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3 years ago

How many mL of 1.76 M sodium hydroxide solution (strong base) must be added to 0.75

1 answer:

5 0

Answer: Special care is required to prepare a solution of sodium hydroxide or NaOH in water because considerable heat is liberated by the exothermic reaction. The solution may splatter or boil. Here is how to make a sodium hydroxide solution safely, along with recipes for several common concentrations of NaOH solution.

Explanation:

Hope this helps!

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Un tecnico di laboratorio deve preparare una soluzione di carbonato di sodio decaidrato, Na2CO3⋅ 10 H2O per eseguire alcune anal

fiasKO [112]

Answer:

1. 3.70 g Na₂CO₃·10H₂O

2. 50.0 mL of the first solution

Explanation:

1. Prepare the solution

(a) Calculate the molar mass of Na₂CO₃·10H₂O

$\begin{array}{rrr}\textbf{Atoms} &\textbf{M}_{\textbf{r}} & \textbf{Mass/u}\\\text{2Na} & 2\times22.99 & 45.98\\\text{1C} & 1\times 12.01 & 12.01\\\text{13O}&13 \times16.00 & 208.00\\\text{20H}&20 \times 1.008 & 20.16\\&\text{TOTAL =} & \mathbf{286.15}\\\end{array}$

The molar mass of Na₂CO₃·10H₂O is 286.15 g/mol.

(b) Calculate the moles of Na₂CO₃·10H₂O

$\text{Moles of Na$_{2}$CO$_{3}\cdot$10H$_{2}$O}\\= \text{0.250 L solution} \times \dfrac{\text{0.0500 mol Na$_{2}$CO$_{3}\cdot$10H$_{2}$O}}{\text{1 L solution}}\\\\= \text{0.0125 mol Na$_{2}$CO$_{3}\cdot$10H$_{2}$O}$

(c) Calculate the mass of Na₂CO₃·10H₂O

$\text{Mass of Na$_{2}$CO$_{3}\cdot$10H$_{2}$O }\\= \text{0.012 50 mol Na$_{2}$CO$_{3}\cdot$10H$_{2}$O } \times \dfrac{\text{296.15 g Na$_{2}$CO$_{3}\cdot$10H$_{2}$O}}{\text{1 mol Na$_{2}$CO$_{3}\cdot$10H$_{2}$O}}\\\\= \text{3.70 g Na$_{2}$CO$_{3}\cdot$10H$_{2}$O}\\\text{You need $\large \boxed{\textbf{3.70 g}}$ of Na$_{2}$CO$_{3}\cdot$10H$_{2}$O}$

2. Dilute the solution

We can use the dilution formula to calculate the volume needed.

V₁c₁ = V₂c₂

Data:

V₁ = ?; c₁ = 0.0500 mol·L⁻¹

V₂ = 100 mL; c₂ = 0.0250 mol·L⁻¹

Calculation:

$\begin{array}{rcl}V_{1}c_{1} & = & V_{2}c_{2}\\V_{1}\times \text{0.0500 mol/L} & = & \text{100 mL} \times\text{0.0250 mol/L}\\0.0500V_{1}& = & \text{2.500 mL}\\V_{1}&=& \dfrac{\text{2.500 mL}}{0.0500}\\\\& = & \text{50.0 mL}\\\end{array}\\\text{You need $\large \boxed{\textbf{50.0 mL}}$ of the concentrated solution.}$

7 0

3 years ago

An unknown element, X, has an atomic mass of 107.868 amu. The X-109 isotope (108.905 amu) is 48.16%. What is the amu of the othe

juin [17]

Answer:

106.905 amu is the mass of the other isotope

Explanation:

The atomic mass of an element is the sum of the masses of the isotopes multiplied by its abundance. The atomic mass of an element X with 2 isotopes is:

X = X-109*i + X-107*i

Where X is the atomic mass = 107.868 amu

X-109 = 108.905amu, i = 48.16% = 0.4816

X-107 = ?, i = 1-0.4816 = 0.5184

Replacing:

107.868amu = 108.905amu*0.4816 + X-107*0.5184

55.4194 = X-107*0.5184

106.905 = X-107

<h3>106.905 amu is the mass of the other isotope</h3>

4 0

3 years ago

You are provided with a compound fertilizer, 40-15-10.Calculate the quantity of fertilizer to add to a one hectare field supply

victus00 [196]

Answer:

a. 300 kg of Fertilizer

b. 225 kg of fertilizer

c.400 Kg of fertilizer

d.600 Kg of fertilizer

Explanation:

The percentage composition ratio of Nitrogen, Phosphorus and Potassium bag of the given fertilizer is 40:15:10.

The percentages can be expressed as fractions as follows:

For nitrogen; 40/100 = 0.4

For phosphorus; 15/100 = 0.15

For potassium; 10/100 = 0.1

To find the quantity of fertilizer required to add to a hectare to supply the given amount of nutrients, the amount to be provided is divided by the percentage or fractional compostion of each nutrient.

Quantity of fertilizer required to add to a hectare to supply;

a. Nitrogen at 120 kg/ha = 120/0.4 = 300 Kg of fertilizer

b.. Nitrogen at 90 Kg/ha = 90/0.4 = 225 Kg of fertilizer

c. Phosphorus at 60 kg/ha = 60/0.15 = 400 Kg of fertilizer

d. Potassium at 60 kg/ha = 60/0.1 = 600 Kg of fertilizer

5 0

3 years ago

In the experiment "Beer-Lambert’s Law and Spectrophotometry", you prepared a calibration plot similar to the one pictured below.

dexar [7]

Answer:

0.025M

Explanation:

As you must see in your graph, each concentration of the experiment has an absorbance. Following the Beer-Lambert's law that states "The absorbance of a solution is directely proportional to its concentration".

At 0.35 of absorbance, the plot has a concentration of:

8 0

3 years ago

The use of crucible

Norma-Jean [14]

a ceramic or metal container in which metals or other substances may be melted or subjected to very high temperatures.

"the crucible tipped and the mold filled with liquid metal"

a situation of severe trial, or in which different elements interact, leading to the creation of something new.

"their relationship was forged in the crucible of war"

3 0

3 years ago