QUESTION 2

A 7.591-9 gaseous mixture contains methane (CH4) and butane

mestny [16]

Answer:

65.71%

Explanation:

First, we can write the mass of the mixture, thus:

7.519g = X + Y (1)

Where X is the mass of methane and Y the mass of butane

Also, the reactions of combustion are:

CH₄ + 2O₂ → CO₂ + 2H₂O

2 moles of oxygen react per mole of methane

C₄H₁₀ + 13/2 O₂ → 4CO₂ + 5H₂O

13/2 moles of oxygen react per mole of methane

That means, in therms of moles of oxygen we can write:

0.9050 moles = 2X/16.04 + 13/2Y/ 58.12

0.9050 = 0.12469X + 0.11184Y (2)

Where 16.04 and 58.12 are molar masses of methane and butane

That is because if X is the mass of methane:

X g Methane * (1mol / 16.04g) = Moles methane

Moles methane * (2 moles Oxygen / mole methane) = Moles oxygen

Replacing (1) in (2):

0.9050 = 0.12469X + 0.11184 (7.519 - X)

0.9050 = 0.12469X + 0.841 - 0.11184X

0.0641 = 0.01285X

X = 4.988g = Mass of methane.

And mass percent of methane is:

4.988g / 7.591g * 100

7 0

2 years ago

A 25 gram(m) metal ball is heated to 200C(delta T) with 2330 Joules(q) of energy. What is the specific heat of the metal?

Dominik [7]

Answer:

The specific heat of the metal is 0.466 $\frac{J}{g*C}$

Explanation:

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

The equation that allows calculating heat exchanges is:

Q = c * m * ΔT

where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation.

In this case:

Q= 2330 J
c= ?
m= 25 g
ΔT= 200 °C

Replacing:

2330 J= c*25 g* 200 °C

Solving:

$c=\frac{2330 J}{25 g* 200 C}$

c=0.466 $\frac{J}{g*C}$

The specific heat of the metal is 0.466 $\frac{J}{g*C}$

6 0

2 years ago

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A laboratory utilizes a mixture of 10% dimethyl sulfoxide (DMSO) in the freezing and long-term storage of embryonic stem cells.

Mars2501 [29]

Answer:

The correct answer is "1.0100".

Explanation:

Let the volume of mixture be 100 ml.

then,

The volume of DMSO will be 10 mL as well as that of water will be 90 mL.

DMSO will be:

= $10\times 1.1004$

= $11.004 \ g$

The total mass of mixture will be:

= $90+11.004$

= $101.004 \ g$

Density of mixture will be:

= $\frac{Mass}{Volume}$

= $\frac{101.004}{100}$

= $1.01004 \ g/mL$

hence,

Specific gravity of mixture will be:

= $\frac{Density \ of \ mixture}{Density \ of \ water}$

= $\frac{1.01004}{1}$

= $1.0100$

3 0

2 years ago

A solution contains cr3+ ions. the addition of 0.063 l of 1.50 m naf solution was needed to completely precipitate the chromium

steposvetlana [31]

Cr{3+} + 3 NaF → CrF3 + 3 Na{+} 

First calculate the total mols of NaF.

(0.063 L) x (1.50 mol/L NaF) = 0.0945 mol NaF total

Using stoichiometric ratio:

0.0945 mol NaF * (1 mol Cr3+ / 3 mol NaF) * (51.9961 g Cr3+/mol) = 1.6379 g Cr3+

6 0

2 years ago

Simplify: (100 m)/(26 s)

podryga [215]

100m/26s=50m/13s

50m/13s=3.846m/s

4 0

2 years ago

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