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2 years ago

133Cs contains how many protons, neutrons, and electrons?

Chemistry

2 answers:

Alekssandra [29.7K]2 years ago

7 0

Answer:

Caesium-133 is composed of...

- 55 protons

- 78 neutrons

- 55 electrons.

irina [24]2 years ago

5 0

Answer:

55 protons, 55 electrons and 78 neutrons

Explanation:

55 protons, 55 electrons and 78 neutrons

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How many moles of carbon dioxide gas can be produced when 12.8 moles of benzene (C6H6) react with excess oxygen? Unbalanced equa

swat32

Answer: 76.8 mole

Explanation:

Firstly, the balanced equation of reacting benzene with excess oxygen is:

2C6H6 + 15O2 → 6H2O + 12CO2

It is clear that every two moles of benzene react with 15 mole of oxygen to produce 6 moles of water and 12 mole of CO2

Which means that 2 moles of benzene give 12 moles of CO2

So, n(C6H6)/2 = n(CO2)/12

which give n(C6H6) = n(CO2)/6

n(CO2) = n(C6H6) x 6 = 12.8 x 6 = 76.8 mole

6 0

3 years ago

Read 2 more answers

Round the number 0.005758 to 3 sig figs

kvv77 [185]

Answer:

0.00576

Explanation:

7 0

4 years ago

A diver planning to dive to a depth of 100 ft can choose to breathe air that is a mixture of oxygen, nitrogen, and helium. If th

Elodia [21]

Answer:

1216 mmHg = Partial pressure N₂

Explanation:

In a mixture of n gases, the partial pressure of each compound follows the equation:

Total pressure = Partial pressure n₁ + Partial pressure n₂ + Partial pressure n₃ + Partial pressure n₄ + Partial pressure n₅ + ... + Partial pressure nₙ

In a mixture of O₂, He and N₂, the total pressure = 3040mmHg is defined as:

3040 mmHg = Partial pressure O₂ + Partial pressure He + Partial pressure N₂

Replacing:

3040 mmHg = 304 mmHg + 1520 mmHg + Partial pressure N₂

3040 mmHg = 1824 mmHg + Partial pressure N₂

1216 mmHg = Partial pressure N₂

6 0

3 years ago

What is the molar mass of barium hydroxide

maksim [4K]

Answer:

171.34 g/mol

Explanation:

Ba molar mass = 137.328 g/mol

O molar mass = 15.999 g/mol * 2 = 31.9980 g/mol

H molar mass = 1.008 g/mol * 2 = 2.0160 g/mol

137.328 + 31.9980 + 2.0160 = 171.3420 = 171.34 g/mol

4 0

3 years ago

1. The solubility of lead(II) chloride at some high temperature is 3.1 x 10-2 M. Find the Ksp of PbCl2 at this temperature.

solniwko [45]

Answer:

1) The solubility product of the lead(II) chloride is $1.2\times 10^{-4}$ .

2) The solubility of the aluminium hydroxide is $1.6\times 10^{-10} M$ .

3)The given statement is false.

Explanation:

Solubility of lead chloride = $S=3.1\times 10^-2M$

$PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)$

S 2S

The solubility product of the lead(II) chloride = $K_{sp}$

$K_{sp}=[Pb^{2+}][Cl^-]^2$

$K_{sp}=S\times (2S)^2=4S^3=4\times (3.1\times 10^{-2})^3=1.2\times 10^{-4}$

The solubility product of the lead(II) chloride is $1.2\times 10^{-4}$ .

Concentration of aluminium nitrate = 0.000010 M

Concentration of aluminum ion = $1\timed 0.000010 M=0.000010 M$

Solubility of aluminium hydroxide in aluminum nitrate solution = $S$

$Al(OH)_3(aq)\rightleftharpoons Al^{3+}(aq)+3OH^-(aq)$

S 3S

The solubility product of the aluminium nitrate = $K_{sp}=1.0\times 10^{-33}$

$K_{sp}=[Al^{3+}][OH^-]^3$

$1.0\times 10^{-33}=(0.000010+S)\times (3S)^3$

$S=1.6\times 10^{-10} M$

The solubility of the aluminium hydroxide is $1.6\times 10^{-10} M$ .

$Molarity=\frac{Moles}{Volume (L)}$

Mass of NaCl= 3.5 mg = 0.0035 g

1 mg = 0.001 g

Moles of NaCl = $\frac{0.0035 g}{58.5 g/mol}=6.0\times 10^{-5} mol$

Volume of the solution = 0.250 L

$[NaCl]=\frac{6.0\times 10^{-5} mol}{0.250 L}=0.00024 M$

1 mole of NaCl gives 1 mole of sodium ion and 1 mole of chloride ions.

$[Cl^-]=[NaCl]=0.00024 M$

Moles of lead (II) nitrate = n

Volume of the solution = 0.250 L

Molarity lead(II) nitrate = 0.12 M

$n=0.12 M]\times 0.250 L=0.030 mol$

1 mole of lead nitrate gives 1 mole of lead (II) ion and 2 moles of nitrate ions.

$[Pb^{2+}]=[Pb(NO_2)_3]=0.030 M$

$PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)$

Solubility of lead(II) chloride = $K_{sp}=1.2\times 10^{-4}$

Ionic product of the lead chloride in solution :

$Q_i=[Pb^{2+}][Cl^-]^2=0.030 M\times (0.00024 M)^2=1.7\times 10^{-9}$

$Q_i$ ( no precipitation)

The given statement is false.

3 0

4 years ago