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Maksim231197 [3]
3 years ago
7

What product(s) are formed during the complete combustion reaction that occurs when methane (CH4) and molecular oxygen (O2) reac

t? CO2 and H4 C and H2O CO2 and H2O C(OH)4
Chemistry
1 answer:
Vilka [71]3 years ago
3 0

Answer: CO2 and H2O

Explanation: I already took the test it's right

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At STP, which sample contains the same number of molecules as 11.2 liters of CO2(g) at STP?
Lana71 [14]
The answer is (3) 11.2 L of N2(g). When the choices are all gases, the number of molecules depends on the molar number of the gas. Under the condition of STP, the molar number of a gas is depends on volume only. So the same volume of gas has the same number of molecules.
8 0
4 years ago
Solid aluminum and gaseous oxygen react in a combination reaction to produce aluminum oxide: 4Al (s) + 3O2 (g) → 2Al2O3 (s) In
jek_recluse [69]

Answer: The percent yield of the reaction is 74 %

Explanation:

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

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

\text{Moles of aluminium}=\frac{2.5g}{27g/mol}=0.092mol

For oxygen gas:

\text{Moles of oxygen gas}=\frac{2.5g}{32g/mol}=0.078mol

The chemical equation for the reaction of titanium and chlorine gas follows:

4Al(s)+3O_2(g)\rightarrow 2Al_2O_3(s)

By Stoichiometry of the reaction:

4 moles of aluminium reacts with 3 moles of oxygen.

So, 0.092 moles of aluminium reacts with = \frac{3}{4}\times 0.092=0.069mol of oxygen

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

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

By Stoichiometry of the reaction:

4 moles of aluminium produce = 2 moles of Al_2O_3

So, 0.092 moles of aluminium will produce = \frac{2}{4}\times 0.092=0.046moles of Al_2O_3

Now, calculating the mass of aluminium oxide:

\text{Mass of aluminium oxide}=moles\times {\text {molar mas}}=0.046mol\times 102g/mol=4.7g

To calculate the percentage yield of titanium (IV) chloride, we use the equation:

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

Experimental yield  = 3.5 g

Theoretical yield = 4.7 g

Putting values in above equation, we get:

\%\text{ yield of reaction}=\frac{3.5g}{4.7g}\times 100\\\\\% \text{yield of reaction}=74\%

Hence, the percent yield of the reaction is 74 %

6 0
3 years ago
Hello, please help tysmmmm!!
Fed [463]

Answer:

2.07

Explanation:

1870=(20.01)(c)(45.2)

1870 = 904.452c

c = 2.07 J/g C (to 3 sf]

3 0
2 years ago
Can somebody help me this one? i have thr answer but i'm not sure
Brilliant_brown [7]
I believe your answer is correct. A) Solar Cell
7 0
3 years ago
Read 2 more answers
Who was the first person to show strong empirical evidence for the existence of atoms?.
Mekhanik [1.2K]

Given what we know, we can confirm that John Dalton was the first person to show strong empirical evidence for the existence of atoms.

<h3>Who was John Dalton?</h3>
  • He was a renowned scientist with knowledge in many fields.
  • He was known to be a chemist, meteorologist, and physicist.
  • He proposed the atomic theory and carried out experiments to provide supporting evidence.

Therefore, given his proposal of the atomic theory and the experiments he carried out to provide evidence to support his claims, we can confirm that John Dalton was the first person to show strong empirical evidence for the existence of atoms.

To learn more about atoms visit:

brainly.com/question/13981855?referrer=searchResults

6 0
2 years ago
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