Density is an example of what type of property?Intensive

Extensive

Chemical

Biological

Answers

Answer 1

Answer: Density is

Mass/Volume

both these are extensive because they change with different sizes and shapes of mass.

Density is a intensive property because the change had/had already been made so it is intensive. Density would be an intensive property.

Answer 2

Answer:

Density is an intensive property, meaning it does not change as the amount of substance changes. It only depends on the type of substance, not its quantity.

Density is an example of an intensive property. Intensive properties don't change as the amount of the substance changes. They depend only on the nature of the substance, not its quantity. Other examples of intensive properties are boiling point, color, and temperature. On the other hand, extensive properties, such as volume and mass, change when the quantity of the substance changes.

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At atom that has only one electron in its valence electron shell is ________ chemically reactive.a. not
b. mildly
c. highly
d. The name of the element is needed to figure this out.

Answers

The correct answer is C. An atom having only one valence electron is highly chemically reactive. The valence electron says a lot to the reactivity of an atom. For example, sodium has only one valence electron which makes it so reactive and unstable. The alkali metals have only 1 valence electron.

Answer:The correct answer is C. Highly

Explanation:

its correct because I got it correct on my test sooo..yuup.!

-hope this helps.!!

Salt water is classified as a1) compound because the proportion of its atoms is fixed
2) compound because the proportion of its atoms can vary
3) mixture because the proportion of its components is fixed
4) mixture because the proportion of its components can vary

Answers

Salt Water is classified as a:

4) mixture because the proportion of its components can vary.

Salt water is composed of salt and water. Regardless of the amount of salt or water put together, these two still form salt water.

Answer:

4) mixture

Explanation:

The answer is 4) because salt water has 2 properties which are salt and water. Never the less the amount of salt or water you keep adding the substance would always remain as salt water.

If 62.0 grams of magnesium metal (Mg) react with 55.5 grams of oxygen gas (O2) in a synthesis reaction, how many grams of the excess reactant will be left over when the reaction is complete? Be sure to write out the balanced equation for this reaction and to show all of your work.

Answers

The balanced chemical equation would be as follows:

Mg + O2 → MgO2

We are given the amounts of the reactants. We need to determine first which one is the limiting reactant. We do as follows:

62.0 g Mg (1 mol / 24.31 g ) = 2.55 mol Mg
55.5 g O2 ( 1 mol / 32 g ) = 1.74 mol O2 -----> consumed completely and therefore the limiting reactant

2.55 mol - 1.74 mol O2( 1 mol Mg / 1 mol O2 ) = 0.81 mol Mg excess

Answer:

14.1648 grams of oxygen gas will be left.

Explanation:

$2Mg +O_2\rightarrow 2MgO$

Moles of magnesium metal = $(62.0 g)/(24 g/mol)=2.5833 mol$

Moles of oxygen gas = $(55.5 g)/(32 g/mol)=1.7343 mol$

According to reaction, 2 mol of magnesium react with 1 mol of oxygen gas .

Then 2.5833 moles of magnesium will react with:

$(1)/(2)* 2.5833 mol=1.29165 mol$ of oxygen gas.

Moles of oxygen left unreacted =1.7343 mol - 1.29165 mol = 0.44265 mol

Oxygen gas is an excessive reagent.

Mass of 0.44265 moles of oxygen gas:

0.44265 mol × 32 g/mol = 14.1648 g

14.1648 grams of oxygen gas will be left.

What is the molarity of a solution of 14.0 g NH4Br in enough H2O to make 150 mL of solution?

Answers

Explanation:

Molarity is defined as the number of moles per liter of solution.

Mathematically, Molarity = $(no. of moles)/(volume of solution in liter)$

Since it is given that the molarity of a solution of 14.0 g and volume is 150 mL or 0.15 L.

Whereas number of moles = $(mass)/(molar mass)$

So, molar mass of $NH_(4)Br$ is 97.94 g/mol.

Thus, number of moles = $(14.0 g)/(97.94 g/mol)$

= 0.142 mol

Therefore, calculate the molarity as follows.

Molarity = $(no. of moles)/(volume of solution in liter)$

= $(0.142 mole)/(0.15 L)$

= 0.946 mol/L

Hence, we can conclude that molarity of the solution is 0.946 mol/L.

Final answer:

The molarity of the solution is approximately 0.952 M.

Explanation:

To calculate the molarity of the solution, we need to convert the given mass of NH4Br to moles. The molar mass of NH4Br is 97.94 g/mol. So, 14.0 g of NH4Br is equal to 0.143 mol. Next, we convert the given volume of the solution to liters, which is 0.150 L. Finally, we divide the number of moles by the volume in liters to find the molarity.

Molarity (M) = moles/volume (L)

Therefore, the molarity of the solution is approximately 0.952 M.

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how many milliliters of 1.50 m hno3 contain enough nitric acid to dissolve an old copper penny with a mass of 3.94 g?

Answers

Molar mass HNO₃ = 63.0 g/mol

number of moles = 3.94 / 63.0 => 0.0625 moles

Volume = moles / molarity

V = 0.0625 / 1.50

V = 0.04166 L x 1000 = 41.66 mL

hope this helps!

Final answer:

To dissolve a 3.94g copper penny, we would need approximately 331 milliliters of 1.50 M HNO3, taking into consideration the stoichiometry of the copper-nitric acid reaction.

Explanation:

Stoichiometry is required to solve this chemistry problem. The balanced equation for the reaction between copper and nitric acid is: 3Cu(s) + 8HNO3(aq) → 2NO(g) + 3Cu(NO3)2(aq) + 4H2O(l)

From this, we can conclude that every mole of copper reacts with eight moles of nitric acid. The molecular weight of copper is approximately 63.546 g/mole, so the moles of copper in an old penny can be calculated by mass (3.94 g) divided by the molar mass (63.546 g/mol).

This calculation shows that there are approximately 0.062 moles of copper in a penny. Since we determined that eight moles of nitric acid are required to react with one mole of copper, eight times the moles of copper gives the moles of nitric acid required, which is 0.496 moles.

Since molarity (M) = moles/liters, we can find out the volume of 1.50 M HNO3 required by rearranging the formula to be Liters= moles/Molarity. So, 0.496 moles divided by 1.50 M gives roughly 0.331 liters, which can be converted to approximately 331 milliliters.