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2 years ago

Determine the empirical formula of a

Chemistry

1 answer:

k0ka [10]2 years ago

3 0

Answer:

AuCl

Explanation:

Given parameters:

Mass of Gold = 2.6444g

Mass of Chlorine = 0.476g

Unknown:

Empirical formula = ?

Solution:

Empirical formula is the simplest formula of a compound. Here is the way of determining this formula.

Elements Au Cl

Mass 2.6444 0.476

Molar mass 197 35.5

Number of moles 2.6444/197 0.476/35.5

0.013 0.013

Divide by the

smallest 0.013/0.013 0.013/0.013

1 1

The empirical formula of the compound is AuCl

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What is the pressure in a 245 L tank that contains 5.21 kg of helium at 27 Celsius

Kitty [74]

<h2><u>Answer:</u></h2>

(These are not rounded to the correct decimal)

130.94 atm

13,266.6 kPa

99,571.4 mmHg

<h2><u>Explanation:</u></h2>

<u></u>

PV = nRT

V = 245L

P = ?

R = 0.08206 (atm) , 8.314 (kPa) , 62.4 (mmHg)

T = 273.15 + 27 = 300.15K

n = 1302.5 moles

How I found (n).

5.21kg x 1000g/1kg x 1 mole/4.0g = 1302.5 moles

Now, plug all the numbers into the equation.

Pressure in atm = (1302.5)(0.08206)(300.15) / 245 = 130.94 atm (not rounded to the correct decimal)

Pressure in kPa = (1302.5)(8.314)(300.15) / 245 = 13,266.6 kPa (not rounded to the correct decimal)

Pressure in mmHg = (1302.5)(62.4)(300.15) / 245 = 99,571.4 mmHg (not rounded to the correct decimal)

5 0

3 years ago

Can animals detect radio waves

Svetach [21]

Answer:

No.

Explanation:

No organism can detect X-rays or radio waves

4 0

2 years ago

Read 2 more answers

You are on the Titanic and want to find something that will float because you didn't

damaskus [11]

Find volume of pillow

L=78cm

B=55cm

H=25cm

$\\ \bull\tt\longrightarrow V=LBH$

$\\ \bull\tt\longrightarrow V=25(78)(55)$

$\\ \bull\tt\longrightarrow V=107250cm^3$

$\\ \bull\tt\longrightarrow V=10.72m^3$

Now

Mass=5.5kg

$\\ \bull\tt\longrightarrow Density=\dfrac{Mass}{Volume}$

$\\ \bull\tt\longrightarrow Density=\dfrac{5.5}{10.72}$

$\\ \bull\tt\longrightarrow Density=0.5kg/m^3$

Density of water=1000kg/m^3

As it is less than density of water it will float on water

8 0

3 years ago

What is the specific heat of a substance that absorbs 2500 joules of heat when a sample of 100g of the substance increases in te

Ierofanga [76]

Answer:

0.417 J/gºC

Explanation:

From the question given above, the following data were obtained:

Heat (Q) absorbed = 2500 J

Mass (M) of substance = 100 g

Initial temperature (T1) = 10 °C

Final temperature (T2) = 70 °C

Specific heat capacity (C) =?

Next, we shall determine the change in temperature (ΔT). This can be obtained as follow:

Initial temperature (T1) = 10 °C

Final temperature (T2) = 70 °C

Change in temperature (ΔT) =?

Change in temperature (ΔT) = T2 – T1

Change in temperature (ΔT) = 70 – 10

Change in temperature (ΔT) = 60 °C

Finally, we shall determine the specific heat capacity of the substance as follow:

Heat (Q) absorbed = 2500 J

Mass (M) of substance = 100 g

Change in temperature (ΔT) = 60 °C

Specific heat capacity (C) =?

Q = MCΔT

2500 = 100 × C × 60

2500 = 6000 × C

Divide both side by 6000

C = 2500 / 6000

C = 0.417 J/gºC

Therefore, the specific heat capacity of substance is 0.417 J/gºC

5 0

3 years ago

If 250.0 mL of a 0.96 M solution of acetic acid (C 2H 4O 2) are diluted to 800.0 mL, what will be the approximate molarity of t

adoni [48]

Answer:

0.30M HOAc

Explanation:

Given 250.0ml (0.96M HOAc) => 800ml(??M HOAc)

Use the dilution equation...

(Molarity x Volume)concentrated soln = (Molarity x Volume)dilute soln

(0.96M)(250.0ml) = (Molarity diluted soln)(800ml)

Molarity diluted soln = (0.96M)(250.0ml)/(800ml) = 0.30M HOAc

7 0

3 years ago