In the reaction of 6 moles of CH4 with excess oxygen how many moles of water could be produced

1 answer:

6 0

The answer lies in the stoichiometry of the reaction. If u look at the number BEFORE the reagent u will see the ratios of the reagents.

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Probability is a number that describes how likely it is that a certain event will occu O True O False

lara31 [8.8K]

Answer:

Between 0 and 1

Explanation:

Hope this helps :)

6 0

2 years ago

If the vapor's volume were to be incorrectly recorded as 125ml, how will this error affect the calculated molar mass of the unkn

pishuonlain [190]

Since you didn't give the actual volume (or any of the experimental values) I can only tell you how to do it. Do the calculation using the real (determined) volume of the flask. Then, re-do the calculation with v = 125ml. Take the two values and calculate % error; m = measured vol; g = guessed vol.

<span>[mW (m) - mW (g)]/mW (m) x 100% </span>

<span>(they want % error so, if it is negative, just get rid of the sign) </span>

3 0

2 years ago

White vinegar is a 5.0% by mass solution of acetic acid in water. If the density of white

Alex_Xolod [135]

The pH = 2.41

<h3>Further explanation</h3>

Given

5.0% by mass solution of acetic acid

the density of white vinegar is 1.007 g/cm3

Required

Solution

Molarity of solution :

$\tt M=\dfrac{\%mass\times \rho\times 10}{MW~acetic~acid}\\\\M=\dfrac{5\times 1.007\times 10}{60}\\\\M=0.839$

Ka for acetic acid = 1.8 x 10⁻⁵

[H⁺] for weak acid :

$\tt [H^+]=\sqrt{Ka.M}$

Input the value :

$\tt [H^+]=\sqrt{1.8\times 10^{-5}\times 0.839}\\\\(H^+]=0.00388=3.88\times 10^{-3}\\\\pH=3-log~3.88=2.41$

7 0

2 years ago

Name 5 metals used to make the penny over the years.

Irina18 [472]

These 5 metals are: zinc,copper,tin,bronze,nickel

6 0

3 years ago

Cuando se quema 1 mol de metano –o sea, 16 g–, se desprenden 802

Anvisha [2.4K]

Answer:

1 gramo de metano aporta 50.125 kilojoules.

1 gramo de metano aporta 48.246 kilojoules.

Explanation:

La cantidad de energía liberada por la combustión de una unidad de masa del hidrocarburo ( $Q$ ), en kilojoules por mol, es igual a la cantidad de energía liberada por mol de compuesto ( $\bar {Q}$ ), en kilojoules por mol, dividido por su masa molar ( $M$ ), en gramos por mol: