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3 years ago

Draw 2,3-dichloro octane

Chemistry

1 answer:

Deffense [45]3 years ago

6 0

Answer:

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What does light do in the photoelectric effect?

xz_007 [3.2K]

Answer:

D. light knocks electrons off metal atoms

Explanation: just took it

8 0

3 years ago

Read 2 more answers

The empirical formula of a compound is CH. At 200 degree C, 0.145 g of this compound occupies 97.2 mL at a pressure of 0.74 atm.

Vladimir [108]

Answer:

The molecular formula = $C_{6}H_{6}$

Explanation:

Given that:

Mass of compound, m = 0.145 g

Temperature = 200 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (200 + 273.15) K = 473.15 K

V = 97.2 mL = 0.0972 L

Pressure = 0.74 atm

Considering,

$n=\frac{m}{M}$

Using ideal gas equation as:

$PV=\frac{m}{M}RT$

where,

P is the pressure

V is the volume

m is the mass of the gas

M is the molar mass of the gas

T is the temperature

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

Applying the values in the above equation as:-

$0.74\times 0.0972=\frac{0.145}{M}\times 0.0821\times 473.15$

$M=78.31\ g/mol$

The empirical formula is = $CH$

Molecular formulas is the actual number of atoms of each element in the compound while empirical formulas is the simplest or reduced ratio of the elements in the compound.

Thus,

Molecular mass = n × Empirical mass

Where, n is any positive number from 1, 2, 3...

Mass from the Empirical formula = 12 + 1 = 13 g/mol

Molar mass = 78.31 g/mol

So,

Molecular mass = n × Empirical mass

78.31 = n × 13

⇒ n ≅ 6

The molecular formula = $C_{6}H_{6}$

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3 years ago

How many moles of aluminum do 4.40×1024 aluminum atoms represent?

zhannawk [14.2K]

For this problem we use the Avogadro's number which is an empirical value that relates the number of particles to the number of moles. Its approximated value is 6.022×10²³ atoms/mole. The solution is as follows:

4.40×10²⁴ aluminum atoms * 1 moles/6.022×10²³ atoms = 7.306

Thus, there are 7.306 moles of aluminum.

3 0

4 years ago

When the temperature of a rigid hollow sphere containing 685 L of helium gas is held at 621 K, the pressure of the gas is 1.89 x

Leno4ka [110]

You have to use the equation PV=nRT.
P=pressure (in this case 1.89x10^3 kPa which equals 18.35677 atm)
1V=volume (in this case 685L)
n=moles (in this case the unknown)
R=gas constant (0.08206 (L atm)/(mol K))
T=temperature (in this case 621 K)
with the given information you can rewrite the ideal gas law equation as n=PV/RT.
n=(18.35677atm x 685L)/(0.08206atmL/molK x 621K)
n=246.8 moles

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3 years ago

if scientist had tried to repeat Thomson's experiment and found that they could not, would Thomson's conclusion still have been

Mariulka [41]

It wouldn’t, it must be at least 96% close to what Thomson did to make the Conclusion valid, it must be repeated the same and tested by another person at least 3 times. Hope this helps, mate! :)

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3 years ago