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4 years ago

What makes alkali metals a family?

Chemistry

1 answer:

navik [9.2K]4 years ago

3 0

Answer:

The metals in this group are lithium, sodium, potassium, rubidium, cesium, and francium. The gas hydrogen is also put in this group because it shares similar reactivity with the alkali metals.

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onsider the following cyclic process carried out in two steps on a gas: Step 1: 42 J of heat is added to the gas, and 14 J of ex

Tema [17]

Answer:

W = -25 J

Explanation:

For a cyclic process, the variation of internal energy (ΔU) must be 0. It means that ΔU from step 1 must be equal to ΔU from step 2. For the first law of the thermodynamics, the energy must be conserved, so:

ΔU = Q - W, from each step. Q is the heat and W the work

When the heat is gained by the system, the process is endothermic, so Q >0, when the system loses heat, the process is exothermic, and Q<0. When the system is expanding W>0, and when it is compressing, W<0.

From step 1:

ΔU = 42 - 14

ΔU = 28 J

From step 2:

28 = -53 - W

W = -53 + 28

W = -25 J

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3 years ago

Choose the CORRECT statement.

devlian [24]

Answer:

Paper has been changed chemically when it is torn into pieces.

Particles of matter move more quickly when their temperature gets cooler.

(this two are correct)

Explanation:

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3 years ago

Hii:) how to do (c)?

Vesna [10]

Answer:

Explanation:

C shows a mixture of CO2 and H2O, 2 compounds

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4 years ago

Read 2 more answers

Draw the Lewis structure of N₂O₄ and then choose the appropriate pair of hybridization states for the two central atoms. Your an

marissa [1.9K]

Answer:

See explanation

Explanation:

In this case, we have to keep in mind the valence electrons for each atom:

N => 5 electrons

O => 6 electrons

If the formula is $N_2O_4$ , we will have in total:

$(5*2)+(6*4)=34~electrons$

Additionally, we have to remember that each atom must have 8 electrons. So, for oxygens 5 and 3 we will have 3 lone pairs and 1 bond (in total 8 electrons. For oxygens, 6 and 4 we will have 2 lone pairs and 2 bonds (in total 8 electrons) and for nitrogens 1 and 2 we will have 4 bonds (in total 8 electrons).

To find the hybridization, we have to count the atoms and the lone pairs around the nitrogen. We have 3 atoms and zero lone pairs. If we take into account the following rules:

$Sp^3~=~4$