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3 years ago

Give the ion symbol for the the ion that has 15 protons and 18 electrons

2 answers:

Montano1993 [528]3 years ago

7 0

By referring to a periodic table or table of elements, we see that phosphorus (symbol P) has an atomic number of 15. Thus, each atom has 15 protons. The mass number of the ion is 15 + 16 = 31. Because the ion has 15 protons and 18 electrons (three more electrons than protons), its net charge is 3-.

lana66690 [7]3 years ago

7 0

Answer: The symbol of the ion is $P^{3-}$

Explanation:

An ion is formed when a neutral atom looses or gains electrons.

When an atom looses electrons, it results in the formation of positive ion known as cation.
When an atom gains electrons, it results in the formation of negative ion known as anion.

Atomic number is defined as the number of protons or electrons that are present in a neutral atom.

Atomic number = number of protons = number of electrons

We are given:

Number of protons = 15

Number of electrons = 18

Charge = Number of protons - Number of electrons = 15 - 18 = -3

The element having 15 protons is phosphorus. So, the symbol of the ion is $P^{3-}$ ion

Hence, the symbol of the ion is $P^{3-}$

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Can you discount alcohol in Texas?.

iragen [17]

According to TABC laws on alcohol use, it is illegal to sell alcohol in many places in texas thus no you can't discount alcohol in Texas.

Alcohol is prohibited in many parts of Texas thus anyone under the age of 21 may not be sold alcohol, including adults. Beer and wine may be purchased from licensed establishments Monday through Friday from 7:00 a.m. to midnight. On Saturday, they may sell them from 7:00 a.m. until 1:00 p.m. Additionally, they may sell on Sundays from noon to midnight. Only liquor stores, nevertheless, are allowed to sell distilled alcohol. On Sundays before noon, restaurants are allowed to serve alcoholic beverages with food. On the other hand, it is prohibited to sell any alcoholic beverages with a volume of more than 17% on Sunday (ABV). On Thanksgiving Day, New Year's Day, or Christmas Day, no establishment may offer any alcoholic beverages. Alcohol from a single producer cannot be sold by a restaurant or other retailer. And alcohol cannot be sold by food trucks.

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7 0

1 year ago

What else is produced during the combustion of butane, C4H10? 2C4H10 + 13O2 → + 10H2O

Neko [114]

<h3>Answer:</h3>

8CO₂

<h3>Explanation:</h3>

We are given;

Butane, C₄H₁₀

Butane is a hydrocarbon in the homologous series known as alkane.

We are required to determine the other product produced in the combustion of butane apart from water.

We know that the complete combustion of alkane yields carbon dioxide and water.
Therefore, combustion of butane will yield carbon dioxide and water.

The balanced equation for the complete combustion of butane will be;

2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O

8 0

3 years ago

Read 2 more answers

If 1 mol of XO₂ contains the same number of atoms as 60 g of XOs, what is the molar mass of XO₂?

Lena [83]

Answer:

44 grams/mole

Explanation:

If 1 mol of XO₂ contains the same number of atoms as 60 g of XO3, what is the molar mass of XO₂?

60 grams of XO3 is one mole XO3, since it has the same number of atoms as 1 mole of XO2.

Let c be the molar mass of X. The molar mass of XO3 is comprised of:

X: c

3O: 3 x 16 = 48

Total molar mass of XO3 is = 48 + c

We know that the molar mass of XO3 = 60 g/mole, so:

48 + c = 60 g/mole

c = 12 g/mole

The molar mass of XO2 would be:

1 X = 12

2 O = 32

Molar mass = 44 grams/mole, same as carbon dioxide. Carbon's molar mass is 12 grams.

5 0

2 years ago

At equilibrium, the concentrations of the products and reactants for the reaction, H2 (g) + I2 (g)  2 HI (g), are [H2] = 0.106

lana [24]

Answer:

The new equilibrium concentration of HI: [HI] = 3.589 M

Explanation:

Given: Initial concentrations at original equilibrium- [H₂] = 0.106 M; [I₂] = 0.022 M; [HI] = 1.29 M

Final concentrations at new equilibrium- [H₂] = 0.95 M; [I₂] = 0.019 M; [HI] = ? M

Given chemical reaction: H₂(g) + I₂(g) → 2 HI(g)

The equilibrium constant ( $K_{c}$ ) for the given chemical reaction, is given by the equation:

$K_{c} = \frac {[HI]^{2}}{[H_{2}]\: [I_{2}]}$

At the original equilibrium state:

$K_{c} = \frac {(1.29\: M)^{2}}{(0.106\: M) \times (0.022\: M)}$

$K_{c} = \frac {1.6641}{0.002332} = 713.59$

Therefore, at the new equilibrium state:

$K_{c} = \frac {[HI]^{2}}{(0.95\: M) \times (0.019\: M)}$

$\Rightarrow K_{c} = 713.59 = \frac {[HI]^{2}}{0.01805}$

$\Rightarrow [HI]^{2} = 713.59 \times 0.01805 = 12.88$

$\Rightarrow [HI] = \sqrt {12.88} = 3.589 M$

Therefore, the new equilibrium concentration of HI: [HI] = 3.589 M

6 0

3 years ago

Pls help me with this

Zina [86]

Answer:

$[I_2]=[Br]=0.31M$

Explanation:

Hello there!

In this case, according to the given information, it is possible for us to set up the following chemical equation at equilibrium:

$I_2+Br_2\rightleftharpoons 2IBr$

Now, we can set up the equilibrium expression in terms of x (reaction extent) whereas the initial concentration of both iodine and bromine is 0.5mol/0.250L=2.0M:

$K=\frac{[IBr]^2}{[I_2][Br_2]} \\\\1.2x10^2=\frac{(2x)^2}{(2.0-x)^2}$

Thus, we solve for x as show below:

$\sqrt{1.2x10^2} =\sqrt{\frac{(2x)^2}{(2.0-x)^2}} \\\\10.95=\frac{2x}{2.0-x}\\\\21.91-10.95x=2x\\\\21.91=12.95x\\\\x=\frac{21.91}{12.95} \\\\x=1.69M$

Therefore, the concentrations of both bromine and iodine are:

$[I_2]=[Br]=2.0M-1.69M=0.31M$

Regards!

8 0

3 years ago