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3 years ago

What are 5 steps of balancing a chemical equation plz help me

Chemistry

1 answer:

nlexa [21]3 years ago

6 0

Answer:

Explanation:

Identify each element found in the equation. The number of atoms of each type of atom must be the same on each side of the equation once it has been balanced.

What is the net charge on each side of the equation? The net charge must be the same on each side of the equation once it has been balanced.

If possible, start with an element found in one compound on each side of the equation. Change the coefficients (the numbers in front of the compound or molecule) so that the number of atoms of the element is the same on each side of the equation. Remember, to balance an equation, you change the coefficients, not the subscripts in the formulas.

Once you have balanced one element, do the same thing with another element. Proceed until all elements have been balanced. It's easiest to leave elements found in pure form for last.

Check your work to make certain the charge on both sides of the equation is also balanced.

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The bond between sodium and chlorine in the compound sodium chloride (NaCl) is a/an:

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3 years ago

A compound distributes between benzene (solvent 1) and water (solvent 2) with a distribution coefficient, K = 2.7. If 1.0g of th

mote1985 [20]

Explanation:

The given data is as follows.

Solvent 1 = benzene, Solvent 2 = water

$K_{p}$ = 2.7, $V_{S_{2}}$ = 100 mL

$V_{S_{1}}$ = 10 mL, weight of compound = 1 g

Extract = 3

Therefore, calculate the fraction remaining as follows.

$f_{n} = [1 + K_{p}(\frac{V_{S_{2}}}{V_{S_{1}}})]^{-n}$

= $[1 + 2.7(\frac{100}{10})]^{-3}$

= $(28)^{-3}$

= $4.55 \times 10^{-5}$

Hence, weight of compound to be extracted = weight of compound - fraction remaining

= 1 - $4.55 \times 10^{-5}$

= 0.00001

or, = $1 \times 10^{-5}$

Thus, we can conclude that weight of compound that could be extracted is $1 \times 10^{-5}$ .

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3 years ago

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Which sentence describes an example of sublimation?

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4 years ago

why would it not be necessary to heat the air in the balloon as much if the air outside the balloon is colder?

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Consider a pot of water at 100 C. If it took 1,048,815 J of energy to vaporize the water and heat it to 135 C, how many grams of

jeka57 [31]

Answer:

There was 450.068g of water in the pot.

Explanation:

Latent heat of vaporisation = 2260 kJ/kg = 2260 J/g = L

Specific Heat of Steam = 2.010 kJ/kg C = 2.010 J/g = s

Let m = x g be the weight of water in the pot.

Energy required to vaporise water = mL = 2260x