All categories

4 years ago

1. Write the complete balanced equation for the following reaction:

1 answer:

6 0

To completely balance the equation, we do elemental balance. First for C, the C in the left side is 5 so there must be 5 CO2. Balancing the H, H20 must be 9/2 to balance the left side. The total O on the right is 29/2. In the left side, O2 must be 27/4. Multiplying by 4 to get whole number stoich.coeff, we get 4 C5H9O + 27O2 = 20 CO2 + 18 H2O.

You might be interested in

A chemical reaction is the process by which a ___ takes place

Verizon [17]

A chemical reaction is the process by which a CHANGE takes place (?)

4 0

3 years ago

Read 2 more answers

Study the diagram below and answer these questions 1-6:

grandymaker [24]

N=1 shell contains 2 electrons.
n=2 shell contains 5 electrons.
n=2 is the outermost shell. Therefore there were 5 electrons before sharing.
6 electrons are being shared between 2 nitrogen atoms.
Pair of electrons form a covalent bond. Therefore 3 pair of electrons will form 3 bonds.

6 0

4 years ago

Read 2 more answers

What is true of enzymes? They only speed up reactions. They speed up or slow down any reaction in a living thing. They act as ca

muminat

They act as catalysts

7 0

3 years ago

Read 2 more answers

NaCIO3 compound name

Black_prince [1.1K]

Answer:

Sodium chlorate

Explanation:

I searched it up

Can I pls have a brainliest Pls

:) :) :) :)

6 0

3 years ago

G determine the concentration of an hbr solution if a 45.00 ml aliquot of the solution yields 0.6485 g agbr when added to a solu

Sunny_sXe [5.5K]

The molecular weight of silver bromide (AgBr) is 187.77 g/mole. The presence of the ions in solution can be shown as- AgBr (insoluble) ⇄ $Ag^{+}$ + $Br^{-1}$ .

45.00 mL of the aliquot contains 0.6485 g of AgBr. Thus 1000 mL of the aliquot contains $\frac{0.6485}{45}$ ×1000 = 14.411 gm-mole. Thus the solubility product $K_{sp}$ of AgBr = [ $Ag^{+}$ ]× $Br^{-}$ .

Or, 5.0× $10^{-13}$ = $S^{2}$ (the given value of solubility product of AgBr is 5.0× $10^{-13}$ and the charge of the both ions are same).

Thus S = (5.00× $10^{-13}$ ) $^{1/2}$ = 7.071× $10^{-7}$ g/mL.

Thus the concentration of Br $^{-1}$ or HBr is 7.071× $10^{-7}$ g/mL.

4 0

4 years ago