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3 years ago

How many grams of Aluminum Oxide are produced when 3.64 g of Aluminum reacts with an excess of Oxygen?

Chemistry

2 answers:

Vladimir79 [104]3 years ago

6 0

Answer:

<h2>6.877649 or <u>6.88 grams of Aluminum Oxide</u></h2>

Explanation:

- If there are 4 moles of Al that equals 107.926156 grams of 4Al.

- If there is 2 moles of Al2O3 that equals 203.9226 grams.

So...

If 107.926156 grams of 4Al equals 203.9226 grams of 2Al2O3 then, 3.64g of Al equals x.

fgiga [73]3 years ago

4 0

55664 grams of aluminum oxide gag

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MaRussiya [10]

Answer:

Molecules in the hot water bottle are

moving faster than molecules in the skin.

Explanation:

If the molecules in the skin were to be moving faster than the water then the water would seem to be cold.

5 0

3 years ago

1. Which of the combinations in the lab activity had indications that a chemical change occured? Defend your argument with evide

ArbitrLikvidat [17]

Answer:The green growing on the penny of copper and the rust forming on the nail of iron are chemical changes. Boiling away salt water, scraping iron filings from a mixture of sand with a magnet, and breaking a rock with a hammer, are physical changes.

Explanation:

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3 years ago

A rectangle has a volume of 395cm3. It has a mass of 147g. What is its density? *

LekaFEV [45]

The density of a rectangle : ρ = 0.372 g/cm³

<h3>Further explanation</h3>

Given

The volume of rectangle : 395 cm³

Mass : 147 g

Required

The density

Solution

Density is a quantity derived from the mass and volume

Density is the ratio of mass per unit volume

Density formula:

$\large {\boxed {\bold {\rho ~ = ~ \frac {m} {V}}}}$

ρ = density

m = mass

v = volume

Input the value :

ρ = 147 g : 395 cm³

ρ = 0.372 g/cm³

6 0

3 years ago

Calculate the unit cell edge length for an 85 wt% fe-15 wt% v alloy. All of the vanadium is in solid solution, and, at room temp

Lady bird [3.3K]

Answer is 0.289nm.

Explanation: The wt % of Fe and wt % of V is given for a Fe-V alloy.

wt % of Fe in Fe-V alloy = 85%

wt % of V in Fe-V alloy = 15%

We need to calculate edge length of the unit cell having bcc structure.

Using density formula,

$\rho_{ave}=\frac{Z\times M_{ave}}{a^3\times N_A}$

For calculating edge length,

$a=(\frac{Z\times M_{ave}}{\rho_{ave}\times N_A})^{1/3}$

For calculating $M_{ave}$ , we use the formula

$M_{ave}= \frac{100}{\frac{(wt\%)_{Fe}}{M_{Fe}}+\frac{(wt\%)_{V}}{M_V}}$

Similarly for calculating $(\rho)_{ave}$ , we use the formula

$\rho_{ave}= \frac{100}{\frac{(wt\%)_{Fe}}{\rho_{Fe}}+\frac{(wt\%)_{V}}{\rho_V}}$

From the periodic table, masses of the two elements can be written

$M_{Fe}= 55.85g/mol$

$M_{V}=50.941g/mol$

Specific density of both the elements are

$(\rho)_{Fe}=7.874g/cm^3\\(\rho)_{V}=6.10g/cm^3$

Putting $M_{ave}$ and $\rho_{ave}$ formula's in edge length formula, we get

$a=\left [\frac{Z\left (\frac{100}{\frac{(wt\%)_{Fe}}{M_{Fe}}+\frac{(wt\%)_{Fe}}{M_{Fe}}} \right )}{N_A\left (\frac{100}{\frac{(wt\%)_V}{\rho_V}+\frac{(wt\%)_V}{\rho_V}} \right )} \right ]^{1/3}$

$a=\left [\frac{2atoms/\text{unit cell}\left (\frac{100}{\frac{85\%}{55.85g/mol}+\frac{15\%}{50.941g/mol}} \right )}{(6.023\times10^{23}atoms/mol)\left (\frac{100}{\frac{85\%}{7.874g/cm^3}+\frac{15\%}{6.10g/cm^3}} \right )} \right ]^{1/3}$

By calculating, we get

$a=2.89\times10^{-8}cm=0.289nm$

7 0

3 years ago

A science club made pine wood cars. Each car was set on the same track and then released. The distance traveled was measured. Li

Minchanka [31]

Isn't this a math problem?
If it is the the answer should be 102.

10 decimeters=1 meter

27x10=270
270-168=102

7 0

3 years ago