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2 years ago

Element X reacts with element Y to give a product containing X3+ ions and Y2− ions.

Chemistry

1 answer:

Maksim231197 [3]2 years ago

5 0

Answer:

X₂Y₃

Explanation:

X⇒X³⁺ + 3e⁻ /×2

Y + 2e⁻ ⇒ Y²⁻ /×3

2X⇒2X³⁺ 6e⁻

3Y + 6e⁻⇒ 3Y²⁻

2X + 3Y ⇒2X³⁺ + 3Y²⁻ ⇒ X₂Y₃

Electron from one side and from other side can be shortened so we multiply half equations to get equivalent number of electrons on both side. Next step is summing these two half equations. Element Y takes two electrons from element X, so X become positive charged, and Y become negatively charged. It is very likely that element Y have greater electronegativity.

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A. what are substances called whose water solutions conduct electricity?

erastova [34]

A. If a substance dissociates in water solution on ions, its water solution conduct electricity
B. salt dissociates in water solution on ions,so its water solution conduct electricity
C. sugar does not dissociates in water solution on ions,so its water solution does not conduct electricity

3 0

2 years ago

How many liters of SO2 will be produced from 26.9L O2?

Alexeev081 [22]

Answer:

26.9 L SO₂

Explanation:

Step 1: Write the balanced equation

S(s) + O₂(g) = SO₂(g)

Step 2: Establish the appropriate volume ratio

For gases at the same conditions, the volume ratio is equal to the molar ratio. The volume ratio of O₂(g) to SO₂(g) is 1:1.

Step 3: Calculate the liters of SO₂ produced from 26.9 L of O₂

We will use the previously established volume ratio.

26.9 L O₂ × 1 L SO₂/1 L O₂ = 26.9 L SO₂

5 0

1 year ago

What is the oxidation state of an individual bromine atom in kbro2?

san4es73 [151]

The oxidation state of potassium ion K = +1
The oxidation state of oxygen ion O = -2
So, the oxidation state of O2 is = -2 x 2 = -4
Since, KBrO2 is neutral so,
(+1) + (x) + (-4) = Zero
-3 + X = Zero
So, X = +3
The oxidation state of individual bromine atom in KBrO2 is +3

6 0

2 years ago

0.9775 grams of an unknown compound is dissolved in 50.0 ml of water. Initially the water temperature is 22.3 degrees Celsius. A

elena-14-01-66 [18.8K]

Answer:

The enthlapy of solution is -55.23 kJ/mol.

Explanation:

Mass of water = m

Density of water = 1 g/mL

Volume of water = 50.0 mL

m = Density of water × Volume of water = 1 g/mL × 50.0 mL=50.0 g

Change in temperature of the water ,ΔT= 27.0°C - 22.3°C = 4.7°C

Heat capacity of water,c =4.186 J/g°C

Heat gained by the water when an unknown compound is dissolved be Q

Q= mcΔT

$Q=50.0 g\times 4.186 J/g^oC\times 4.7^oC=983.71 J$

heat released when 0.9775 grams of an unknown compound is dissolved in water will be same as that heat gained by water.

Q'=-Q

Q'= -983.71 J =-0.98371 kJ

Moles of unknown compound = $\frac{0.9975 g}{56 g/mol}=0.01781 mol$

The enthlapy of solution :

$\frac{Q'}{moles}$

$=\frac{-0.98371 kJ}{0.01781 mol}=-55.23 kJ/mol$

The enthlapy of solution is -55.23 kJ/mol.

8 0

2 years ago

In a laboratory experiment, the freezing point of an aqueous solution of glucose is found to be -0.325°C, What is the molal conc

Hatshy [7]

0.17 M is the is the molal concentration of this solution

Explanation:

Data given:

freezing point of glucose solution = -0.325 degree celsius

molal concentration of the solution =?

solution is of glucose=?

atomic mass of glucose = 180.01 grams/mole

freezing point of glucose = 146 degrees

freezing point of water = 0 degrees

Kf of glucose = 1.86 °C

ΔT = (freezing point of solvent) - (freezing point of solution)

ΔT = 0.325 degree celsius

molality =?

ΔT = Kfm

rearranging the equation:

m = $\frac{0.325}{1.86}$

m= 0.17 M

molal concentration of the glucose solution is 0.17 M

3 0

2 years ago