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2 years ago

Is krypton a anion or cation?

Chemistry

2 answers:

Zarrin [17]2 years ago

7 0

Answer:

krypton is a cation

magnesium is also a cation

aluminium is a anion

and , silicon is cation

cluponka [151]2 years ago

7 0

Answer:

• Krypton: cation
• Magnesium: cation
• Aluminum: anion
• Silicone: cation

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How many moles of HCl are present in 1.30 L of a 0.50 M solution

Shtirlitz [24]

Answer:

0.65moles of HCl was produced

4 0

3 years ago

A compound is 44.82% Potassium, 18.39% Sulfur, and 36.79% Oxygen. Write the empirical formula and name the compound.

sleet_krkn [62]

Answer:

Empirical formula = K₂SO₄

The name of the compound is Potassium tetraoxosulfate (vi)

Explanation:

Percentage composition of the elements:

Potassium = 44.82%

Sulfur = 18.39%

Oxygen = 36.79%

Mole ratio of the elements:

Potassium = 44.82/39 =1.15

Sulfur = 18.39/32 = 0.57

Oxygen = 36.79/16 = 2.29

Dividing by the smallest ratio in order to obtain the simplest ratio

Potassium = 1.15/0.57 =2

Sulfur = 0.57/0.57 = 1

Oxygen = 2.29/0.57 = 4

Therefore, the mole ratio of Potassium : Sulfur : Oxygen = 2:1:4

Empirical formula = K₂SO₄

The name of the compound is Potassium tetraoxosulfate (vi)

5 0

3 years ago

"acid is responsible for the odor in rancid butter. a solution of 0.25 m butyric acid has a ph of 2.71. what is the ka for"

Salsk061 [2.6K]

Answer:- The Ka for the acid is $1.53*10^-^5$ .

Solution:- In general, monoprotic acid could be represented by HA. The dissociation equation for the ionization of HA is written as:

HA(aq)\rightarrow H^+(aq) + A^-(aq)

Now, we make the ice table for this equation as:

HA(aq)\rightarrow H^+(aq) + A^-(aq)

I 0.25 0 0

C -X +X +X

E (0.25 - X) X X

where, I stands for initial concentration, C stands for change in concentration and E stands for equilibrium concentration.

X is the change in concentration and from ice table it's same as the concentration of hydrogen ion that is calculated from given pH.

$Ka = [H^+][A^-]\frac{1}{HA}$

Where, Ka is the acid ionization constant. Let's plug in the values.

$Ka = \frac{X^2}{0.25-X}$

Let's calculate the value of X first using the equation:

$pH = -log[H^+]$ [/tex]

on taking antilog ob above equation we get:

$[H^+]=10^-^p^H$

$[H^+]=10^-^2^.^7^1$

$[H^+]$ = 0.00195

So, X = 0.001195

Let's plug in this value of X in the equation:-

$Ka=\frac{(0.00195)^2}{0.25-0.00195}$

$Ka=1.53*10^-^5$

So, the value of Ka for butyric acid is $1.53*10^-^5$ .

8 0

3 years ago

Read 2 more answers

A 1.00 g sample of a metal X (that is known to form X ions in solution) was added to 127.9 mL of 0.5000 M sulfuric acid. After a

Semenov [28]

<u>Answer:</u> The metal having molar mass equal to 26.95 g/mol is Aluminium

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ .....(1)

Molarity of NaOH solution = 0.5000 M

Volume of solution = 0.03340 L

Putting values in equation 1, we get:

$0.5000M=\frac{\text{Moles of NaOH}}{0.03340L}\\\\\text{Moles of NaOH}=(0.5000mol/L\times 0.03340L)=0.01670mol$

The chemical equation for the reaction of NaOH and sulfuric acid follows:

$2NaOH+H_2SO_4\rightarrow Na_2SO_4+H_2O$

By Stoichiometry of the reaction:

2 moles of NaOH reacts with 1 mole of sulfuric acid

So, 0.01670 moles of NaOH will react with = $\frac{1}{2}\times 0.01670=0.00835mol$ of sulfuric acid

Excess moles of sulfuric acid = 0.00835 moles

Calculating the moles of sulfuric acid by using equation 1, we get:

Molarity of sulfuric acid solution = 0.5000 M

Volume of solution = 127.9 mL = 0.1279 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

$0.5000M=\frac{\text{Moles of }H_2SO_4}{0.1279L}\\\\\text{Moles of }H_2SO_4=(0.5000mol/L\times 0.1279L)=0.06395mol$

Number of moles of sulfuric acid reacted = 0.06395 - 0.00835 = 0.0556 moles

The chemical equation for the reaction of metal (forming $M^{3+}$ ion) and sulfuric acid follows:

$2X+3H_2SO_4\rightarrow X_2(SO_4)_3+3H_2$

By Stoichiometry of the reaction:

3 moles of sulfuric acid reacts with 2 moles of metal

So, 0.0556 moles of sulfuric acid will react with = $\frac{2}{3}\times 0.0556=0.0371mol$ of metal

To calculate the molar mass of metal for given number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Mass of metal = 1.00 g

Moles of metal = 0.0371 moles

Putting values in above equation, we get:

$0.0371mol=\frac{1.00g}{\text{Molar mass of metal}}\\\\\text{Molar mass of metal}=\frac{1.00g}{0.0371mol}=26.95g/mol$

Hence, the metal having molar mass equal to 26.95 g/mol is Aluminium

6 0

3 years ago

Select the correct answer. The gas in a sealed container has an absolute pressure of 9.25 atmospheres. If the air around the con

N76 [4]

Answer:

Explanation:

8.25 atm

7 0

3 years ago

Read 2 more answers