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jek_recluse [69]
4 years ago
15

Write the equation, Aluminum metal reacts with oxygen gas to produce aluminum oxide solid.

Chemistry
1 answer:
Alenkasestr [34]4 years ago
8 0
The equation is:4Al+3O2----2Al2O3
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igor_vitrenko [27]
B- 90 grams bc 45 mL times 2 equals 90
3 0
3 years ago
The Concentration of C6H12O6 May be represented as?
Amiraneli [1.4K]

Answer:

A 12 oz Coca Cola contains 39g of sugar or C6H12O6. 

To calculate for the molarity of sugar in the soda, convert 39 grams of sugar to moles sugar:

39g/ 180.16 g/mol = 0.216 mol sugar

then, convert 12 oz to L:

12oz / (1oz/0.02957L) = 0.35484 L

therefore the concentration of sugar in the soda is:

M = mol sugar / L sol'n

   = 0.216 mol sugar / 0.35484 L

   = 0.609 M

Explanation:

6 0
3 years ago
Read 2 more answers
The solubility of o2 at 20c is 1.38 x10^-3. the partial presure of o2 in the air at sea level is 0.27 atm. using henery;s law, c
netineya [11]

<u>Answer:</u> The solubility of oxygen at 682 torr is 4.58\times 10^{-3}M

<u>Explanation:</u>

To calculate the molar solubility, we use the equation given by Henry's law, which is:

C_{A}=K_H\times p_{A}

Or,

\frac{C_{1}}{C_{2}}=\frac{p_{1}}{p_2}

where,

C_1\text{ and }p_1 are the initial concentration and partial pressure of oxygen gas

C_2\text{ and }p_2 are the final concentration and partial pressure of oxygen gas

We are given:

Conversion factor used:  1 atm = 760 torr

C_1=1.38\times 10^{-3}M\\p_1=0.27atm\\C_2=?\\p_2=682torr=0.897atm

Putting values in above equation, we get:

\frac{1.38\times 10^{-3}}{C_2}=\frac{0.27atm}{0.897atm}\\\\C_2=\frac{1.38\times 10^{-3}\times 0.897atm}{0.27atm}=4.58\times 10^{-3}M

Hence, the solubility of oxygen gas at 628 torr is 4.58\times 10^{-3}M

4 0
3 years ago
What is the electron configuration of bromine whose atomic number is 35
Vanyuwa [196]
1s22s22p63s23p64s23d104p5 is the answer
5 0
3 years ago
If 15 grams of Carbon dioxide is produced in a chemical reaction, how many grams of Carbon must be consumed in the reaction if w
irinina [24]

Answer:

4.13 g

Explanation:

Data Given:

Amount of CO₂ Produced = 15 g

Amount of Oxygen = 11 g

Amount of Carbon used = ?

Solution:

Suppose Carbon dioxide (CO₂) is formed by the reaction of carbon and oxygen then the reaction will be as below

                            C   +   O₂    -------------> CO₂

                          1 mol    1 mol                  1 mol

we come to know from the above reaction that

1 mole of carbon react with 1 mole of oxygen to produce 1 mol of carbon dioxide.

molar mass of C = 12 g/mol

molar mass of O₂ = 32 g/mol

molar mass of CO₂ = 12 + 2(16) = 44 g/mol

if we represent mole in grams then

           C               +                        O₂                     ------------->        CO                 1 mol (12 g/mol)                      1 mol (32 g/mol)                      1 mol (44 g/mol)

                   

              C   +   O₂    -------------> CO₂

            12 g       32 g                   44 g

So,

we come to know that 32 g of Oxygen combine with 12 g  of oxygen produce 44 g CO₂

So now how much of Carbon will be combine with 11 g of oxygen

apply unity formula

                32 g of  O₂ ≅ 12 g of  C

                  11 g of O₂  ≅  g of  C

by doing cross multiplication

           g of C = 12 g x 11 g / 32 g

           g of C = 132 g / 32 g

           g of C = 4.13 g

So,

4.13 g of carbon will consume to produce 15 g of Carbon dioxide.

to check this answer

we use the above information

                     12 g of  C ≅ 44 g of CO₂

                     4.13 g of C ≅  g of  CO₂

by doing cross multiplication

                    g of  CO₂ = 44 g x 4.13 g / 12 g

                    g of CO₂ = 15g

So it is confirmed that

4.13 g of carbon will consume to produce 15 g of Carbon dioxide.

4 0
3 years ago
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