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statuscvo [17]
3 years ago
5

The reacted side of a balanced chemical equation is shown below. C3H8 + 5O2 How many oxygen atoms should there be on the product

side in the equation?
Chemistry
1 answer:
atroni [7]3 years ago
5 0
10 atoms. If there are 10 in the reactants you need the same number in the products
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27/13Al+4/2He >>>> 30/15P+
Veseljchak [2.6K]

Answer

Explanation:

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

27/13Al+4/2He >>>> 30/15P+

5 0
3 years ago
Consider a closed containing a solid in equilibrium with its vapor. The volume of the solid is much less than that of the contai
Furkat [3]

Answer:

Explanation:

check the attachment below

5 0
2 years ago
The most common source of copper (cu) is the mineral chalcopyrite (cufes2). how many kilograms of chalcopyrite must be mined to
tigry1 [53]

Answer : 0.8663 Kg of chalcopyrite must be mined to obtained 300 g of pure Cu.

Solution : Given,

Mass of Cu = 300 g

Molar mass of Cu = 63.546 g/mole

Molar mass of CuFeS_2 = 183.511 g/mole

  • First we have to calculate the moles of Cu.

\text{ Moles of Cu}=\frac{\text{ Given mass of Cu}}{\text{ Molar mass of Cu}}= \frac{300g}{63.546g/mole}=4.7209moles

The moles of Cu = 4.7209 moles

From the given chemical formula, CuFeS_2 we conclude that the each mole of compound contain one mole of Cu.

So, The moles of Cu = Moles of CuFeS_2 = 4.4209 moles

  • Now we have to calculate the mass of CuFeS_2.

Mass of CuFeS_2 = Moles of CuFeS_2 × Molar mass of CuFeS_2 = 4.4209 moles × 183.511 g/mole = 866.337 g

Mass of CuFeS_2 = 866.337 g = 0.8663 Kg         (1 Kg = 1000 g)

Therefore, 0.8663 Kg of chalcopyrite must be mined to obtained 300 g of pure Cu.


3 0
3 years ago
Calculate number of moles of 4.4g of CO2 and 5.6L of NH3 . Pls help . Its urgent
kupik [55]
Moles of CO₂ = mass / molecular weight
Moles of CO₂ = 4.4 / (12 + 16 x 2)
Moles of CO₂ = 0.1 mol

Each mole of gas occupies 22.4 L at STP. Therefore,
Moles of NH₃ = 5.6 / 22.4
Moles of NH₃ = 0.25 mol
4 0
3 years ago
Please, I need at least accurate answers for my Sound Knowledge Check for my Science class.
Alla [95]

Answer:

1.A

2.A

3.C

4.C

5.B

6.A

5 0
3 years ago
Read 2 more answers
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