As the temperature increases from 0°C to 25°C the amount of NH3 that can be dissolved in 100 grams of water.

How many electrons are there in the valence shell of the oxygen atom of water? O a. 2 O 6.4 OC.6 O d. 8

denpristay [2]

Answer:

d. 8

Explanation:

Valence electrons of oxygen = 6

Valence electrons of hydrogen = 1

The total number of the valence electrons = 6 + 2(1) = 8

The Lewis structure is drawn in such a way that the octet of each atom and duet for the hydrogen in the molecule is complete.

In the water molecule,

Oxygen has 2 bond pairs and two lone pairs which means that the total electrons in the valence shell is 8.

7 0

3 years ago

a sample that has a half life of 45 years take ____ for the count rate to drop from 600 cpm to 150 cpm

Bond [772]

Answer:

90yrs

Explanation:

Half life is the time taken for half of an original radioactive material to decay by half.

For the sample;

Half - life = 45yrs

Original amount = 600cpm

Start Number of atoms

0yrs 600cpm

First half life

45yrs 300cpm

Second half life

90yrs 150cpm

5 0

3 years ago

Using standard heats of formation, calculate the standard enthalpy change for the following reaction. 2NH3(g) 3N2O(g)4N2(g) 3H2O

trasher [3.6K]

Answer: $\Delta H^{0} = -879.15 kJ/mol$

Explanation: Heats of formation is the amount of heat necessary to create 1 mol of a compound from its molecular constituents. The basic conditions the substance is formed is at standard conditions: 1 atm and 25°C. Each compound has its own heat of formation per mol of compound (kJ/mol), but to an element is assigned a value of zero.

Standard Enthalpy Change is defined as the heat absorbed or released when a reaction takes place. It can be positive or negative, which means reaction is endothermic or exothermic, respectively.

Enthalpy change is calculated as the difference between the sum of heat formation of products and the sum of heat formation of the reactants:

$\Delta H^{0}=\Sigma H^{0}_{f}_{(products)}-\Sigma H^{0}_{f}_{(reactants)}$

For the reaction

2NH₃ + 3N₂O → 4N₂ + 3H₂O

2(-46.2) + 3(82.05) 4(0) + 3(-241.8)

$\Delta H^{0}=3(-241.8)-[ 2(-46.2)+3(82.05)]$

$\Delta H^{0}=-725.4-153.75$

$\Delta H^{0}=-879.15$

The standard enthalpy change for the reaction is $\Delta H^{0}=-879.15$ kJ

8 0

3 years ago

Identify the type of bonding, molecular geometry (shape), and intermolecular forces experienced by the compounds HF and HBr. Whi

Anestetic [448]

Im not sure but i think its hbr. i hope im right and i hope this helped,

8 0

3 years ago

Use your periodic table to answer the following questions. Pay attention to the spelling of the element names and make sure to w

Novay_Z [31]

Explanation:

1. K (Potassium)

2. Te (Tellurium)

3. F (Flourine)

4. group 11, period 4

5. group 18, period 4

6. group 12, period 6

7. Cl

8. Pb

9. W

10. Sb

11. Na

12. period 6, group 11

13. period 5, group 11

14. period 2, group 16

15. Groups 13–16 of the periodic table contain one or more metalloids

16. Hydrogen

17. Helium

18. Calcium

19. Cobalt

20. Carbon

3 0

3 years ago