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3 years ago

How many moles of methane will react to form 3.2 moles of carbon dioxide?

Chemistry

1 answer:

Hunter-Best [27]3 years ago

6 0

Answer:

3.2 moles

Explanation:

First, we'll begin by writing a balanced equation for the Combustion of methane to produce carbon dioxide. This is illustrated below:

CH4 + 2O2 —> CO2 + 2H2O

From the balanced equation above,

1 mole of methane (CH4) reacted to produced 1 mole of carbon dioxide (CO2).

Therefore, 3.2 moles of methane (CH4) will react to produce 3.2 moles of carbon dioxide (CO2).

From the illustration above, 3.2 moles of methane is needed to produce 3.2 moles of carbon dioxide.

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#12 please! worth 40 points

Serga [27]

Okay so because of the difference in density a simple method for telling the difference between the two is to put a sample in a container with oil, because water has a higher density than the oil it would sink to the bottom but alcohol on the other hand is lighter than oil and would float on top of the oil.

However with this question I think that what you would do is use the ice to find out what the substance is, it would float on top of the liquid if it were water because the water is denser than ice but the ice would sink if it was alcohol because the alcohol is less dense than ice.

I hope this helps you, good luck : )

3 0

3 years ago

Consider the following reaction between mercury(II) chloride and oxalate ion:

siniylev [52]

Answer : The reaction rate will be, $1.9\times 10^{-4}M/s$

Explanation :

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

$2HgCl_2(aq)+C_2O_2^{4-}(aq)\rightarrow 2Cl^-(aq)+2CO_2(g)+HgCl_2(s)$

Rate law expression for the reaction:

$\text{Rate}=k[HgCl_2]^a[C_2O_2^{4-}]^b$

where,

a = order with respect to $HgCl_2$

b = order with respect to $C_2O_2^{4-}$

Expression for rate law for first observation:

$3.2\times 10^{-5}=k(0.164)^a(0.15)^b$ ....(1)

Expression for rate law for second observation:

$2.9\times 10^{-4}=k(0.164)^a(0.45)^b$ ....(2)

Expression for rate law for third observation:

$1.4\times 10^{-4}=k(0.082)^a(0.45)^b$ ....(3)

Expression for rate law for fourth observation:

$4.8\times 10^{-5}=k(0.246)^a(0.15)^b$ ....(4)

Dividing 1 from 2, we get:

$\frac{2.9\times 10^{-4}}{3.2\times 10^{-5}}=\frac{k(0.164)^a(0.45)^b}{k(0.164)^a(0.15)^b}\\\\9=3^b\\(3)^2=3^b\\b=2$

Dividing 3 from 2, we get:

$\frac{2.9\times 10^{-4}}{1.4\times 10^{-4}}=\frac{k(0.164)^a(0.45)^b}{k(0.082)^a(0.45)^b}\\\\2=2^a\\a=1$

Thus, the rate law becomes:

$\text{Rate}=k[HgCl_2]^1[C_2O_2^{4-}]^2$

Now, calculating the value of 'k' by using any expression.

Putting values in above rate law, we get:

$3.2\times 10^{-5}=k(0.164)^1(0.15)^2$

$k=8.7\times 10^{-3}M^{-2}s^{-1}$

Now we have to determine the reaction rate when the concentration of $HgCl_2$ is 0.135 M and that of $C_2O_2^{-4}$ is 0.40 M.

$\text{Rate}=k[HgCl_2]^1[C_2O_2^{4-}]^2$

$\text{Rate}=(8.7\times 10^{-3})\times (0.135)^1\times (0.40)^2$

$\text{Rate}=1.9\times 10^{-4}M/s$

Therefore, the reaction rate will be, $1.9\times 10^{-4}M/s$

6 0

3 years ago

How many grams of cupric sulfate pentahydrate are needed to prepare 50.00 mL of 0.0800M CuSO4× 5H2O?

shepuryov [24]

Explanation:

Molarity is defined as number of moles per liter of solution.

Mathematically, molarity = $\frac{no. of moles}{Volume (in L) of solution}$

It is given that molarity is 0.0800 M and volume is 50.00 mL or 0.05 L.

molarity = $\frac{no. of moles}{Volume of solution in liter}$

0.0800 M = $\frac{no. of moles}{0.05 L}$

no. of moles = 1.6 mol

Therefore, molar mass of cupric sulfate pentahydrate is 249.68 g/mol. So, calculate the mass as follows.

No. of moles = $\frac{mass in grams}{molar mass}$

mass in grams = $no. of moles \times molar mass of CuSO_{4}.5H_{2}O$

= $1.6 mol \times 249.68 g/mol$

= 399.488 g

Thus, we can conclude that 399.488 g of cupric sulfate pentahydrate are needed to prepare 50.00 mL of 0.0800M CuSO4× 5H2O.

4 0

3 years ago

What are the moles of silver metal produced from 0.0999 mol of copper?

julia-pushkina [17]

Answer:

1234567i9812345678912121212121

6 0

3 years ago

Which is a statement of cell theory? All cells are made up of living molecules. All plants are made of cells. All animals are ma

erastovalidia [21]

Answer:

all cells are produced from other preexisting cells through cell division

4 0

3 years ago