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storchak [24]
3 years ago
15

How do you balance the equation Hydrogen plus Oxygen yield water?

Chemistry
1 answer:
Andre45 [30]3 years ago
5 0
Hey there! Hello!

Not sure if you still need the answer to this question, but I'd love to help out if you do. 

So, the way to balance this equation is pretty simple. First, you need to keep in mind that molecules of hydrogen and oxygen do not come in single molecules, but in bonded pairs, represented by H2 and O2. 

H2+O2=H2O

But, that's incorrect. The combination of 2 hydrogen molecules with 1 oxygen molecule yields water, but that leaves one oxygen molecule leftover. When broken down, this is how many of each molecule is on each side of the previously stated equation:

Left:
H: 2
O: 2

Right:
H: 2
O: 1

So we have to multiply H2O on the right side by 2 in order to get this:

H2+O2=2(H2O)

Left: 
H: 2
O: 2

Right: 
H: 4
O: 2

The last step is to multiply H2 on the left by two to make it match up with the right side, balancing the equation:

2(H2)+O2=2(H2O)

Left: 
H: 4
O: 2

Right: 
H: 4
O: 2

That makes our equation balanced! I hope this helped you out, feel free to ask any additional questions if you need further clarification. :-)
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How old is a bone if it now .3125 of C-14 when it originally had 80.0g of C-14
Taya2010 [7]
<h3>Answer:</h3>

42960 years

<h3>Explanation:</h3>

<u>We are given;</u>

  • Remaining mass of C-14 in a bone is 0.3125 g
  • Original mass of C-14 on the bone is 80.0 g
  • Half life of C-14 is 5370 years

We are required to determine the age of the bone;

  • Using the formula;
  • Remaining mass = Original mass × 0.5^n , where n is the number of half lives.

Therefore;

0.3125 g = 80.0 g × 0.5^n

3.90625 × 10^-3 = 0.5^n

  • Introducing logarithm on both sides;

log 3.90625 × 10^-3 = n log 0.5

Solving for n

n = log 3.90625 × 10^-3 ÷ log 0.5

   = 8

  • Therefore, the number of half lives is 8
  • But, 1 half life is 5370 years
  • Therefore;

Age of the rock = 5370 years × 8

                          = 42960 years

Thus, the bone is 42960 years old

7 0
3 years ago
A sample of bleach was analyzed as in this procedure. The only procedural difference is that the student weighed out the bleach
Bogdan [553]

Answer:

% = 5.69%

Explanation:

To do this, we need to write the equations taking place here. First, this is a REDOX reaction where the hypoclorite and thiosulfate solution reacts. The balanced equations are:

ClO⁻ + 2I⁻ + 2H⁺ -------> Cl⁻ +  I₂ + H₂O

I₂ + 2S₂O₃²⁻ -----------> 2I⁻ + S₄O₆²⁻

We already have the required volume and concentration of the thiosulfate solution, so we can calculate the moles of thiosulfate. With this moles, we can calculate the moles of hypochlorite, then the mass and finally the %.

The moles of thiosulfate would be:

moles S₂O₃²⁻ = V * M

moles S₂O₃²⁻ = 0.01324 * 0.0732 = 9.69x10⁻⁴ moles

Now according to the above reactions, we can see that

moles I₂ = moles ClO⁻

and

moles I₂ / moles S₂O₃²⁻ = 1/2

Therefore, let's calculate the moles of ClO⁻:

moles ClO⁻ = 9.69x10⁻⁴ / 2 = 4.845x10⁻⁴ moles

Now, we can calculate the mass of these moles, using the molar mass of sodium hypochlorite which is 74.44 g/mol:

m = 74.44 * 4.845x10⁻⁴

m = 0.036 g

Finally the % of this, in the bleach sample would be:

% = 0.036 / 0.634 * 100

<h2>% = 5.69%</h2>
6 0
3 years ago
Be sure to answer all parts. Show the overall reaction for formation of racemic 3−bromohexane from (E)−3−hexene by enterin
Viktor [21]

Answer:

See explanation

Explanation:

The mechanism for the formation of bromohexane from hexene has been clearly shown in the image attached to this answer.

Hexene is attacked by HBr and a carbocation is first formed as shown. The carbocation is flat and planar. it can be attacked on either face by the bromide ion.

Attack on either faces yields a racemic mixture of the R and S enantiomer as shown in the image.

You can use the mechanism shown to fill in the structures.

8 0
2 years ago
What are ways heat can spread
bearhunter [10]
Added on wood, gasoline, wind
3 0
3 years ago
Read 2 more answers
PDF on Unit operation 1 and 2​
krok68 [10]
https://mgdic.files.wordpress.com/2016/12/unit-operations.pdf this
6 0
2 years ago
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