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bekas [8.4K]
2 years ago
6

You need to make 10.0 L of 1.2 M KNO3. What molar ( concentration) would the potassium nitrate solution need to be if you were t

o use only 2.5 L of it? *​
Chemistry
1 answer:
solniwko [45]2 years ago
6 0

Answer:

2.5L [NaCl] concentrate needs to be 4.8 Molar solution before dilution to prep 10L of 1.2M KNO₃ solution.

Explanation:

Generally, moles of solute in solution before dilution must equal moles of solute after dilution.

By definition Molarity = moles solute/volume of solution in Liters

=> moles solute = Molarity x Volume (L)

Apply moles before dilution = moles after dilution ...

=> (Molarity X Volume)before dilution = (Molarity X Volume)after dilution

=> (M)(2.5L)before = (1.2M)(10.0L)after

=> Molarity of 2.5L concentrate = (1.2M)(10.0L)/(2.5L) = 4.8 Molar concentrate

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Which macromolecules break apart by hydrolysis?
ankoles [38]

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Dehydration synthesis reactions build molecules up and generally require energy, while hydrolysis reactions break molecules down and generally release energy. Carbohydrates, proteins, and nucleic acids are built up and broken down via these types of reactions, although the monomers involved are different in each case.

Explanation:

4 0
2 years ago
Consider the balanced equation for the following reaction:
Bad White [126]

<u>Answer:</u> The theoretical yield of the lithium chlorate is 1054.67 grams

<u>Explanation:</u>

To calculate the mass for given number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Actual moles of lithium chlorate = 9.45 moles

Molar mass of lithium chlorate = 90.4 g/mol

Putting values in above equation, we get:

9.45mol=\frac{\text{Actual yield of lithium chlorate}}{90.4g/mol}\\\\\text{Actual yield of lithium chlorate}=(9.45mol\times 90.4g/mol)=854.28g

To calculate the theoretical yield of lithium chlorate, we use the equation:

\%\text{ yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100

Actual yield of lithium chlorate = 854.28 g

Percentage yield of lithium chlorate = 81.0 %

Putting values in above equation, we get:

81=\frac{854.28g}{\text{Theoretical yield of lithium chlorate}}\times 100\\\\\text{Theoretical yield of lithium chlorate}=\frac{854.28\times 100}{81}=1054.67g

Hence, the theoretical yield of the lithium chlorate is 1054.67 grams

7 0
2 years ago
How many liters of CO2 gas can be produced at 30.0 °C and 1.50 atm from the reaction of 5.00 mol of C3H8 and an excess of O2 acc
lbvjy [14]

Answer:

249 L

Explanation:

Step 1: Write the balanced equation

C₃H₈(g) + 5 O₂(g) → 3 CO₂(g) + 4 H₂O(g)

Step 2: Calculate the moles of CO₂ produced from 5.00 moles of C₃H₈

The molar ratio of C₃H₈ to CO₂ is 1:3. The moles of CO₂ produced are 3/1 × 5.00 mol = 15.0 mol

Step 3: Convert "30.0°C" to Kelvin

We will use the following expression.

K = °C + 273.15

K = 30.0°C + 273.15 = 303.2 K

Step 4: Calculate the volume of carbon dioxide

We will use the ideal gas equation.

P × V = n × R × T

V = n × R × T/P

V = 15.0 mol × 0.0821 atm.L/mol.K × 303.2 K/1.50 atm

V = 249 L

5 0
3 years ago
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