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DerKrebs [107]
3 years ago
15

Calculate the quantitative concentration of HCl in the solution if in the reaction of 500 cm3 the solution with AgNO3 produces 0

.7 g of AgCl AgNO3 + HCl -------> AgCl + HNO3
Chemistry
1 answer:
ss7ja [257]3 years ago
8 0

Answer:

9.77x10⁻³M HCl

Explanation:

<em>Assuming the solution of AgNO₃ is in excess:</em>

<em />

Based on the chemical equation:

AgNO₃ + HCl → AgCl + HNO₃

We must find the moles of AgCl that will be produced. With the moles and knowing 1 mole of AgCl is produced from 1 mole of HCl we can find the moles of HCl and its concentration as follows:

<em>Moles AgCl -Molar mass: 143.32g/mol-:</em>

0.7g AgCl * (1mol / 143.32g) = 4.88x10⁻³moles AgCl = Moles HCl

As the volume of the solution is 500cm³ = 0.500L, the concentration of HCl is:

4.88x10⁻³moles HCl / 0.500L =

<h3>9.77x10⁻³M HCl</h3>
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Answer:

The answer to your question is below

Explanation:

Propane is a hydrocarbon that has only single bonds in its structure so it is an alkane.

Propane formula = C₃H₈

Oxygen formula = O₂

Carbon dioxide formula = CO₂

Water formula = H₂O

<u>Reaction:</u>

                         C₃H₈  +   5O₂   ⇒    3CO₂   +   4H₂O

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                          3                          C                    3

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6 0
3 years ago
What mass of Al(OH)3 is needed to completely react with 17.9g of HCl?
musickatia [10]

Answer:

12.8 g

Explanation:

Step 1: Write the balanced neutralization reaction

Al(OH)₃ + 3 HCl ⇒ AlCl₃ + 3 H₂O

Step 2: Calculate the moles corresponding to 17.9 g of HCl

The molar mass of HCl is 36.46 g/mol.

17.9 g × 1 mol/36.46 g = 0.491 mol

Step 3: Calculate the moles of Al(OH)₃ that completely react with 0.491 moles of HCl

The molar ratio of Al(OH)₃ to HCl is 1:3. The reacting moles of Al(OH)₃ are 1/3 × 0.491 mol = 0.164 mol

Step 4: Calculate the mass corresponding to 0.164 moles of Al(OH)₃

The molar mass of Al(OH)₃ is 78.00 g/mol.

0.164 mol × 78.00 g/mol = 12.8 g

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3 years ago
How much cesium (half-life = 2 years) would remain from a 10 g sample after 4 years?
Ksenya-84 [330]
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7 0
2 years ago
Omg pls help i dunno what the frick frack this is
snow_lady [41]

Answer:

1. Mass of KCl produced = 774.8 g of KCl

2. Mass of KNO₃ produced = 13.837g

3. Moles of NaOH made = 0.846 moles

4. Moles of LiCl produced = 0.846 moles

5. Moles of CO₂ produced = 207.6 moles

Explanation:

1. From the equation of reaction, 1 mole of ZnCl₂ produces, 2 moles of KCl.

5.02 moles of ZnCl₂ will produce, 2 × 5.02 moles of KCl = 10.4 moles of KCl

Molar mass of KCl = (39 + 35.5) g/mol = 74.5 g/mol

10.4 moles of KCl = 10.4 × 74.5 g

Mass of KCl produced = 774.8 g of KCl

2. Mole ratio of KNO₃ and KOH = 1:1

O.137 moles of KOH will produce 0.137 moles of KNO₃

Molar mass of KNO₃ = 101 g/mol

Mass of KNO₃ produced = 0.137 × 101 g = 13.837g

3. Molar mas of Ca(OH)₂ = 74.0 g

Moles of Ca(OH)₂ in 31.3 g = 31.3/74.0 = 0.423 moles of Ca(OH)₂

Mole ratio of NaOH and Ca(OH)₂ in the reaction = 2 : 1

Moles of NaOH made = 2 × 0.423 = 0.846 moles

4. Molar mass of MgCl₂ = 95.0 g

Moles of MgCl₂ in 40.2 g = 40.2/95.0 = 0.423 moles

From the reaction equation, mole ratio of MgCl₂ and LiCl = 1:2

Moles of LiCl produced = 2 × 0.423 = 0.846 moles

5. From the equation of reaction, 1 mole of C₆H₁₀O₅ produces 6 moles of cO₂

34.6 moles of C₆H₁₀O₅ will produce 34.6 × 6 moles of CO₂

Moles of CO₂ produced = 207.6 moles

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3 years ago
Which gas laws apply to each of the steps of the 4 stroke engine?
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Answer:

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Explanation:

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3 years ago
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