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3 years ago

What is the difference between a mixture and compound

1 answer:

3 0

A compound is made up of a single type of molecule. A mixture, on the other hand, is made up of different substances that can be distinguished from one another.
As an example:
Water (pure water) is a compound since it's only made up of water molecules
Seawater is a mixture since it is made up of water molecules and soluble salts

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BRAINLIEST AWARD PLZ HELP :)

Vera_Pavlovna [14]

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High melting points
High density
Ductile
Malleable
Good conductors of electricity
Good conductors of heat

I think if you added a proton you would have chlorine.

The noble gasses are the he chemical elements in group 18 of the periodic table. The gasses in this family include helium, neon, argon, krypton, xenon, and radon. All these gasses are colorless are oderless, elements in this family have atoms with a full outer shell of electrons. They are also called inert gasses.

Six valence electrons

6 0

3 years ago

Read 2 more answers

Identify the statement that correctly describes light and how it travels? (2 points)

schepotkina [342]

Light does not travel at a constant speed in a vacuum, compared to in air, because the light is being absorbed by atoms and molecules in the air. But light does travel at a constant speed in a vacuum.
So I agree with A
All that talk about moving forward is irrelevant (I think)

4 0

4 years ago

Read 2 more answers

How many formula units are there in 450 g of Na2S04?

Mashutka [201]

Formula units in 450 g of $Na_{2} So_{4}$ is 1.93 × 10²⁴ formula units.

Explanation:

First we have to find the number of moles in the given mass by dividing the mass by its molar mass as,

$$\frac{450 g}{142.04 g/mol} = 3.2 moles$

Now, we have to multiply the number of moles of Na₂SO₄ by the Avogadro's number, 6.022 × 10²³ formula units/mol, so we will get the number of formula units present in the given mass of the compound.

3.2 mol × 6.022 × 10²³ = 1.93 × 10²⁴ formula units.

So, 1.93 × 10²⁴ formula units is present in 450g of Na₂SO₄.

4 0

3 years ago

What is the volume of 0.80 grams of O2 gas at STP? (5 points) Group of answer choices 0.59 liters 0.56 liters 0.50 liters 0.47 l

Anarel [89]

Answer:

0.56 liters

Explanation:

First we convert 0.80 grams of O₂ into moles, using its molar mass:

0.80 g ÷ 32 g/mol = 0.025 mol

At STP, 1 mol of any given mass occupies 22.4 L. With that information in mind we calculate the volume that 0.025 moles of O₂ gas would occupy:

0.025 mol * 22.4 L/mol = 0.56 L

Thus the answer is 0.56 liters.

3 0

3 years ago

How many moles of sodium carbonate are contained by 57.3g of sodium carbonate

Lady_Fox [76]

Answer:

$\boxed {\boxed {\sf 0.541 \ mol \ Na_2CO_3}}$

Explanation:

We are asked to find how many moles of sodium carbonate are in 57.3 grams of the substance.

Carbonate is CO₃ and has an oxidation number of -2. Sodium is Na and has an oxidation number of +1. There must be 2 moles of sodium so the charge of the sodium balances the charge of the carbonate. The formula is Na₂CO₃.

We will convert grams to moles using the molar mass or the mass of 1 mole of a substance. They are found on the Periodic Table as the atomic masses, but the units are grams per mole instead of atomic mass units. Look up the molar masses of the individual elements.

Na: 22.9897693 g/mol
C: 12.011 g/mol
O: 15.999 g/mol

Remember the formula contains subscripts. There are multiple moles of some elements in 1 mole of the compound. We multiply the element's molar mass by the subscript after it, then add everything together.

Na₂ = 22.9897693 * 2= 45.9795386 g/mol
O₃ = 15.999 * 3= 47.997 g/mol
Na₂CO₃= 45.9795386 + 12.011 + 47.997 =105.9875386 g/mol

We will convert using dimensional analysis. Set up a ratio using the molar mass.

$\frac {105.9875386 \ g \ Na_2CO_3}{1 \ mol \ Na_2CO_3}$

We are converting 57.3 grams to moles, so we multiply by this value.

$57.3 \ g \ Na_2CO_3} *\frac {105.9875386 \ g \ Na_2CO_3}{1 \ mol \ Na_2CO_3}$

Flip the ratio so the units of grams of sodium carbonate cancel.

$57.3 \ g \ Na_2CO_3} *\frac {1 \ mol \ Na_2CO_3}{105.9875386 \ g \ Na_2CO_3}$

$57.3 } *\frac {1 \ mol \ Na_2CO_3}{105.9875386 }$

$\frac {57.3 }{105.9875386 } \ mol \ Na_2CO_3$

$0.5406295944 \ mol \ Na_2CO_3$

The original measurement of moles has 3 significant figures, so our answer must have the same. For the number we found that is the thousandth place. The 6 in the ten-thousandth place to the right tells us to round the 0 up to a 1.

$0.541 \ mol \ Na_2CO_3$

There are approximately 0.541 moles of sodium carbonate in 57.3 grams.

6 0

3 years ago