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3 years ago

Aluminum can react with oxygen gas to produce aluminum oxide, what time of reaction is this?

Chemistry

1 answer:

Helga [31]3 years ago

3 0

This is known as a synthesis reaction. A+B->C
Hope this helps!

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A material found in air, water, or soil that is harmful to humans or other organisms

UNO [17]

The answer is pollutant

3 0

3 years ago

How many types of hybridization does carbon undergo?

Novosadov [1.4K]

Answer:

For carbon the most important forms of hybridization are the sp2- and sp3- hybridization. Besides these structures there are more possiblities to mix dif- ferent molecular orbitals to a hybrid orbital. An important one is the sp- hybridization, where one s- and one p-orbital are mixed together.

3 0

3 years ago

A certain sample of coal contains 1.60 percent sulfur by mass. when the coal is burned, the sulfur is converted to sulfur dioxid

alina1380 [7]

<span>1.40 x 10^5 kilograms of calcium oxide The reaction looks like SO2 + CaO => CaSO3 First, determine the mass of sulfur in the coal 5.00 x 10^6 * 1.60 x 10^-2 = 8.00 x 10^4 Now lookup the atomic weights of Sulfur, Calcium, and Oxygen. Sulfur = 32.065 Calcium = 40.078 Oxygen = 15.999 Calculate the molar mass of CaO CaO = 40.078 + 15.999 = 56.077 Since 1 atom of sulfur makes 1 atom of sulfur dioxide, we don't need the molar mass of sulfur dioxide. We merely need the number of moles of sulfur we're burning. divide the mass of sulfur by the atomic weight. 8.00 x 10^4 / 32.065 = 2.49 x 10^3 moles Since 1 molecule of sulfur dioxide is reacted with 1 molecule of calcium oxide, just multiply the number of moles needed by the molar mass 2.49 x 10^3 * 56.077 = 1.40 x 10^5 So you need to use 1.40 x 10^5 kilograms of calcium oxide per day to treat the sulfur dioxide generated by burning 5.00 x 10^6 kilograms of coal with 1.60% sulfur.</span>

5 0

3 years ago

25 grams of radon-222 remains after 15.28 days. How much radon was in the original sample? Write an equation for the decay of ra

Crazy boy [7]

Answer:

half-life = 3.8 days

total time of decay = 15.2 days

initial amount = 100. g

number of half-lives past: 15.2/3.8 = 4 half-lives

4 half-lives = 1/16 remains

100. g x 1/16 = 6.25 g

8 0

3 years ago

Two samples of sodium chloride with different masses were decomposed into their constituent elements. One sample produced 1.731.

Amanda [17]

Answer:

The answer to your question is letter c) 6.09 g of sodium and 9.38 g of chlorine.

Explanation:

This problem is solve using rule of three

We know that the proportion Sodium to Chloride is 1 to 1 in sodium chloride, so we have to look for this proportion in the options

AM Sodium = 23 g

AM Chlorine = 35.5 g

Sodium Chlorine

23 g ---------------- 1 mol 35.5 g -------------- 1 mol

1713.73g ------------- x 2666.6 g ------------- x

x = 1713.73/23 = 74.51 x = 2666.6/35.5 = 75.12

These values are very similar, we have to look for the proportion in the options

a) 6.09g of sodium = 0.26 mol

4,87 g of chlorine = 0.14 mol These numbers are not very similar

b) We have 0.26 mol of Na

0.037 mol of Cl This is not the answer

c) We have 0.26 mol of Na

0.26 mol of Cl These numbers are the same, the proportion is 1:1, this is the answer

d) We have 0.26 mol of Na

0.36 mol of Cl This is not the answer

8 0

3 years ago