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umka21 [38]
4 years ago
7

Can someone help me understand how to solve mole problems in chemistry?

Chemistry
2 answers:
AURORKA [14]4 years ago
5 0
I can if you give man example I'll show you how to do it.
stepladder [879]4 years ago
5 0
<span>I start the reaction below with 1.5g of Al, how many moles of Cl can I make" (Al + Cl2 ➡ NaCl)

First thing to do is find out what the amu of Al is.
You can find that on a periodic table. 26.98grams per mole
1.5g Al x </span>\frac{1 mole Al}{26.98g Al} = <span>0.05559673832 = 0.056 moles of Al
Now ratio the moles.
0.056 moles Al x </span>\frac{1 mole Cl}{1 mole Al} = 0.056 moles of Cl
The answer is 0.056 moles of Cl
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A 10.50 gram sample of a compound is decomposed to yield 3.40 g Na, 2.37 g S, and 4.73 g O. What is the mass percentage of each
9966 [12]

Answer:

Na = 32.4% , % S = 22.6% and %O = 45.0%

Explanation:

% Na = 3.4/10.5. × 100%

= 32.4%

%S = 2.37/10.5 × 100%

= 22.6%

% O= 4.73/10.5 × 100%

= 45.0%xplanation:

6 0
3 years ago
At a given temperature, 300m of carbon dioxide has a mass of 5.94kg. What is the density of carbon dioxide at this temperature
erica [24]
I think the would have to be 0.198kg/g
4 0
3 years ago
What does condensation mean in the water cycle?
tester [92]
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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
Suppose you carry out a titration involving 2.50 molar NaOH and an unknown concentration of HBr. To bring the reaction to its en
givi [52]

Answer:

B

Explanation:

Brainly Patrol ima stop you here

7 0
2 years ago
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