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2 years ago

Which of the following regions of the periodic table tends to prefer a –1 charge and occupies Group 17?

Chemistry

1 answer:

irina [24]2 years ago

6 0

Answer:

b. halogens

Explanation:

The elements of group 17 are called halogens. These are six elements Fluorine, Chlorine, Bromine, Iodine, Astatine. Halogens are very reactive these elements cannot be found free in nature. Their chemical properties are resemble greatly with each other. As we move down the group in periodic table size of halogens increases that's way fluorine is smaller in size as compared to other halogens elements. Their boiling points also increases down the group which changes their physical states.

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What type of reaction is represented by the following chemical equation? Explain your answer. Thanks!:)

Rashid [163]

DISPLACEMENT REACTION

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3 years ago

The mass and volume of each box is given in the chart below. Calculate the density of each box. Density = Mass/ volume which box

stira [4]

Box C will have the greatest density.

All boxes have the same volume.

Explanation:

We calculate the density using the following formula:

density = mass / volume

density of Box A = 10 g / 20 cm³ = 0.5 g/cm³

density of Box B = 30 g / 20 cm³ = 1.5 g/cm³

density of Box C = 170 g / 20 cm³ = 8.5 g/cm³

Box C will have the greatest density.

All boxes have the same volume.

Learn more about:

density

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2 years ago

12.23 g of ammonia (NH3) are dissolved in enough water to make 560.0 mL of solution. How many moles of NH3 are added to the wate

Karolina [17]

Answer:

0.719 moles of NH₃

Explanation:

Molar mass of ammonia 17 g/mol

Mass of amonia = 12.23 g

Mass / Molar mass = Moles

12.23 g / 17 g/mol = 0.719 moles

4 0

2 years ago

Which is the electron configuration for lithium

creativ13 [48]

Answer:

The electron configuration is [HE] 2s1

7 0

3 years ago

Calculate the standard enthalpy change for the reaction at 25 ∘ C. Standard enthalpy of formation values can be found in this li

meriva

Answer:

The standard enthalpy change for the reaction at $25^{0}\textrm{C}$ is -2043.999kJ

Explanation:

Standard enthalpy change ( $\Delta H_{rxn}^{0}$ ) for the given reaction is expressed as:

$\Delta H_{rxn}^{0}=[3mol\times \Delta H_{f}^{0}(CO_{2})_{g}]+[4mol\times \Delta H_{f}^{0}(H_{2}O)_{g}]-[1mol\times \Delta H_{f}^{0}(C_{3}H_{8})_{g}]-[5mol\times \Delta H_{f}^{0}(O_{2})_{g}]$

Where $\Delta H_{f}^{0}$ refers standard enthalpy of formation

Plug in all the given values from literature in the above equation:

$\Delta H_{rxn}^{0}=[3mol\times (-393.509kJ/mol)]+[4mol\times (-241.818kJ/mol)]-[1mol\times (-103.8kJ/mol)]-[5mol\times (0kJ/mol)]=-2043.999kJ$

4 0

2 years ago