

= 7 - ( 5 + 2 )

=0

 Formal charge of ICl₂⁺

In this case , central iodine will have only 6 valence electrons due to absence one electron.

So formal charge of chlorine in  ICl₂⁺

= group no - ( no of non bonding electrons +no of bonds)

= 7 - ( 6 + 1 )

= 0

formal charge of iodine in  in   ICl₂⁺

7 - ( 4 + 2)

= 1