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2 years ago

A sample of gas occupies a volume of 350.0 mL at 840mm Hg and 33°C. Determine the volume of this sample at 600 mm By and 52°C

Chemistry

1 answer:

insens350 [35]2 years ago

7 0

Answer:

V₂ = 520.42 mL

Explanation:

Given data:

Initial volume = 350.0 mL

Initial pressure = 840 mmHg

Initial temperature = 33°C (33 +273 = 306 K)

Final temperature = 52°C (52+273 = 325 K)

Final volume = ?

Final pressure = 600 mmHg

Formula:

P₁V₁/T₁ = P₂V₂/T₂

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

Solution:

V₂ = P₁V₁ T₂/ T₁ P₂

V₂ = 840 mmHg × 350.0 mL × 325 K / 306 K × 600 mmHg

V₂ = 95550000 mmHg.mL.K /183600 K.mmHg

V₂ = 520.42 mL

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Inessa [10]

Answer:

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Explanation:

bcoz it is solid at room temperature and does not have double bonds between carbons

8 0

2 years ago

Given the following balanced equation, if the rate of O2 loss is 3.64 × 10-3 M/s, what is the rate of formation of SO3? 2 SO2(g)

Fynjy0 [20]

Answer:

Rate of formation of SO₃ $[\frac{d[SO_{3}] }{dt}]$ = 7.28 x 10⁻³ M/s

Explanation:

According to equation 2 SO₂(g) + O₂(g) → 2 SO₃(g)

Rate of disappearance of reactants = rate of appearance of products

⇒ $-\frac{1}{2} \frac{d[SO_{2} ]}{dt} = -\frac{d[O_{2} ]}{dt}=\frac{1}{2} \frac{d[SO_{3} ]}{dt}$ -----------------------------(1)

Given that the rate of disappearance of oxygen = $-\frac{d[O_{2} ]}{dt}$ = 3.64 x 10⁻³ M/s

So the rate of formation of SO₃ $[\frac{d[SO_{3}] }{dt}]$ = ?

from equation (1) we can write

$\frac{d[SO_{3}] }{dt} = 2 [-\frac{d[O_{2}] }{dt} ]$

⇒ $\frac{d[SO_{3}] }{dt}$ = 2 x 3.64 x 10⁻³ M/s

⇒ $[\frac{d[SO_{3}] }{dt}]$ = 7.28 x 10⁻³ M/s

∴ So the rate of formation of SO₃ $[\frac{d[SO_{3}] }{dt}]$ = 7.28 x 10⁻³ M/s

7 0

3 years ago

An unknown compound contains only carbon, hydrogen, and oxygen (CxHyOz). Combustion of 2.50g of this compound produced 3.67g of

REY [17]

The solution to the problem is as follows:

<span>a)

C + O2 = CO2

Molar mass CO2 = 44 g/mol

3.67 g CO2 * 1 mol / 44 g =

=0.0834 mol CO2 = 0.0834 mol C

I hope my answer has come to your help. God bless and have a nice day ahead!
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4 0

3 years ago

Chromium metal is plated onto a car bumper at a current of 10.0 amperes for 100 seconds. The plating bath contains chromium in t

Phoenix [80]

Answer:

0.54g of Cr

Explanation:

Current (I) = 10A

Time (t) = 100s

Molecular mass of Cr = 51.996 amu

Faraday's first law of electrolysis states that

The mass of the substance (m) of a given substance deposited at an electrode is directly proportional to the quantity of electricity or charge (Q) passed

m = nQ

M = mass of the substance

n = electrochemical constant

Q = charge passed through it

Q = IT

Q = (10 * 100) = 1000C

1 moles = molarmass = Faraday's constant (96500C)

Molar mass = Faraday's constant (96500C)

51.996 g = 96500C

How many grams will be liberated with 1000C

51.996g = 96500C

Xg = 1000C