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3 years ago

O2 + C3H60 → H2O + CO2 Balance please

Chemistry

1 answer:

irinina [24]3 years ago

7 0

Answer:

C3H6O + 4O2 → 3CO2 + 3H2O

Explanation:

I think that's the answer

Hope this helps :)

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Which of the following statements is TRUE about the loose rocks on the surface

Ivan

Answer:B

Explanation:

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3 years ago

Read 2 more answers

If the solubility of a gas in water at 25°C is 22.25 mg/L at 2.5 atm of pressure, what is its solubility (in mg/L) at 25°C

tamaranim1 [39]

Answer:

8.9 mg/l

Explanation: Temp doesnt matter so throw that out automatically then your equation is;

S1/P1=S2/P2

We are looking for S2 and that equation is;

S2=S1*P2/P1 and that is S2=22.25*1/2.5

A little bit of simple math and you get your answer: 8.9 mg/l

8 0

2 years ago

Write a balanced chemical equation for the reaction that occurs when calcium metal undergoes a combination reaction with O2(g)?

ivann1987 [24]

When calcium joins with calcium its follows the following equation:

2Ca + O2 = 2CaO

hope that helps

8 0

3 years ago

Find percent yield:

saveliy_v [14]

Answer: The percent yield of the reaction is 91.8 %

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For $B_5H_9$ :

Given mass of $B_5H_9$ = 4.0 g

Molar mass of $B_5H_9$ = 63.12 g/mol

Putting values in equation 1, we get:

$\text{Moles of }B_5H_9=\frac{4g}{63.12g/mol}=0.0634mol$

For oxygen gas:

Given mass of oxygen gas = 10.0 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

$\text{Moles of oxygen gas}=\frac{10g}{32g/mol}=0.3125mol$

The chemical equation for the reaction of $B_5H_9$ and oxygen gas follows:

$2B_5H_9+12O_2\rightarrow 5B_2O_3+9H_2O$

By Stoichiometry of the reaction:

12 moles of oxygen gas reacts with 2 moles of $B_2H_5$

So, 0.3125 moles of oxygen gas will react with = $\frac{2}{12}\times 0.3125=0.052mol$ of $B_2H_5$

As, given amount of $B_2H_5$ is more than the required amount. So, it is considered as an excess reagent.

Thus, oxygen gas is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

12 moles of oxygen gas produces 5 moles of $B_2O_3$

So, 0.3125 moles of oxygen gas will produce = $\frac{5}{12}\times 0.3125=0.130moles$ of water

Now, calculating the mass of $B_2O_3$ from equation 1, we get:

Molar mass of $B_2O_3$ = 69.93 g/mol

Moles of $B_2O_3$ = 0.130 moles

Putting values in equation 1, we get:

$0.130mol=\frac{\text{Mass of }B_2O_3}{69.63g/mol}\\\\\text{Mass of }B_2O_3=(0.130mol\times 69.63g/mol)=9.052g$

To calculate the percentage yield of $B_2O_3$ , we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of $B_2O_3$ = 8.32 g

Theoretical yield of $B_2O_3$ = 9.052 g

Putting values in above equation, we get:

$\%\text{ yield of }B_2O_3=\frac{8.32g}{9.052g}\times 100\\\\\% \text{yield of }B_2O_3=91.8\%$

Hence, the percent yield of the reaction is 91.8 %

6 0

3 years ago

Help pls! Answer all of it. Idk if what I wrote is correct.

trasher [3.6K]

The reaction is actually endothermic because delta H is positive, indicating that it absorbing heat.

8 0

3 years ago