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Lana71 [14]
4 years ago
6

the student determines the molar mass of the gas to be 64 g mol-1. write the expression (set-up) for calculating the percent err

or in the experimental value, assuming that the unknown gas is butane (molar mass 58 g mol-1). calculations are not required
Chemistry
1 answer:
sleet_krkn [62]4 years ago
7 0

Answer:

Percentage error = \frac{(64 - 58)}{58} x 100%

Explanation:

In order to calculate the percentage error in the student's calculation, the following formula will be used:

Percentage error = \frac{(experimental molar mass - theoretical molar mass)}{theoretical molar mass} x 100%

experimental molar mass = 64 g/mol

theoretical molar mass = 58 g/mol

Hence, the expression becomes:

Percentage error = \frac{(64 - 58)}{58} x 100%

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Two moles of an ideal gas are placed in a container whose volume is 2.3 x 10^-3 m3. The absolute pressure of the gas is 6.9 x 10
PtichkaEL [24]

Answer:

K.E.=1.97\times 10^{-21}\ J

Explanation:

Given that:-

Pressure = 6.9\times 10^5\ Pa

The expression for the conversion of pressure in Pascal to pressure in atm is shown below:

P (Pa) = \frac {1}{101325} P (atm)

Given the value of pressure = 43,836 Pa

So,  

6.9\times 10^5\ Pa = \frac{6.9\times 10^5}{101325} atm

Pressure = 6.80977 atm

Volume = 2.3\times 10^{-3}\ m^3 = 2.3 L ( 1 m³ = 1000 L)

n = 2 mol

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

6.80977 atm × 2.3 L = 2 mol × 0.0821 L.atm/K.mol × T

⇒T = 95.39 K

The expression for the kinetic energy is:-

K.E.=\frac{3}{2}\times K\times T

k is Boltzmann's constant = 1.38\times 10^{-23}\ J/K

T is the temperature

So, K.E.=\frac{3}{2}\times 1.38\times 10^{-23}\times 95.39\ J

K.E.=1.97\times 10^{-21}\ J

3 0
3 years ago
Where did the longest tsunami form?
max2010maxim [7]
July 9, 1958: in Lituya Bay<span>, Alaska was caused by a landslide triggered by an 8.3 magnitude earthquake.</span>
8 0
3 years ago
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Cassady? you there ?
iren [92.7K]

Answer:

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Explanation:

8 0
3 years ago
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A 1.2 kg block of iron at 32 ∘C is rapidly heated by a torch such that 12 kJ is transferred to it. What temperature would the bl
lbvjy [14]

Answer:

For iron

Final temperature = 54,22°C

For copper

Final Temperature = 63.67 °C

Explanation

Hello,

You are using a torch to warm up a block of iron that has an initial temperature of 32°C.

The first you have to know is that the "heat capacity" could simply define as the heat required to go from an initial temperature to a final temperature.

So you need to use the heat capacity equation as follow in the paper.

The equation has to have all terms in the same units, so:

q = 12000 J

s = 0.450 J / g °C

m = 1200 g

Ti = 32 °C

Download odt
3 0
3 years ago
Consider the equilibrium: HCOOH(aq) + F-(aq) &lt;----&gt; HCOO-(aq) + HF (aq) Given that the Ka of HCOOH = 1.8 x 10-4 and the Ka
Aleks [24]

Hey there!:

K = Ka * Kb / Kw

Ka = 1.8*10⁻⁴

Kb = 10⁻¹⁴ / 6.8*10⁻⁴

K =  1.8*10⁻⁴ * ( 10⁻¹⁴/ 6.8*10⁻⁴ ) * ( 1 / 10⁻¹⁴ )

K =  = 1.8 / 6.8

K = 0.265

Answer A

Therefore:

K is less than on the forward reaction is not favorable .


Hope That helps!

8 0
3 years ago
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