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4 years ago

How many moles are In 30 grams of H2O

Chemistry

2 answers:

sergiy2304 [10]4 years ago

7 0

One gram of H20 is close to 0.06 moles.

So,
$0.06 \times 30$
is equal to 1.8 moles

EXACT ANSWER: 1.665

DanielleElmas [232]4 years ago

6 0

The answer is 1.665253

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Krypton-79 has a half-life of 35 hours. how many half-lives have passed after 105 hours?

Nookie1986 [14]

The number of half -lives that has passed after 105 hours for krypton-79 that has half-life of 35 hours is calculated as below

if 1 half life = 35 hours

what about 105 hours = ? half-lives

= (1 half life x105 hours) /35 hours = 3 half-lives has passed after 105 hours

6 0

4 years ago

0.350 mol of a solid was dissolved in 260 mL of water at 21.2 oC. After the solid had fully dissolved, the final temperature of

Fittoniya [83]

Answer: Heat of the solution = mass water × specific heat water × change in temperature

mass water = 260ml (1.00g/ml ) = 260g

specific heat of water = c(water) = 4.184J/ g°C

Heat change of water = final temperature - initial temperature

= 26.5 - 21.2

= 5.3 °C

H = 260 g ( 4.184J/g°C ) (5.3°C) = 5765J

Molar heat = $\frac{5765J}{0.350mol}$

= 16473J/mol

Explanation: finding molar heat requires first to look at specific heat of water and the change of water temperature

7 0

3 years ago

No matter where you are on Earth, the center of the Earth is always _______.

FinnZ [79.3K]

Answer:

D) straight down

Hope this helps!

4 0

3 years ago

Read 2 more answers

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

4 years ago

What is the answer please

xeze [42]

Answer:

Selenium

Explanation:

Oxygen is present in the group sixteen.

This group consist of oxygen, sulfur, selenium, tellurium and polonium.

All of these have same number of valance electrons thus they show similar properties.

From the given list of elements only selenium is belong to oxygen family thus it has properties similar to the oxygen.

Electronic configuration of oxygen:

O = [He] 2s² 2p⁴

Electronic configuration of selenium:

Se = [Ar] 3d¹⁰ 4s² 4p⁴

While aluminium, boron and phosphorus are belongs to other groups that's why their properties are not similar to the phosphorus.

phosphorus is present in group 15.

Boron and aluminium are present in group 13.

6 0

4 years ago