Why were gold and copper the first forms of matter investigated by humans

46. Which of the following is not a form of matter?

Mice21 [21]

C cuz it just doesn’t make sense

3 0

3 years ago

The energy required to ionize boron is 801 kJ/mol. You may want to reference (Pages 93 - 98) Section 2.5 while completing this p

earnstyle [38]

Answer:

The frequency is $f = 2,01 * 10^{15} \ Hz$

Explanation:

From the question we are told that

The energy required to ionize boron is $E_b = 801 KJ/mol$

Generally the ionization energy of boron pre atom is mathematically represented as

$E_a = \frac{E_b}{N_A}$

Here $N_A$ is the Avogadro's constant with value $N_A = 6.022*10^{23}$

So

$E_a = \frac{801}{6.022*10^{23}}$

=> $E_a = 1.330 *10^{-18} \ J/atom$

Generally the energy required to liberate one electron from an atom is equivalent to the ionization energy per atom and this mathematically represented as

$E = hf = E_a$

=> $hf = E_a$

Here h is the Planks constant with value $h = 6.626 *10^{-34}$

So

$f = \frac{1.330 *10^{-18}}{ 6.626 *10^{-34}}$

=> $f = 2,01 * 10^{15} \ Hz$

8 0

2 years ago

Be sure to answer all parts. A concentration cell consists of two Sn/Sn2+ half-cells. The electrolyte in compartment A is 0.23 M

kolbaska11 [484]

Answer:

Part A. The half-cell B is the cathode and the half-cell A is the anode

Part B. 0.017V

Explanation:

Part A

The electrons must go from the anode to the cathode. At the anode oxidation takes place, and at the cathode a reduction, so the flow of electrons must go from the less concentrated solution to the most one (at oxidation the concentration intends to increase, and at the reduction, the concentration intends to decrease).

So, the half-cell B is the cathode and the half-cell A is the anode.

Part B

By the Nersnt equation:

E°cell = E° - (0.0592/n)*log[anode]/[cathode]

Where n is the number of electrons being changed in the reaction, in this case, n = 2 (Sn goes from S⁺²). Because the half-reactions are the same, the reduction potential of the anode is equal to the cathode, and E° = 0 V.

E°cell = 0 - (0.0592/2)*log(0.23/0.87)

E°cell = 0.017V

3 0

3 years ago

Consider the reaction of magnesium metal with hydrochloric acid to produce magnesium chloride and hydrogen gas. If 3.29 mol of m

gregori [183]

Answer:

1.645 moles of excess reactant that is of magnesium metal are left over.

Explanation:

Moles of magnesium metal = 3.29 mol

Moles of HCl = 3.29 mol

$Mg(s)+2HCl(aq)\rightarrow MgCl_2(aq)+H_2(g)$

According to recation, 2 moles of HCl reacts with 1 mol of magnesium metal, then 3.29 moles of HCl will react with :

$\frac{1}{2}\times 3.29 mol=1.645 mol$ of magnesium metal

Moles of HCl left = 3.29mol - 3.29 mol = 0

Moles of magnesium metal left = 3.29 mol - 1.645 mol = 1.645 mol

1.645 moles of excess reactant that is of magnesium metal are left over.

7 0

3 years ago

If you go from the Earth to the Moon, will your MASS change?

Solnce55 [7]

Answer:

No matter if you are on Earth, the moon or just chilling in space, your mass does not change. But your weight depends on the gravity force; you would weigh less on the moon than on Earth, and in space you would weigh almost nothing at all.

8 0

2 years ago

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