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kkurt [141]
3 years ago
12

I need help bad bro i have a 30 cs this is missing

Chemistry
1 answer:
Nat2105 [25]3 years ago
4 0

Answer:  11.02 moles of Ag can be formed.

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of} Cu=\frac{350.00g}{63.55g/mol}=5.51moles

The balanced chemical equation is:

Cu+2AgNO_3\rightarrow Cu(NO_3)_2+2Ag  

According to stoichiometry :

1 moles of Cu produces =  2 moles of Ag

Thus 5.51 moles of Cu will produce=\frac{2}{1}\times 5.51=11.02moles  of Ag  

Thus 11.02 moles of Ag can be formed.

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Carbon needs four more valence electrons, or a total of eight valence electrons, to fill its outer energy level.
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Consider the following unbalanced equation. N2+ O2=N2O. When the equation is completely balanced using smallest whole numbers, t
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Answer:

2

Explanation:

Since you are balancing the equation you need to have equal amounts of each element on each side. Right now there is 2n and 2o on one side, and on the other 2n and 1o. Although the number of n's on each side is equal the number of o's are not. First I balance the number of oxygen's on each below, but doing that made the number of nitrogen's unbalanced. So then I balanced the nitrogens in step 2.

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Combustion of hydrocarbons such as propane ( C3H8 ) produces carbon dioxide, a "greenhouse gas." Greenhouse gases in the Earth's
Naddika [18.5K]

Answer:

1. C₃H₈(g) + 5O₂ (g)  → 3CO₂ (g) + 4H₂O (g)

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Explanation:

Hi there!

The chemical equation is the following:

C₃H₈(g) + 5O₂ (g)  → 3CO₂ (g) + 4H₂O (g)

The molar mass of propane is calculated as follows:

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mass of 1 mol propane 3 · (12 g) + 8 · (1 g) = 44 g

If we have 150 g of propane, we will have (150 g · 1 mol / 44 g) 3.4 mol propane.

Looking at the chemical equation, notice that 1 mol propane produces 3 mol CO₂. Then 3.4 mol of propane will produce:

(3.4 mol C₃H₈ ·  3 mol CO₂ / mol C₃H₈ ) 10.2 mol CO₂

Using the Ideal Gas Law we can obtain the volume of CO₂ produced:

P · V = n · R · T

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P = pressure.

V = volume

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R = gas constant (0.082 atm · l / ( K · mol))

T = temperature in kelvin.

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V = 10.2 mol · 0.082 atm · l/ (K mol) · 285 K / 1 atm

Notice that 12 °C = 273  + 12  = 285 K

V = 238 l

The volume of CO₂ produced is 238 l.

Have a nice day!

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