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2 years ago

Chemical equation=CH4 + O2 = CO2 + H2O.

Chemistry

1 answer:

tankabanditka [31]2 years ago

4 0

Answer:

1.The substance(s) to the left of the arrow in a chemical equation are called reactants. A reactant is a substance that is present at the start of a chemical reaction. The substance(s) to the right of the arrow are called products . A product is a substance that is present at the end of a chemical reaction.The combustion of methane or octane is exothermic; it releases energy. CH4 + 2 O2 → CO2 + 2 H2O + energy The energies of the products are lower than the energiies of the reactants.

Explanation:

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Weight of one mole of carbon = 12.01 g Weight of one mole of oxygen = 16.00 g The molecular weight (gram formula weight) for CO

murzikaleks [220]

Answer:

28.01g

Explanation:

Given the weight of one mole of Cabon as 12.01g and that of oxygen as 16.00g.

The molecular weight of a compound can be gotten by adding the molar weights of the elements that constitutes the compound .

The molecular weight of the compound CO is therefore

equal to the sum of the weight of both elements.

That’s = 12.01g + 16.00g

= 28.01g

Therefore, the molecular weight of CO is 28.01g

4 0

3 years ago

PLZ HELP IM GIVING 50 FRICKING POINTS. NO WRONG ANSWERS PLZ

Liula [17]

Answer:

I think it's B

Explanation:

I dont have much experience with the periodic table, but I just think its B.

3 0

2 years ago

Upon balancing the equation what is the smallest possible integers that goes in front O₂? HBr + O₂ ➞ H₂O + Br₂ *

Leokris [45]

Answer:

Explanation:

4 HBr + O2 → 2H 20 + 2Br 2

...............

7 0

2 years ago

Read 2 more answers

1.this substance is a natural non-living material with a uniform structure throughout

vovikov84 [41]

1.Mineral
2.Igneous
3.Composition

7 0

3 years ago

Read 2 more answers

From the value Kf=1.2×109 for Ni(NH3)62+, calculate the concentration of NH3 required to just dissolve 0.016 mol of NiC2O4 (Ksp

Nina [5.8K]

<u>Answer:</u> The concentration of $NH_3$ required will be 0.285 M.

<u>Explanation:</u>

To calculate the molarity of $NiC_2O_4$ , we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Moles of $NiC_2O_4$ = 0.016 moles

Volume of solution = 1 L

Putting values in above equation, we get:

$\text{Molarity of }NiC_2O_4=\frac{0.016mol}{1L}=0.016M$

For the given chemical equations:

$NiC_2O_4(s)\rightleftharpoons Ni^{2+}(aq.)+C_2O_4^{2-}(aq.);K_{sp}=4.0\times 10^{-10}$

$Ni^{2+}(aq.)+6NH_3(aq.)\rightleftharpoons [Ni(NH_3)_6]^{2+}+C_2O_4^{2-}(aq.);K_f=1.2\times 10^9$

Net equation: $NiC_2O_4(s)+6NH_3(aq.)\rightleftharpoons [Ni(NH_3)_6]^{2+}+C_2O_4^{2-}(aq.);K=?$

To calculate the equilibrium constant, K for above equation, we get:

$K=K_{sp}\times K_f\\K=(4.0\times 10^{-10})\times (1.2\times 10^9)=0.48$

The expression for equilibrium constant of above equation is:

$K=\frac{[C_2O_4^{2-}][[Ni(NH_3)_6]^{2+}]}{[NiC_2O_4][NH_3]^6}$

As, $NiC_2O_4$ is a solid, so its activity is taken as 1 and so for $C_2O_4^{2-}$

We are given:

$[[Ni(NH_3)_6]^{2+}]=0.016M$

Putting values in above equations, we get:

$0.48=\frac{0.016}{[NH_3]^6}}$

$[NH_3]=0.285M$

Hence, the concentration of $NH_3$ required will be 0.285 M.

7 0

2 years ago