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3 years ago

What is the valence e- (outermost e-) for the following elements:

Chemistry

1 answer:

shutvik [7]3 years ago

6 0

Answer:

A options hydrogen

Explanation:

valence outermost element is hydrogen.

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Calculate the values of ΔU, ΔH, and ΔS for the following process:

ladessa [460]

Answer:

ΔU = 45.814 KJ

ΔH = 46.4375 KJ

ΔS = 18.76 J/K

Explanation:

H2O(l) → H2O(l) → H2O(steam)

298.15K, 1atm ΔHp 373.15K,1atm ΔHv 373.15K,1 atm

∴ ΔHp = Qp = nCpΔT

∴ n H2O = 1 mol

∴ Cp,n = 75.3 J/mol.K

∴ ΔT = 373.25 - 298.15 = 75 K

⇒ Qp = (1 mol)*(75.3 J/mol.K)*(75K) = 5647.5 J

⇒ ΔHp = 5647.5 J = 5.6475 KJ

⇒ ΔH = ΔHp + ΔHv

∴ ΔHv = 40.79 KJ/mol * 1 mol = 40.79 KJ

⇒ ΔH = 5.6475 KJ + 40.79 KJ = 46.4375 KJ

ideal gas:

∴ ΔH = ΔU + PΔV

∴ V1 = nRT1/P1 = ((1)*(0.082)*(298.15))/1 = 24.45 L

∴ V2 = nRT2/P2 = ((1)*(0.082)*(373.15))/ 1 = 30.59 L

⇒ ΔV = V2 - V1 = 6.15 L * (m³/1000L) = 6.15 E-3 m³

∴ P = 1 atm * (Pa/ 9.86923 E-6 atm) = 101325.027 Pa

⇒ ΔU = ΔH - PΔV = 46.4375 KJ - ((101325.027 Pa*6.15 E-3m³)*(KJ/1000J))

⇒ ΔU = 46.4375 KJ - 0.623 KJ

⇒ ΔU = 45.814 KJ

∴ ΔS = Cv,n Ln (T2/T1) + nR Ln (V2/V1)

⇒ ΔS = (75.3) Ln(373.15/298.15) + (1)*(8.314) Ln (30.59/24.45)

⇒ ΔS = 16.896 J/K + 1.863 J/K

⇒ ΔS = 18.76 J/K

3 0

3 years ago

At a birthday pool party, the temperature is 28.50°C and the atmospheric pressure is 755.4 mmHg. One of the decoration helium ba

Iteru [2.4K]

Answer:

The volume of the balloon will be 5.11L

Explanation:

An excersise to solve with the Ideal Gases Law

First of all, let's convert the pressure in mmHg to atm

1 atm = 760 mmHg

760 mmHg ___ 1 atm

755.4 mmHg ____ (755.4 / 760) = 0.993 atm

922.3 mmHg ____ ( 922.3 / 760) = 1.214 atm

T° in K = 273 + °C

28.5 °C +273 = 301.5K

26.35°C + 273= 299.35K

P . V = n . R .T

First situation: 0.993atm . 6.25L = n . 0.082 . 301.5K

(0.993atm . 6.25L) / 0.082 . 301.5 = n

0.251 moles = n

Second situation:

1.214 atm . V = 0.251 moles . 0.082 . 301.5K

V = (0.251 moles . 0.082 . 301.5K) / 1.214 atm

V = 5.11L

7 0

3 years ago

Read 2 more answers

A gas at 42. 0°c occupies a volume of 1. 32 l. if the volume increases to 2. 24 l, what is the new temperature in kelvin?

Nina [5.8K]

the answer is 534.54 kelvin

8 0

2 years ago

At what step of glycolysis one NADH+ H+ is formed?

ElenaW [278]

Answer:

During Glycolysis, there is one step where NADH + H+ is formed from NAD+ O F-6-P - F- 1,6-BP O PGAL – 1,3-BPGA O 2-PGA - PEP O 3-PGA → 2-PGA.

3 0

2 years ago

A certain radioactive isotope has a half-life of 5 days. If one is to make a table showing the half-life decay of a sample of th

Pepsi [2]

It will take 25 days for this substance to become 1gram from 32 grams.

<h3><u>Explanation</u>:</h3>

Radioactivity or radioactive decay is the process of formation of different element from one Radioactive element due to decay of the radioactive nucleus. The radioactive decay is a first order kinetic reaction. This reaction has a straight log C vs T graph.

Here in this particular reaction, the half life of the radioactive substance is 5 days. So within 5 days, the concentration of the substance becomes halved.

Now, in the sample the amount of substance present = 32 grams

The final amount of this sample = 1 grams.

So number of half lives needed for this decay = 5 half lives.

So number of days required for this decay = $5\times5$ =25.

So within 25 days, 32 grams of the substance becomes 1grams.

6 0

3 years ago