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2 years ago

If 1.3 L of C3H8 combusts according to the equation below, how much CO2 will be produced?

Chemistry

1 answer:

SCORPION-xisa [38]2 years ago

6 0

Answer:

0.162 moles of CO₂ are produced by this reaction

Explanation:

The reaction is:

C₃H₈(g) + 5O₂(g) → 3CO₂(g) +4H₂O(g)

As we have the volume of propane, we need to know the mass that has reacted, so we apply density's concept.

Density = Mass / Volume → Density . Volume = Mass

0.00183 g/mL . 1300 mL = Mass → 2.379 g

We determine the moles → 2.379 g . 1mol / 44 g = 0.054 moles

Ratio is 1:3. 1 mol of propane can produce 3 moles of dioxide

Then, 0.054 moles may produce (0.054 .3)/1 = 0.162 moles

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Consider 0.01 m aqueous solutions of each of the following. a) NaI; b) CaCl2; c) K3PO4; and d) C6H12O6 (glucose) Arrange the sol

stealth61 [152]

Answer:

The solutions are ordered by this way (from lowest to highest freezing point): K₃PO₄ < CaCl₂ < NaI < glucose

Option d, b, a and c

Explanation:

Colligative property: Freezing point depression

The formula is: ΔT = Kf . m . i

ΔT = Freezing T° of pure solvent - Freezing T° of solution

We need to determine the i, which is the numbers of ions dissolved. It is also called the Van't Hoff factor.

Option d, which is glucose is non electrolyte so the i = 1

a. NaI → Na⁺ + I⁻ i =2

b. CaCl₂ → Ca²⁺ + 2Cl⁻ i =3

c. K₃PO₄ → 3K⁺ + PO₄⁻³ i=4

Potassium phosphate will have the lowest freezing point, then we have the calcium chloride, the sodium iodide and at the end, glucose.

7 0

2 years ago

The maximum amount of nickel(II) cyanide that will dissolve in a 0.220 M nickel(II) nitrate solution is...?

sweet [91]

Answer : The maximum amount of nickel(II) cyanide is $5.84\times 10^{-12}M$

Explanation :

The solubility equilibrium reaction will be:

$Ni(CN)_2\rightleftharpoons Ni^{2+}+2CN^-$

Initial conc. 0.220 0

At eqm. (0.220+s) 2s

The expression for solubility constant for this reaction will be,

$K_{sp}=[Ni^{2+}][CN^-]^2$

Now put all the given values in this expression, we get:

$3.0\times 10^{-23}=(0.220+s)\times (2s)^2$

$s=5.84\times 10^{-12}M$

Therefore, the maximum amount of nickel(II) cyanide is $5.84\times 10^{-12}M$

7 0

2 years ago

Which nonmetal would an ion of Ca2+ MOST LIKELY ionically bond with in a 1:1 ratio? A) one atom of neon B) one atom of carbon C)

GalinKa [24]

Ca2+ would bond to any element in a 1 to 1 ratio that had an equal and opposite charge.

Neon is a noble gas, and doesn’t form bonds m

Carbon isn’t typically found in ion state, but if it did, it would likely by C4+

Flouring in ionic state is F1-, so you would need 2 flourines to cancel the 2+ charge of Calcium

Then the only option left would be Oxygen which, when in ion form is found be 2-

5 0

2 years ago

Read 2 more answers

In the space provided, for each element, type the ionic charge of an atom with a full set of valence electrons. Then, type the n

kobusy [5.1K]

11. ionic charge +1, helium.

12. ionic charge 2-, neon.

13. ionic charge 3+, neon.

3 0

2 years ago

Hydrazine (N2H4) is used as rocket fuel. It reacts with oxygen to form nitrogen and water.

Marina86 [1]

Answer:

See explanation below for answers

Explanation:

This is a stochiometry reaction. LEt's write the overall reaction again:

N₂H₄ + O₂ ---------> N₂ + 2H₂O

This reaction is taking place at Standard temperature and pressure conditions (STP) which are P = 1 atm and T = 273 K. To know the volume of N₂ formed, we need to know first how many moles are formed, and this can be calculated with the reagents and the limiting reagent. Let's calculate the moles first of the reagents:

MM N₂H₄ = 32 g/mol; MM O₂ = 32 g/mol

mol N₂H₄ = 2000 / 32 = 62.5 moles

mol O₂ ? 2100 / 32 = 65.63 moles

Now that we have the moles, we need to apply the stochiometry and calculate the limiting reagent. According to the overall reaction we have a mole ratio of 1:1 between N₂H₄ and O₂, therefore:

1 mole N₂H₄ ---------> 1 mole O₂

62.5 moles ----------> X

X = 62.5 moles of O₂

But we have 65.63 moles, therefore, the limiting reactant is the N₂H₄.

We also have a 1:1 mole ratio with the N₂, so:

moles N₂H₄ = moles N₂ = 62.5 moles

Now that we have the moles, we can calculate the volume with the ideal gas equation:

PV = nRT

V = nRT / P

R: gas constant (0.082 L atm / K mol)

Replacing we have:

v = 62.5 * 0.082 * 273 / 1

V = 1399.13 L of N₂

Now, how many grams of the excess remains?, we know how many moles are reacting so, let's see how much is left:

moles remaining = 65.63 - 62.5 = 3.12 moles

then the mass of oxygen:

m = 3.12 * 32 = 100.16 g of O₂

7 0

2 years ago

Read 2 more answers