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3 years ago

A single atom of an element has 11. protons, 11. electrons, and 12 neutrons. Which element is it?

Chemistry

2 answers:

8_murik_8 [283]3 years ago

5 0

Answer:

Sodium (Na)

Explanation:

The atomic number (it's order in the periodic table) is the same as the number of protons. Since there are 11 protons, it is the 11th element. Referencing the periodic table, that is Sodium, represented by Na.

Bumek [7]3 years ago

4 0

Answer: B

Explanation:

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The answer is: variable

sergeinik [125]

Answer:

variable: not consistent or having a fixed pattern; liable to change.

Explanation:

6 0

3 years ago

Rank the following elements by increasing atomic radius: carbon, aluminum, oxygen, potassium. Rank the following elements by inc

Lapatulllka [165]

The trend of atomic radius in the periodic table is increasing from right to left and from top to bottom. In reference to the table, oxygen has smaller radius compared to carbon, which is smaller than aluminum and which is also smaller than potassium. Thus, the trend from smallest to biggest is oxygen carbon, aluminum and potassium.

8 0

4 years ago

If the first energy level can hold 2 electrons, the second holds 8, and the third can hold 8, what if the atomic number is bigge

sergejj [24]

You're wrong because the third energy level can contain 18 electrons. And the 4th have up to 32 elections.

The pattern go by 2-8-18-32

4 0

3 years ago

Using Charles law if the volume of a gas is 5.0 L when the temperature is 5.0C what will the new volume be if the temperature is

Vikentia [17]

Answer:

The answer to your question is V2 = 5.09 l

Explanation:

Data

Volume 1 = V1 = 5.0 L

Temperature 1 = T1 = 5°C

Volume 2 = V2 = ?

Temperature 2 = T2 = 10°C

Formula (Charles law)

V1/T1 = V2/T2

-Solve for V2

V2 = V1T2 / T1

-Convert temperature to °K

T1 = 5 + 273 = 278°K

T2 = 10 + 273 = 283°K

-Substitution

V2 = (5)(283) / (278)

-Simplification

V2 = 1415 / 278

-Result

V2 = 5.09 l

4 0

3 years ago

Write the molecular equation and the net inonic equation for each of the following aqueous reactions. If no reaction occurs, wri

SIZIF [17.4K]

Answer: The equations are given below.

Explanation:

For the given options:

Option 1: $AgNO_3+NaI$

The molecular equation for this follows:

$AgNO_3(aq.)+NaI(aq.)\rightarrow AgI(s)+NaNO_3(aq.)$

Ionic form of the above equation follows:

$Ag^+(aq.)+NO_3^-(aq.)+Na^+(aq.)+I^-(aq.)\rightarrow AgI(s)+Na^+(aq.)+NO_3^-(aq.)$

As, sodium and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation.

The net ionic equation for the above reaction follows:

$Ag^+(aq.)+I^-(aq.)\rightarrow AgI(s)$

Option 2: $Ba(NO_3)_2+K_2SO_4$

The molecular equation for this follows:

$Ba(NO_3)_2(aq.)+K_2SO_4(aq.)\rightarrow 2KNO_3(aq.)+BaSO_4(s)$

Ionic form of the above equation follows:

$Ba^{2+}(aq.)+2NO_3^-(aq.)+K^+(aq.)+SO_4^{2-}(aq.)\rightarrow 2K^+(aq.)+2NO_3^-(aq.)+BaSO_4(s)$

As, potassium and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation.

The net ionic equation for the above reaction follows:

$Ba^{2+}(aq.)+SO_4^{2-}(aq.)\rightarrow BaSO_4(s)$

Option 3: $NH_4NO_3+K_2SO_4$

The molecular equation for this follows:

$2NH_4NO_3(aq.)+K_2SO_4(aq.)\rightarrow 2KNO_3(aq.)+H_2SO_4(aq.)+NH_3(g)$

Ionic form of the above equation follows:

$2NH_4^+(aq.)+2NO_3^-(aq.)+2K^+(aq.)+SO_4^{2-}(aq.)\rightarrow 2NH_3(g)+2H^+(aq.)+SO_4^{2-}(aq.)+2K^+(aq.)+2NO_3^-(aq.)$

As, nitrate, potassium and sulfate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation.

The net ionic equation for the above reaction follows:

$2NH_4^+(aq.)\rightarrow 2NH_3(g)+2H^+(aq.)$

Option 4: $LiCl+Al(NO_3)_3$

The molecular equation for this follows:

$3LiCl(aq.)+Al(NO_3)_3(aq.)\rightarrow 3LiNO_3(aq.)+AlCl_3(s)$

Ionic form of the above equation follows:

$3Li^+(aq.)+3Cl^-(aq.)+Al^{3+}(aq.)+NO_3^-(aq.)\rightarrow 3Li^+(aq.)+3NO_3^-(aq.)+AlCl_3(s)$

As, lithium and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation.

The net ionic equation for the above reaction follows:

$Al^{3+}(aq.)+3Cl^-(aq.)\rightarrow AlCl_3(s)$

Hence, the molecular and ionic equations are given above.

5 0

3 years ago