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givi [52]
3 years ago
8

Give the nuclear symbol for the isotope of gallium, ga, that contains 40 neutrons per atom. Replace the question marks with the

proper integers.
Chemistry
1 answer:
Cerrena [4.2K]3 years ago
7 0

The nuclear symbol of an element have three parts namely:  

  1. the symbol of an element
  2. mass number of the element
  3. atomic number of the element

The general representation is as:

_{Z}^{A}X

where X is the chemical symbol of the element, A is mass number of the element that is total number of neutrons and protons in the nucleus of an atom, and Z is atomic number of the element that is number of protons in the atom.

Element = Gallium, Ga    (given)

Number of neutrons = 40 (given)

Let the nuclear symbol be _{?}^{?}Ga. So, we need to determine the values of "?" in the symbol.

Atomic number of Gallium, Ga = 31

Since, atomic number = Number of protons.

So, number of protons of Gallium, Ga = 31

Mass number = number of protons + number of neutrons

Substituting the values:

Mass number = 31+40 = 71

So, the nuclear symbol is _{31}^{71}Ga.

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Which reactant (X) is missing from the equation shown?<br> X+ PO4 → H3PO4
insens350 [35]

Answer:

H

Explanation:

b/c the polyatomic PO4¯³

have 3 valance number

3 0
3 years ago
Mercury is the only metal that is a liquid at room temperature. When mercury vapor is inhaled, it is readily absorbed by the lun
Lapatulllka [165]

Answer:

P = 0.0166 mm Hg

Explanation:

To solve this question, we need to use the Clausius Clapeyron equation, which is a commonly used expression to calculate vapour pressure at a given temperature. We have the enthalpy of vaporization of the mercury, so, let's write the equation:

Clausius Clapeyron equation:

Ln (P₂ / P₁) = (-ΔHv / R)(1/T₂ - 1/T₁)    (1)

Where:

R: universal constant of gases (8.314 J / K.mol)

P₂: Vapour pressure at 43°C (or 316 K)

P₁: Pressure of mercury at the boiling point (1 atm)

T₂: temperature at 43 °C

T₁: Boiling point of mercury (357 °C or 630 K)

As we are given the boiling point of the mercury, we can safely assume that the pressure at this point is 1 atm, becuase remember that when a sustance boils, is because it's internal pressure has reached the atmospherical pressure of 1 atm. With this clear, all we just need to do is solve for P₂. We are going to do this very slowly so you can understand the process. First let's replace the given data:

Ln (P₂ / 1) = (-59100 J/mol / 8.314 J / K.mol) (1/316 - 1/630)

Ln P₂ = -7108.49 * (3.16x10⁻³ - 1.59x10⁻³)

Ln P₂ = -7108.49 * (1.51x10⁻³)

Ln P₂ = -10.7338

P₂ = 10⁽⁻¹⁰°⁷³³⁸⁾

P₂ = 2.18x10⁻⁵ atm

We can express this value in mm Hg and it will be:

P₂ = 2.18x10⁻⁵ * 760

<h2>P₂ = 0.0166 mm Hg</h2>

Hope this helps

8 0
3 years ago
The diagram below show an enlarged view of the beams of a triple-beam balance
VLD [36.1K]

Answer:

1) 455.2 g. 2) 545.2 g.

Explanation:

i think this is right

6 0
3 years ago
What is that the theoretical yield of aluminum oxide I if 3.20 mol of aluminum metal is exposed to 2.70 mole of oxygen
photoshop1234 [79]

Answer:

163.2g

Explanation:

First let us generate a balanced equation for the reaction. This is shown below:

4Al + 3O2 —> 2Al2O3

From the question given, were were told that 3.2moles of aluminium was exposed to 2.7moles of oxygen. Judging by this, oxygen is excess.

From the equation,

4moles of Al produced 2moles of Al2O3.

Therefore, 3.2moles of Al will produce = (3.2x2)/4 = 1.6mol of Al2O3.

Now, let us covert 1.6mol of Al2O3 to obtain the theoretical yield. This is illustrated below:

Mole of Al2O3 = 1.6mole

Molar Mass of Al2O3 = (27x2) + (16x3) = 54 + 48 =102g/mol

Mass of Al2O3 =?

Number of mole = Mass /Molar Mass

Mass = number of mole x molar Mass

Mass of Al2O3 = 1.6 x 102 = 163.2g

Therefore the theoretical of Al2O3 is 163.2g

8 0
3 years ago
A solution has a [OH-] of 1 × 10-9. What is the pOH of this solution?
zhannawk [14.2K]
POH = - log [ OH⁻ ]

pOH = - log [ 1 x 10⁻⁹ ] 

pOH = 9

Answer C

hope this helps!
3 0
3 years ago
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