How many moles of CO2 are produced from the combustion of 6.40 mol c2h8?

give a general description of the location of metals , non metals and metalloids on the periodic table

zheka24 [161]

Answer:

Metals at the top

nonmetals at the bottom

metalloids in the middle

Don't quote me, i could be wrong. i think this is the correct order.

Explanation:

7 0

3 years ago

Carbon tetrachloride, CCl4, once used as a cleaning fluid and as a fire extinguisher, is produced by heating methane and chlorin

Elena-2011 [213]

Answer:

1.882 g

Explanation:

Data Given

mass of Cl₂ = 33.4 g

mass of CH₄ = ?

Reaction Given:

CH₄+ 4Cl₂ --------→ CCl₄ + HCl

Solution:

First find the mass of CH₄ from the reaction that it combine with how many grams of Chlorin.

Look at the balanced reaction

CH₄ + 4Cl₂ --------→ CCl₄ + 4HCl

1 mol 4 mol

So 1 mole of CH₄ combine with 4 moles of Cl₂

Now

convert the moles into mass for which we have to know molar mass of CH₄ and Cl₂

Molar mass of Cl₂ = 2 (35.5)

Molar mass of Cl₂ = 71 g/mol

mass of Cl₂

mass in grams = no. of moles x molar mass

mass of Cl₂ = 4 mol x 71 g/mol

mass of Cl₂ = 284 g

Molar mass of CH₄= 12+ 4(1)

Molar mass of CH₄= 16 g/mol

mass of CH₄

mass in grams = no. of moles x molar mass

mass of CH₄= 1 mol x 16 g/mol

mass of CH₄ = 16 g

So,

284 g of Cl₂ combine with 16 g of methane ( CH₄ ) then how many grams of CH₄ is needed to combine with 33.4 g of Cl₂

Apply unity Formula

284 g of Cl₂ ≅ 16 g of methane ( CH₄ )

33.4 g of Cl₂ ≅ X g of methane ( CH₄ )

By cross multiplication

X g of methane ( CH₄ ) = 16 g x 33.4 g / 284 g

X g of methane ( CH₄ ) = 1.88 g

1.882 g of methane (CH₄) will needed to combine with 33.4 g of Cl₂

So

methane (CH₄) = 1.882 g

5 0

2 years ago

Which of the following equations is balanced?

stiv31 [10]

Answer:

A) 4P + 5O₂ → 2P₂O₅

Explanation:

A) 4P + 5O₂ → 2P₂O₅

This equation is balanced. There are four phosphorus and ten oxygen atoms are on both side of equation.

B) 5P + 4O₂ → 2P₄O₅

This equation is not balanced. There are five phosphorus and eight oxygen atoms on left, eight phosphorus ten oxygen on right side of equation.

C) 2P + O₂ → P₂O₅

This equation is not balanced. There are two phosphorus, two oxygen atoms on left and two phosphorus five oxygen on right side of equation.

D) 4P + 2O₂ → 2P₄O₅

This equation is not balanced. There are four phosphorus, four oxygen atoms on left and eight phosphorus ten oxygen on right side of equation.

4 0

3 years ago

The density of the fat tristearin is 0.95 g cm−3 . Calculate the change in molar Gibbs energy of tristearin when a deep-sea crea

KatRina [158]

The formula for the change in Gibbs energy of a solid is:

ΔG = Vm ΔP

where, ΔG is change in Gibbs, Vm is molar volume, ΔP is change in pressure

ΔP = P(final) – P(initial)

P(final) = 1 atm = 101325 Pa

P(initial) = ρ_water *g *h = (1030 kg/m^3) * 9.8 m/s^2 * 2000 m = 20188000 kg m/s^2 = 20188000 Pa

Vm = (950 kg/m^3) * (1000 mol / 891.48 kg) = 1065.64 mol/m^3

So,

ΔG = (1065.64 mol/m^3) * (101325 Pa - 20188000 Pa)

8 0

3 years ago

An acid with molar mass 84.48 g/mol is titrated with 0.650 M KOH. What volume of KOH solution is needed to titrate 1.70 grams of

RSB [31]

Answer:

$V=0.0310L=3.10mL$

Explanation:

Hello.

In this case, since the acid is monoprotic and the KOH has one hydroxyl ion only, we can see that at the equivalence point the moles of both of them are the same:

$n_{acid}=n_{KOH}$

Thus, since we are given 1.70 g of the acid, we compute the moles of acid that were titrated:

$n_{acid}=1.70g*\frac{1mol}{84.48g}=0.0201mol$

Which equal the moles of KOH. In such a way, since the molarity is defined as moles over liters (M=n/V), the liters are moles over molarity (V=n/M), thus, the resulting volume is:

$V=\frac{0.0201mol}{0.650mol/L}\\\\V=0.0310L=3.10mL$

Best regards!

7 0

2 years ago