When the equation for the complete combustion of one mole of c3h7oh is balanced, the coefficient for oxygen is?

5. A 5.00 g sample of an unknown substance was heated from 25.2 C to 55.1 degrees * C , and it required 133 to do so. Identify t

Leviafan [203]

Answer:

Aluminum

Explanation:

Given

$T_1 = 25.2^oC$

$T_2 = 55.1^oC$

$m = 5.00g$

$\triangle Q= 133J$

See attachment for chart

Required

Identify the unknown substance

To do this, we simply calculate the specific heat capacity from the given parameters using:

$c = \frac{\triangle Q}{m\triangle T}$

This gives:

$c = \frac{\triangle Q}{m(T_2 - T_1)}$

So, we have:

$c = \frac{133J}{5.00g * (55.1C - 25.2C)}$

$c = \frac{133J}{5.00g * 29.9C}$

$c = \frac{133J}{149.5gC}$

$c = 0.89\ J/gC$

From the attached chart, we have:

$Al(s) = 0.89\ J/gC$ --- The specific heat capacity of Aluminum

Hence, the unknown substance is Aluminum

7 0

3 years ago

When 1.104 grams of iron metal are mixed with 26.023 grams of hydrochloric acid in a coffee cup calorimeter, the temperature ris

Alika [10]

Answer:

a) An exothermic reaction, will release heat. No heat will be absorbed.

b) 903.71 J of heat released

c) reaction is exothermic and ∆H will be negative.

d) ΔHreaction = 818.6 J/g

e) ΔHreaction = 45.6 kJ/mol

f) ΔHreaction = 91.2 kJ

Explanation:

Step 1: Data given

Mass of iron = 1.104 grams

Mass of hydrochloric acid = 26.023 grams

Initial temperature = 25.2°C

Final temperature = 33.5 °C

Temperature change = 8.3 °C

Step 2: The balanced equation

2Fe(s)+6HCl(aq) → 2FeCl3 (aq)+3H2 (g)

(A) Determine the amount of heat (in J) absorbed by the reaction mixture.

Since we have a rise of temperature, this means the reaction is exothermic.

An exothermic reaction, will release heat. No heat will be absorbed.

(B) How much heat (in J) was released by the reaction that occurred?

q = mC∆T

with q = heat released (in J)

with m = the mass (in grams)

with c = the specific heat capacity (in J/g°C)

with ∆T = The change in temperature (in °C)

q = (26.023g)*(4.184 J/g°C)*(8.3 °C) = 903.71 J of heat released

(C) Is this reaction exothermic or endothermic? Is ΔHreaction positive or negative?

Since we have a rise of temperature, this means the reaction is exothermic.

There is heat released so ∆H will be negative.

(D) Under constant pressure conditions (as used in this experiment), the heat released by the reaction equals the reaction enthalpy, qreleased = ΔHreaction. Determine ΔHreaction in Joules per gram of metal used (J/g).

ΔHreaction = 903.71 J/1.104 g = 818.6 J/g

(E) Determine ΔHreaction in kilojoules per mole of metal used (kJ/mol)

Number of moles of iron =1.104 grams / 55.845 g/mol = 0.0198 moles

ΔHreaction = 903.71 J / 0.0198 moles = 45641.9 J/mol = 45.6 kJ/mol

(F) Determine ΔHreaction in kilojoules per mole for the balanced reaction equation provided

Since we have 2moles of Fe in the balanced reaction;

ΔHreaction = 45.6 kJ/mol * 2 mol = 91.2 kJ

7 0

2 years ago

A 0.652-g sample of a pure strontium halide reacts with excess sulfuric acid. the solid strontium sulfate formed is separated, d

seraphim [82]

Answer:

The formula of the original halide is SrCl₂.

Explanation:

The balanced equation of this reaction is:

SrX₂ + H₂SO₄ → SrSO₄ + 2 HX, where X is the halide.

From the equation stichiometry, 1.0 mole of strontium halide will result in 1.0 mole of SrSO₄.

The number of moles of SrSO₄ (n = mass/molar mass) = (0.755 g) / (183.68 g/mole) = 4.11 x 10⁻³ mole.

The number of moles of SrX are 4.11 x 10⁻³ moles from the stichiometry of the balanced equation.

n = mass / molar mass, n = 4.11 x 10⁻³ moles and mass = 0.652 g.

The molar mass of SrX₂ = mass / n = (0.652) / (4.11 x 10⁻³ moles) = 158.62 g/mole.

The molar mass of SrX₂ (158.62 g/mole) = Atomic mass of Sr (87.62 g/mole) + (2 x Atomic mass of halide X).

The atomic mass of halide X = (158.62 g/mole) - (87.62 g/mole) / 2 = 71 / 2 g/mole = 35.5 g/mole.

This is the atomic mass of Cl.

So, the formula of the original halide is SrCl₂.

4 0

3 years ago

A boy with pneumonia has lungs with a volume of 1.7 L that fill with 0.070 mol of air when he inhales. When he exhales, his lung

BlackZzzverrR [31]

Answer:

0.053moles

Explanation:

Hello,

To calculate the number of moles of gas remaining in his after he exhale, we'll have to use Avogadro's law which states that the volume of a given mass of gas is directly proportional to its number of moles provided that temperature and pressure are kept constant. Mathematically,

V = kN, k = V / N

V1 / N1 = V2 / N2= V3 / N3 = Vx / Nx

V1 = 1.7L

N1 = 0.070mol

V2 = 1.3L

N2 = ?

From the above equation,

V1 / N1 = V2 / N2

Make N2 the subject of formula

N2 = (N1 × V2) / V1

N2 = (0.07 × 1.3) / 1.7

N2 = 0.053mol

The number of moles of gas in his lungs when he exhale is 0.053 moles

7 0

3 years ago

A student has a 2.123 g sample of a hydrated salt. she heats it to drive off all the water. after heating to constant mass, the

morpeh [17]

The mass of hydrated salt - 2.123 g
mass of anhydrous salt - 1.861 g
mass that has been reduced is the mass of water that has been heated and lost from the compound thereby making the salt anhydrous.
therefore mass of water lost - 2.123 - 1.861 = 0.262 g
number of moles of water lost - 0.262 g / 18 g/mol = 0.0146 mol
number of moles of salt - 1.861 g / 380.6 g/mol = 0.00490 mol
molar ratio of moles of water to moles of salt
molar ratio = 0.146 mol / 0.00490 mol = 2.98 rounded off to 3
for every 1 mol of salt there are 3 moles of water
therefore empirical formula - Cu₃(PO₄)₂.3H₂O

4 0

3 years ago