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2 years ago

I need help lol, I’ll brainlist u!

1 answer:

8 0

To make an electron jump into a higher level we basically need to provide them with the energy. This energy can be provided by heating or by striking the potassium surface with a high frequency monochromatic wave of light.

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The water lattice: a. is formed from hydrophobic bonds. b. causes ice to be denser than water. c. causes water to have a relativ

Alexus [3.1K]

Answer: Option (d) is the correct answer.

Explanation:

It is known that water loving bonds are called hydrophilic bonds and water hating bonds are called hydrophobic bonds. Since, water is a polar solvent and it is only able to dissolve polar molecules and not non-polar molecules.

Lattice of water and hydrogen bonding are responsible for the various properties of water like cohesion, adhesion, heat of vaporization etc.

Thus, we can conclude that water lattice excludes non-polar substances.

7 0

2 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

2 years ago

I NEED HELP ASAP PLEASE I BEG

zhenek [66]

with what?..................

7 0

2 years ago

Need help ASAP plz. Only gave 10 min

MrRissso [65]

Answer:

the answer is D

Explanation:

5 0

3 years ago

What is the concentration (M) of CH3OH in a solution prepared by dissolving 34.4 g of CH3OH in sufficient water to give exactly

bonufazy [111]

Answer:

4.67M

Explanation:

The concentration of methanol (CH3OH) can be calculated using the following:

Molarity (M) = number of moles(n)/volume(v)

However, mole is not given. It can be obtained by using:

Mole = mass / molar mass

Where; mass = 34.4g

Molar mass (MM) of CH3OH is:

= 12 + 1(3) + 16 + 1

= 12 + 3 + 17

= 32g/mol

mole = 34.4/32

mole = 1.075mol

The volume needs to be converted to L by dividing by 1000

230mL = 230/1000

= 0.230L

Molarity = mol/volume

Molarity = 1.075/0.230

Molarity = 4.6739

Molarity = 4.67M

The concentration of CH3OH in solution is 4.67M

5 0

2 years ago