A ____ has a mass of almost zero<br> A. Proton<br> B.Neutron<br> C. Electron<br> D.lon

Can anyone do my Ap environmental science package I will pay you

Phoenix [80]

Answer:

i have taken environmental science so i can help

4 0

2 years ago

An aqueous solution is listed as being 33.8% solute by mass with a density of 1.15 g/mL, the molar mass of the solute is 145.6 g

vodomira [7]

Answer:

A) 2.69 M

B) 0.059

Explanation:

A) We have:

33.8% solute by mass= 33.8 g solute/100 g solution

molarity = mol solute/ 1 L solution

molarity= $\frac{33.8 g solute}{100 g solution}$ x $\frac{1.15 g solution}{1 ml}$ x $\frac{1 mol solute}{145.6 g solute}$ x $\frac{1000 ml}{1 L}$

molarity= 2.69 mol solute/L solution = 2.69 M

B) We know that there are 33.8 g of solute in 100 g of solution.

As the total solution is compounded by solute+solvent (in this case, solvent is water), the mass of water is the difference between the mass of the total solution and the mass of solute:

mass of water= 100 g - 33.8 g = 66.2 g

Now, we calculate the number of mol of both solute and water:

mol solute= 33.8 g solute x $\frac{1 mol solute}{145.6 g}$ = 0.232 mol

mol H20= 66.2 g H₂O x $\frac{1 mol H2O}{18 g}$

Finally, the mol fraction of solute (Xsolute) is calculated as follows:

Xsolute= $\frac{mol solute}{total mol}= \frac{mol solute}{mol solute + mol H2O}=\frac{0.232 mol}{0.232 mol + 3.677 mol}$

Xsolute= 0.059

4 0

2 years ago

How many atoms are in 52.3 g of lithium hypochlorite (LiClO)?

garri49 [273]

Answer:

1.62 × 10²⁴ atoms are in 52.3 g of lithium hypochlorite.

Explanation:

To find the amount of atoms that are in 52.3 g of lithium hypochlorite, we must first find the amount of moles. We do this by dividing by the molar mass of lithium hypochlorite.

52.3 g ÷ 58.4 g/mol = 0.896 mol

Next we must find the amount of formula units, we do this be multiplying by Avagadro's number.

0.896 mol × 6.02 × 10²³ = 5.39 × 10²³ f.u.

Now to get the amount of atoms we can multiply the amount of formula units by the amout of atoms in one formula unit.

5.39 × 10²³ f.u. × 3 atom/f.u. = 1.62 × 10²⁴ atoms

1.62 × 10²⁴ atoms are in 52.3 g of lithium hypochlorite.

7 0

3 years ago

As a substance changes fees from gas to liquid to solid :

Korolek [52]

I would say that the density increases because the distance between the molecules decreases, and therefore becomes smaller, and vice versa the density increases, and that is a physical change not a chemical change.

3 0

2 years ago

Read 2 more answers

For the following reaction, 53.7 grams of iron(III) oxide are allowed to react with 22.8 grams of aluminum. iron(III) oxide (s)

ZanzabumX [31]

Answer:

34.23 grams is the maximum amount of aluminum oxide that can be formed.

Iron (III) oxide is a limiting reagent i.e $Fe_2O_3$ .

4.6764 grams is the amount of the aluminum which remains after the reaction is complete

Explanation:

iron(III) oxide (s) + aluminum (s) → aluminum oxide (s) + iron (s)

$Fe2O_3(s)+2Al(s)\rightarrow Al_2O_3(s)+2Fe(s)$

Moles of iron(III) oxide : $\frac{53.7 g}{160 g/mol}=0.3356 mol$

Moles of aluminium : $\frac{22.8 g}{27 g/mol}=0.8444 mol$

According to recation, 1 mole of iron(III) oxide reacts with 2 moles of aluminum.

Then 0.3356 moles of iron(III) oxide will react with:

$\frac{2}{1}\times 0.3356 mol=0.6712 mol$ of aluminum.

As we can see that moles of iron(III) are in limiting amount.Hence iron(III) oxide is a limiting reagent i.e $Fe_2O_3$ and aluminum in the an excessive reagent.

Amount of aluminum oxide will depend upon moles of limiting reagent that is iron(III) oxide.

According to reaction , 1 mole iron(III) oxide gives 1 moles of aluminum oxide.

Then 0.3356 moles will give:

$\frac{1}{1}\times 0.3356 mol=0.3356 mol$ of aluminum oxide

Mass of 0.3356 moles of aluminum oxide:

0.3356 mol × 102 g/mol = 34.23 g

34.23 grams is the maximum amount of aluminum oxide that can be formed.

Moles of excessive reagent left = 0.8444 mol - 0.6712 mol = 0.1732 mol

Mass of 0.1732 moles of aluminum :

0.1732 mol × 27 g/mol = 4.6764 g

4.6764 grams is the amount of the aluminum which remains after the reaction is complete

7 0

3 years ago

A ____ has a mass of almost zero A. Proton B.Neutron C. Electron D.lon