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faust18 [17]
2 years ago
14

Give the following number of protons, neutrons, and electrons in the following isotope:

Chemistry
1 answer:
valina [46]2 years ago
3 0

Answer:

47, 68, 47

Explanation:

Let's consider the following isotope: ¹¹⁵Ag.

If we look in the Periodic Table, the atomic number of Ag is 47, that is, silver has 47 protons.

Since ¹¹⁵Ag is neutral it must have the same number of protons that of electrons. Then, it also has 47 electrons.

The number 115 in ¹¹⁵Ag stands for the mass number, which is the sum of protons and neutrons. Then, the number of neutrons is:

p⁺ + n⁰ = 115

n⁰ = 115 - p⁺ = 115 - 47 = 68

The number of protons, neutrons, and electrons in ¹¹⁵Ag is: 47, 68, 47.

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Mg₃N₂ + 6H₂O  =  3Mg(OH)₂ + 2NH₃
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Based on the activity series, which metals could X represent in the reaction below? (Note: The equation is not balanced.)
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Answer this, please!!!!!!!!!!
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It's C

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5 0
3 years ago
Read 2 more answers
Some one help me please I'm stuck Show using two conversion factors how you would convert from 0.020 kg into mg.
julia-pushkina [17]

Answer:

=2.0x10^4mg

Explanation:

Hello there!

In this case, when performing units conversions involving two proportional factors we need to make sure we first convert to the base unit and then to the target one; thus, since 1 kg = 1000 g and 1 g = 1000 mg, we set up the following expression:

=0.020kg*\frac{1000g}{1kg} *\frac{1000mg}{1g} \\\\=2.0x10^4mg

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4 0
2 years ago
The raw water supply for a community contains 18 mg/L total particulate matter. It is to be treated by addition of 60 mg alum (A
s344n2d4d5 [400]

Solution :

Given :

The steady state flow = 8000 $ m^3 /d $

                                    $= 80 \times 10^5 \ I/d $

The concentration of the particulate matter = 18 mg/L

Therefore, the total quantity of a particulate matter in fluid $= 80 \times 10^5 \ I/d \times 18 \ mg/L $

$= 144 \times 10^6 \ mg/g$

$= 144 \ kg/d $

If 60 mg of alum $ [Al_2(SO_4)_3.14 H_2O] $ required for one litre of the water treatment.

So Alum required for  $ 80 \times 10^5 \ I/d $

$= 80 \times 15^5 \ I/d  \times 60 \ mg \ alum /L$

$= 480 \times 10^6 \ mg/d $

or 480 kg/d

Therefore the alum required is 480 kg/d

1 mg of the alum gives 0.234 mg alum precipitation, so 60 mg of alum will give $ = 60 \times 0.234 \text{ of alum ppt. per litre} $

      $= 14.04 $ mg of alum ppt. per litre

480 kg of alum will give = 480 x 0.234 kg/d

                                        = 112.32 kg/d ppt of alum

Daily total solid load is  $= 144 \ kg/d + 112.32 \ kg/d$

                                       = 256.32 kg/d

So, the total concentration of the suspended solid after alum addition $= 18 \ mg/L + 60 \times 0.234 $

= 32.04 mg/L

Therefore total alum requirement = 480 kg/d

b). Initial pH = 7.4

 The dissociation reaction of aluminium hydroxide as follows :

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

After addition, the aluminium hydroxide pH of water will increase due to increase in $ OH^- $ ions.

Therefore, the pH of water will be acceptable range after the addition of aluminium hydroxide.

c). The reaction of $CO_2$ and water as follows :

$CO_2 (g) + H_2O (l) \rightarrow H_2CO_3$

For the atmospheric pressure :

$p_{CO_2} = 3.5 \times 10^{-4} \ atm $

And the pH is reduced into the range of 5.9 to 6.4

6 0
2 years ago
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