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zlopas [31]
2 years ago
6

Hello guys what is good! What are all the elements of the P block?

Chemistry
1 answer:
Ghella [55]2 years ago
6 0

boron, carbon, nitrogen, oxygen and flourine families in addition to the noble gases.

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also to answer the 'hello guys what is good!':nothin u?

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Atomic mass W = 183.84 u.m.a

1 mole --------- 183.84
1.4 moles ---- ?

1.4 x 183.84 / 1 = 257.376 g

hope this helps!
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(4 points) The following lead compound for a pharmaceutical drug contains a rotatable bond. Using the principles of rigidificati
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Explanation:

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1. A student collected the following data for a fixed volume of gas: Temperature (⁰C) Pressure (mm of Hg) 10 726 20 750 40 800 7
zimovet [89]

Answer is: the missing pressure is 1088.66 mmHg.

Gay-Lussac's Law states that the pressure of a given amount of gas held at constant volume is directly proportional to the Kelvin temperature.

p₁/T₁ = p₂/T₂.

p₁ = 960 mmHg; pressure of the gas.

T₁ = 100°C + 273.15.

T₁ = 373.15 K; temperature of the gas.

T₂ = 150°C + 273.15.

T₂ = 423.15 K.

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6 0
3 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
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