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2 years ago

Why are ionic compounds electrically neutral

Chemistry

1 answer:

trasher [3.6K]2 years ago

5 0

Answer:

they are electrically neutral because they form ionic bonds between positively and negatively charged atoms!

hope i helped :)

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HEY I NEED ALL ANSWERS RIGHT NOW IT'S MISSING AND MY TEACHER ARE GRADING THEM NOW YOU WILL BE MARKED BRAINIEST

emmainna [20.7K]

Molar mass of butane:-

4(12)+10=58u

Moles of Butane:-

100/58=1.7mol

#16

2mols of Butane produce 8mol CO_2
1mol of butane produces 4mol CO_2

Moles of CO_2

4(1.7)=6.8mol

Mass of CO_2:-

44(6.8)=299.2g

#17

2mols of butane need 13mol O_2
1mol of butane needs 6.5mol O_2

Moles of O_2

6.5(1.7)=11.05mol

Mass of O_2

11.05(32)=353.6g

7 0

2 years ago

Which statement is true about elements?

ipn [44]

I believe the answer is a

7 0

3 years ago

Compare melting point freezing point and boiling point

Tresset [83]

Melting point- the temperature at which a substance has changed from a solid to a liquid
freezing- the temperature at which a substance chanes from liquid to a solid
boiling point- the temperature at which a substance changes from a liquid to a gas phase

4 0

2 years ago

How many molecules of propane were in the erlenmeyer flask? Avogadro's number is 6. 022 × 10^23 molecules/mol

Ede4ka [16]

3.74× $10^{21}$

3.74 × $10^{21}$ molecules of propane were in the erlenmeyer flask.

number of moles of propane can be calculated as moles of propane.

mass of propane = 0.274 g

molar mass of propane = 44.1

So this gives us the value of 6.21× $10^{-3}$ moles of propane

No one mole of propane As a 6.0-2 × $10^{23}$

so, 6.21 × $10^{-3}$ × 6. 022 × 10^23

= 3.74 × $10^{21}$

Therefore, molecules of propane were in the erlenmeyer flask is found to be 3.74 × $10^{21}$

<h3>What is erlenmeyer flask?</h3>

A laboratory flask with a flat bottom, a conical body, and a cylindrical neck is known as an Erlenmeyer flask, sometimes known as a conical flask or a titration flask.
It bears the name Emil Erlenmeyer after the German chemist.

<h3>What purpose does an Erlenmeyer flask serve?</h3>

Liquids are contained in Erlenmeyer flasks, which are also used for mixing, heating, chilling, incubating, filtering, storing, and other liquid-handling procedures.

For titrations and boiling liquids, their sloped sides and small necks make it possible to whirl the contents without worrying about spills.

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4 0

1 year ago

Write both answers to at least two decimal places. Calculate the pH of a 0.160 M solution of KOH.Part 2 (1 point) Calculate the

Evgen [1.6K]

To calculate the pH of a solution, we first need to find the concentration of hydronium ions in the solution. Since KOH is a strong base, it dissociates completely in water to produce hydroxide ions (OH-) and potassium ions (K+).

The concentration of hydronium ions in a solution of KOH can be calculated using the concentration of hydroxide ions and the equilibrium constant for water, which is equal to 1.00 x 10^-14 at 25 degrees Celsius.

The concentration of hydroxide ions in a 0.160 M solution of KOH is equal to the concentration of KOH, which is 0.160 M. The concentration of hydronium ions in the solution can be calculated using the equation below:

[H3O+] = (1.00 x 10^-14) / [OH-]

Substituting the concentration of hydroxide ions into the equation above, we get:

[H3O+] = (1.00 x 10^-14) / (0.160 M) = 6.25 x 10^-13 M

To calculate the pH of the solution, we need to take the negative logarithm of the concentration of hydronium ions. This can be done using the equation below:

pH = -log([H3O+])

Substituting the concentration of hydronium ions into the equation above, we get:

pH = -log(6.25 x 10^-13) = 12.20

The pH of a 0.160 M solution of KOH is 12.20.

To calculate the pOH of a solution, we first need to find the concentration of hydroxide ions in the solution. Since we already calculated this value above, we can simply use the concentration of hydroxide ions we found earlier: 0.160 M.

To calculate the pOH of the solution, we need to take the negative logarithm of the concentration of hydroxide ions. This can be done using the equation below:

pOH = -log([OH-])

Substituting the concentration of hydroxide ions into the equation above, we get:

pOH = -log(0.160 M) = 1.80

The pOH of a 0.160 M solution of KOH is 1.80.

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7 0

6 months ago