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Strike441 [17]
3 years ago
14

(in the image)

Chemistry
1 answer:
Ad libitum [116K]3 years ago
7 0

Answer:

[Kr] 5s²

Explanation:

From the question given above, the following data were obtained:

Atomic number of strontium (Sr) = 38

Electronic configuration =?

Next, we shall determine the electronic configuration of the noble gas element before strontium (Sr).

The noble gas element before strontium (Sr) is krypton (Kr). Thus, the electronic configuration of krypton (Kr) is given below:

Kr (36) => 1s² 2s²2p⁶ 3s²3p⁶ 3d¹⁰ 4s²4p⁶

Finally, we shall determine the electronic configuration of strontium (Sr). This can be obtained as follow:

Kr (36) => 1s² 2s²2p⁶ 3s²3p⁶ 3d¹⁰ 4s²4p⁶

Sr (38) =>?

Sr (38) => 1s² 2s²2p⁶ 3s²3p⁶ 3d¹⁰ 4s²4p⁶ 5s²

But

Kr (36) => 1s² 2s²2p⁶ 3s²3p⁶ 3d¹⁰ 4s²4p⁶

Therefore,

Sr (38) => [Kr] 5s²

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gThe mole fraction of potassium nitrate in an aqueous solution is 0.0194. The solution's density is 1.0627 g/mL. What is the mol
xxMikexx [17]

Answer:

0.595 M

Explanation:

The number of moles of water in 1L = 1000g/18g/mol = 55.6 moles of water.

Mole fraction = number of moles of KNO3/number of moles of KNO3 + number of moles of water

0.0194 = x/x + 55.6

0.0194(x + 55.6) = x

0.0194x + 1.08 = x

x - 0.0194x = 1.08

0.9806x= 1.08

x= 1.08/0.9806

x= 1.1 moles of KNO3

Mole fraction of water= 55.6/1.1 + 55.6 = 0.981

If

xA= mole fraction of solvent

xB= mole fraction of solute

nA= number of moles of solvent

nB = number of moles of solute

MA= molar mass of solvent

MB = molar mass of solute

d= density of solution

Molarity = xBd × 1000/xAMA ×xBMB

Molarity= 0.0194 × 1.0627 × 1000/0.981 × 18 × 0.0194×101

Molarity= 20.6/34.6

Molarity of KNO3= 0.595 M

8 0
3 years ago
Classify each organic compound based on the functional group it contains.
bixtya [17]

Answer:

H2C2O1

Explanation:

8 0
2 years ago
7. There are 7. 0 ml of 0.175 M H2C2O4 , 1 ml of water , 4 ml of 3.5M KMnO4 what is the molar concentration ofH2C2O4 ?
Illusion [34]

Answer:

7. 0.1021 M

8. 1.167 M

10. Increase in volume of water would lower the rate of reaction

Explanation:

7. What is the molar concentration of H₂C₂O₄ ?

Since we have 7.0 ml of 0.175 M H₂C₂O₄, the number of moles of H₂C₂O₄ present n = molarity of H₂C₂O₄ × volume of H₂C₂O₄ = 0.175 mol/L × 7.0 ml = 0.175 mol/L × 7 × 10⁻³ L = 1.225 × 10⁻³ mol.

Also, the total volume present V = volume of H2C2O4 + volume of water + volume of KMnO4 = 7.0 ml + 1 ml + 4 ml = 12 ml = 12 × 10⁻³ L

So, the molar concentration of H₂C₂O₄, M = number of moles of H₂C₂O₄/volume = n/V

= 1.225 × 10⁻³ mol/12 × 10⁻³ L

= 0.1021 mol/L

= 0.1021 M

8. Using the data from question 7 what is the molar concentration of KMnO₄ ?

Since we have 4.0 ml of 3.5 M KMnO₄, the number of moles of KMnO4 present n' = molarity of KMnO₄ × volume of KMnO₄ = 3.5 mol/L × 4.0 ml = 3.5 mol/L × 4 × 10⁻³ L = 14 × 10⁻³ mol.

Also, the total volume present V = volume of KMnO₄ + volume of water + volume of KMnO₄ = 7.0 ml + 1 ml + 4 ml = 12 ml = 12 × 10⁻³ L

So, the molar concentration of KMnO₄, M' = number of moles of KMnO₄/volume = n'/V

= 14 × 10⁻³ mol/12 × 10⁻³ L

= 1.167 mol/L

= 1.167 M

10. From question number 7, what effect increasing the volume of water has on the reaction rate?

Increase in volume of water would lower the rate of reaction because, the particles of both substances would have to travel farther distances to collide with each other, since there are less particles present in the solution and thus, the concentration of the particles would decrease thereby decreasing the rate of reaction.

3 0
2 years ago
How many joules are needed to warm 675 grams of water (specific heat= 4.186 J/g degrees Celsius) from 12 degrees Celsius to 85 d
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85-12 = 73 degrees needed
4.186 J/degree Celsius, so 
73 degrees * 4.186 J/degree = 305.578 J to raise 1 gram 73 degrees
there are 675 grams, so 305.578 * 675 = 206265.15 J

2.06 x 10^5 J are needed
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