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choli [55]
2 years ago
15

What is the total number of neutrons in the nucleus of a neutral atom that has 19 electrons and a mass number of 39

Chemistry
1 answer:
zvonat [6]2 years ago
7 0

Answer:

NO OF NEUTRON =MASS NO -ATOMIC NO

=39-19

=20

bcz atomic no is equal to no of electrons

mark as brainliest

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Resistance depends on which properties of a wire
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Read 2 more answers
How many moles of Cu(OH)2 are soluble in 1L of sodium hydroxide (NaOH) when the pH is 8.23?
Morgarella [4.7K]

Answer:

4.96E-8 moles of Cu(OH)2

Explanation:

Kps es the constant referring to how much a substance can be dissolved in water. Using Kps, it is possible to know the concentration of weak electrolytes. Then, pKps is the minus logarithm of Kps.

Now, we know that sodium hydroxide (NaOH) is a strong electrolyte, who is completely dissolved in water. Therefore the pH depends only on OH concentration originating from NaOH. Let us to figure out how much is that OH concentration.

pH= -log[H]\\pH= -log (\frac{kw}{[OH]})

8.23 = - log(\frac{Kw}{[OH]} \\10^{-8.23} = Kw/[OH]\\ [OH] = Kw/10^{-8.23}

[OH]=1.69E-6

This concentration of OH affects the disociation of Cu(OH)2. Let us see the dissociation reaction:

Cu(OH)_2 -> Cu^{2+} + 2OH^-

In the equilibrum, exist a concentration of OH already, that we knew, and it will be added that from dissociation, called "s":

The expression for Kps is:

Kps= [Cu^{2+}] [OH]^2

The moles of (CuOH)2 soluble are limitated for the concentration of OH present, according to the next equation.

Kps= s*(2s+1.69E-6)^2

"s" is the soluble quantity of Cu(OH)2.

The solution for this third grade equation is s=4.96E-8 mol/L

Now, let us calculate the moles in 1 L:

moles Cu(OH)_2 = 4.96E-8 mol/L * 1 L = 4.96E-8 moles

7 0
3 years ago
A sample of g of pure aluminum metal is added to mL of M hydrochloric acid. The volume of hydrogen gas produced at standard temp
MArishka [77]

Answer:

V = 11.2L are produced

Explanation:

... <em>Sample of 27g of pure aluminium, 3added to 333 mL of 3.0 M HCl..</em>

Based on the chemical reaction:

2Al(s) + 6HCl(aq) → 2AlC₃(aq) + 3H₂(g)

<em>Where 3 moles of hydrogen are produced when 6 moles of hydrochloric acid reacts with 2 moles of Al.</em>

<em />

To solve this question, we need to determine limiting reactant converting each reactant to moles. With limiting reactant and the chemical reaction we can find moles of hydrogen and its volume at STP (T=273.15K; P=1atm), thus:

<em>Moles Al-Molar mass: 26.98g/mol-:</em>

27g * (1mol / 26.98g) = 1mol of Al

<em>Moles HCl:</em>

333mL = 0.333L * (3mol/L) = 1mol HCl

For a complete reaction of 1 mole of HCl are required:

1mol HCl * (2mol Al / 6mol HCl) = 0.333 moles of Al. As there is 1 mole of Al, Al is in excess and <em>HCl is limiting reactant.</em>

<em />

Moles of Hydrogen produced are:

1mol HCl * (3 moles H₂ / 6 mol HCl) = 0.5moles H₂ are produced.

Using ideal gas law:

PV = nRT

V = nRT/P

<em>Where V is volume</em>

<em>n are moles: 0.5mol</em>

<em>R is gas constant: 0.082atmL/molK</em>

<em>T is absolute temperature: 273.15K</em>

<em>P is pressure: 1atm.</em>

<em />

Solving for V:

V = 0.5mol*0.082atmL/molK*273.15K / 1atm

<h3>V = 11.2L are produced</h3>
3 0
2 years ago
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