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Mama L [17]
3 years ago
11

which laboratory process is used to determine the concentration of one solution by using a volume of another solution of known c

oncentration?
Chemistry
1 answer:
astra-53 [7]3 years ago
7 0

Answer: Titration : It is a laboratory process where we have to use the volume of a solution of known concentration to determine the concentration of unknown solution.

Explanation:

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2c4H10 +13O2 —>8CO2+10H2O to find how many moles of oxygen would react with 4.3 mol C4H10
Elena-2011 [213]

Answer:

\large\boxed{\text{28 mol of O$_{2}$}}

Explanation:

             2C₄H₁₀ + 13O₂ ⟶ 8CO₂ + 10H₂O

n/mol:      4.3  

13 mol of O₂ react with 2 mol of 2C₄H₁₀

\text{Moles of O}_{2} = \text{4.3 mol C$_{4}$H$_{10}$} \times \dfrac{\text{13 mol O}_{2}}{\text{2 mol C$_{4}$H$_{10}$}} = \textbf{28 mol O}_{2}\\\\\text{The reaction requires $\large\boxed{\textbf{28 mol of O$_{2}$}}$}

5 0
3 years ago
Explain what a scientific theory is, and what is necessary to make a theory strong and well supported?
Anna35 [415]

in order for a scientific theory to become a scientific law it needs to be tested with generations of data to confirm that it is really true.

7 0
3 years ago
According to the rate law (rate = k[a]m[b]n), what does the rate of a reaction depend on?
Andreyy89
Concentration of the reactant,pressure,surface
 area of the reactant  and temperatur
8 0
3 years ago
How many moles are in 1.5L of 0.40M Na2SO4?​
anzhelika [568]

Answer:

0.60 moles of Na₂SO₄

Explanation:

Molarity is an unit of concentration defined as the ratio between moles of solute and liters of solution.

A solution of Na₂SO₄ 0.40M contains 0.40 moles of solute (Na₂SO₄) per liter of solution.

As you have 1.5L of solution, moles of Na₂SO₄:

1.5L × (0.40mol / L) = <em>0.60 moles of Na₂SO₄</em>

7 0
3 years ago
1826.5g of methanol (CH3OH), molar mass = 32.0 g/mol is added to 735 g of water, what is the molality of the methane 0.0348 m 1.
Snowcat [4.5K]

Answer:

Molality = 1.13 m

Explanation:

Molality is defined as the moles of the solute present in 1 kilogram of the solvent.

Given that:

Mass of CH_3OH = 26.5 g

Molar mass of CH_3OH = 32.04 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{26.5\ g}{32.04\ g/mol}

Moles\ of\ CH_3OH= 0.8271\ moles

Mass of water = 735 g = 0.735 kg ( 1 g = 0.001 kg )

So, molality is:

m=\frac {0.8271\ moles}{0.735\ kg}

<u>Molality = 1.13 m</u>

4 0
3 years ago
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