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Mazyrski [523]
2 years ago
11

A student measures the absorbance of a solution containing FeSCN2 ion using a spectrophotometer. The cuvette used by the student

has two frosted walls and two transparent walls. The student properly orients the cuvette so that the path of the light goes through the transparent sides of the cuvette when calibrating the spectrophotometer. How will the measured absorbance of the FeSCN2 be affected if the student incorrectly orients the cuvette so that the path of the light is through the frosted sides of the cuvette
Chemistry
1 answer:
suter [353]2 years ago
6 0

Answer:

The measured absorbance will be too large.  

Explanation:

Fe³⁺(aq) + SCN⁻(aq) ⟶Fe(SCN)²⁺(aq)

A = log₁₀(I₀/I)

If the student orients the cuvette so that the path of the light is through the frosted sides of the cuvette, little light will be able to reach the detector.

The measured intensity (I) will be quite small, so the absorbance (A) will be unusually large.

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A student uses a dye to change the colour of his tee-shirt. he fills a metal bucket with water. When the bucket is heated, what
Tanya [424]

Answer:

The particles begin to vibrate faster and more.

Explanation:

Adding heat to matter increases the energy, thus creating more movement. Eventually, the bucket will melt, turning to a liquid. While it is a sold, it still has particle movement, just not enough to break volume or shape.

8 0
1 year ago
How many moles of ions are in 285 ml of 0.0150 m mgcl2?
Solnce55 [7]
MgCl₂)= Mg²⁺ + 2Cl⁻
V(MgCl₂)=285cm³=0,285dm³
c(MgCl₂)=0,015 mol/dm³
n(MgCl₂)=c·V= 0,015 mol/dm³ · 0,285dm³ = 0,0042 mol
n(Mg²⁺)=n(MgCl₂)=0,0042 mol
n(Cl⁻)=2n(MgCl₂)=0,0084 mol
7 0
2 years ago
Read 2 more answers
a drop of water weighing 0.48 g condenses on the surface of a 55-g block of aluminum that is initially at 25C. if the heat durin
Gwar [14]

Mass of water vapor is 0.48 gms.

Weight of the aluminium block is 55 gms.

Heat of vaporization of water at 25 degree Celsius is 44.0 kJ/mol.

The amount of heat given by the condensation is:

q = Heat of vaporization × Mass of vapor / Molar mass of steam

= 44.0 kJ/mol ×0.48 g / 18 g/mol

= 1.173 kJ

Now the final temperature of the metal block is calculated by the formula:

q = m × c × ΔT

q = m × c × (T₂ - T₁)

Here, q is the amount of heat, m is the mass of the metal, c is the specif heat of the metal, T₁ is initial temperature, and T₂ is the final temperature.

Now, substituting the values we get,

1.173 kJ = 55 g × 0.903 J/g. ° C * (T₂ - 25°C)

1.173 × 10³ J = 55 g × 0.903 J/g. degree C × (T₂ - 25°C)

1.173 × 10³ = 49.665 ° C * (T₂ - 25° C)

T₂ = 49° C.

Thus, the final temperature of the metal block is 49° C.

8 0
1 year ago
How many moles of al2o3 can be produced from the reaction of 10.0 g of al and 19.0 of o2?
Advocard [28]

Answer:

0.185moles of Al₂O₃

Explanation:

Mass of Al = 10g

Mass of O₂ = 19g

Equation of the reaction: 4Al + 3O₂ → 2Al₂O₃

This is the balanced reaction equation.

Solution

From the given parameters, the reactant that would determine the extent of the reaction is Aluminium. It is called the limiting reagent. Oxygen is in excess and it is in an unlimited supply.

Working from the known mass to the unknown, we simply solve for the number of moles of Al using the mass given.

Then from the equation, we can relate the number of moles of Al to that of Al₂O₃ produced:

 Number of moles of Al = \frac{mass}{molar mass}

                                        =   \frac{10}{27}

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From the equation:

         4 moles of Al produced 2 moles of Al₂O₃

    0.37 mole will yield:  \frac{2 x 0.37}{4} = 0.185moles of Al₂O₃

8 0
1 year ago
A 0.89% (w/v) sodium chloride solution is referred to as physiological saline solution because it has the same concentration of
maks197457 [2]
1) 0.89% m/v = 0.89 grams of NaCl / 100 ml of solution

=> 8.9 grams of NaCl in 1000 ml of solution = 8.9 grams of NaCl in 1 liter of solution

2) Molarity = M = number of moles of solute / liters of solution

=> calculate the number of moles of 8.9 grams of NaCl

3) molar mass of NaCl = 23.0 g /mol + 35.5 g/mol = 58.5 g / mol

4) number of moles of NaCl = mass / molar mass = 8.9 g / 58.5 g / mol = 0.152 mol

5) M = 0.152 mol NaCl / 1 liter solution = 0.152 M

Answer: 0.152 M
4 0
1 year ago
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