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Ede4ka [16]
3 years ago
10

Calculation the ionisation energy of rubidium per atom, if light of wavelength 5.84 × 10^-8 m produces electrons with a speed of

2.450 × 10^6 ms^1 .
Chemistry
1 answer:
Masja [62]3 years ago
3 0
E = energy in the light rays = h ν = h c /λ = 6.67 * 10×⁻³⁴ * 3 * 10⁸ / (5.84*10⁻⁸) J = 3.42 * 10⁻18 JKinetic energy of electrons = 1/2 m u^2 = 1/2 * 9.1 * 10⁻³¹ * (2.45 * 10⁶)² J = 2.73 * 10⁻¹⁸ J The ionization energy = (3.42 - 2.73) * 10⁻¹⁸ J = 6.9 * 10⁻¹⁹ J
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3 0
3 years ago
If 15.00 mL of 0.0100 M Ca(IO3)2 solution are mixed with 0.500 g KI, what is the theoretical yield (in grams) of I2?
n200080 [17]

The theoretical yield of I2 in the reaction would be 0.23 g

<h3>Theoretical yield</h3>

This refers to the stoichiometric yield of a reaction.

From the equation of the reaction:

Ca(IO3)2 + 10 KI + 12 HCl → 6 I2 + CaCl2 + 10 KCl + 6 H2O

The mole ratio of Ca(IO3)2 and I2 is 1: 6

Mole of 15.00 mL, 0.0100 M Ca(IO3)2 = 15/1000 x 0.0100

                                                              = 0.00015 mole

Equivalent mole of I2 = 0.00015 x 6

                                      = 0.009 mole

mass of 0.0009 I2 = 0.0009 x 253.809

                                = 0.23 g

More on stoichiometric calculations can be found here: brainly.com/question/6907332

8 0
3 years ago
Multiple Choice Questions 1. Which of the following statement best describes an atom? A. Protons and electrons are grouped toget
Arturiano [62]

Answer:

C is Correct

Explanation:

Protons and Neutrons make up the Nucleus while Electrons surround the nucleus in circular orbits with continuous rotation.

5 0
3 years ago
Read 2 more answers
Balance the following reaction:
kvv77 [185]
  • C_5H_8+13/2O_2—»5CO_2+4H_2O

Balanced one

  • 2C_5H_8+13O_2—»10CO_2+8H_2O

Moles of Pentyne

  • Given mass/Molarmass
  • 34/68
  • 0.5mol

Moles of H_2O

  • 8/2(0.5)
  • 4(0.5)
  • 2mol

1mol releases 241.8KJ

2mol releases 241.8(2)=483.6KJ

8 0
2 years ago
How many molecules are in 7.62 L of CH4, at 87.5°C and 722 torr
pickupchik [31]

Answer: There are 1.469 \times 10^{23} molecules present in 7.62 L of CH_4 at 87.5^{o}C and 722 torr.

Explanation:

Given : Volume = 7.62 L

Temperature = 87.5^{o}C = (87.5 + 273) K = 360.5 K

Pressure = 722 torr

1 torr = 0.00131579

Converting torr into atm as follows.

722 torr = 722 torr \times \frac{0.00131579 atm}{1 torr}\\= 0.95 atm

Therefore, using the ideal gas equation the number of moles are calculated as follows.

PV = nRT

where,

P = pressure

V = volume

n = number of moles

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the values into above formula as follows.

PV = nRT\\0.95 atm \times 7.62 L = n \times 0.0821 L atm/mol K \times 360.5 K\\n = \frac{0.95 atm \times 7.62 L}{0.0821 L atm/mol K \times 360.5 K}\\= \frac{7.239}{29.59705}\\= 0.244 mol

According to the mole concept, 1 mole of every substance contains 6.022 \times 10^{23} atoms. Hence, number of atoms or molecules present in 0.244 mol are calculated as follows.

0.244 mol \times 6.022 \times 10^{23}\\= 1.469 \times 10^{23}

Thus, we can conclude that there are 1.469 \times 10^{23} molecules present in 7.62 L of CH_4 at 87.5^{o}C and 722 torr.

5 0
3 years ago
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