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2 years ago

Which is an example of making a quantitative observation?

Chemistry

1 answer:

Elanso [62]2 years ago

4 0

Answer:

A) measuring mass of metal used in a reaction

Explanation:

Quantitive observations is data involving statistics and numerical values.

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You measure the absorbance of your extracted curcumin solution and realize that the solution is too concentrated. What made you

zepelin [54]

If the absorbance of a solution of curcumin which is too concentrated is measured, the absorbance will be unusually high.

Spectrometry measures the interaction of light with molecules. The absorbance refers to how much light that interacts with molecules of the substance. The more the concentration of the substance the higher the absorbance of the solution.

Hence, if the absorbance of a solution of curcumin which is too concentrated is measured, the absorbance will be unusually high. An unusually high absorbance tells us that the solution is too concentrated.

Learn more: brainly.com/question/13440572

8 0

2 years ago

What is a caliper used for

photoshop1234 [79]

To measure the distance between two opposite sides of an object

8 0

3 years ago

Identify the balanced equation for the following reaction: SO2(g) + O2(g) → SO3(g)

sergiy2304 [10]

Answer: The balanced equation for the given reaction is

$2SO_{2}(g) + O_{2}(g) \rightarrow 2SO_{3}(g)$ .

Explanation:

A chemical equation which contains same number of atoms on both reactant and product side.

For example, $SO_{2}(g) + O_{2}(g) \rightarrow SO_{3}(g)$

Here, number of atoms on reactant side are as follows.

S = 1

O = 4

Number of atoms on product side are as follows.

S = 1

O = 3

To balance this equation, multiply $SO_{2}$ by 2 on reactant side and multiply $SO_{3}$ by 2. Hence, the equation will be re-written as follows.

$2SO_{2}(g) + O_{2}(g) \rightarrow 2SO_{3}(g)$

Here, number of atoms on reactant side are as follows.

S = 2

O = 6

Number of atoms on product side are as follows.

S = 2

O = 6

Now, there are same number of atoms on both reactant and product side. So, this equation is balanced.

Thus, we can conclude that the balanced equation for the given reaction is $2SO_{2}(g) + O_{2}(g) \rightarrow 2SO_{3}(g)$ .

3 0

2 years ago

Type the correct answer in the box. Express your answer to three significant figures.

VladimirAG [237]

Given:

Mass of calcium nitrate (Ca(NO3)2) = 96.1 g

To determine:

Theoretical yield of calcium phosphate, Ca3(PO4)2

Explanation:

Balanced Chemical reaction-

3Ca(NO3)2 + 2Na3PO4 → 6NaNO3 + Ca3(PO4)2

Based on the reaction stoichiometry:

3 moles of Ca(NO3)2 produces 1 mole of Ca3(PO4)2

Now,

Given mass of Ca(NO3)2 = 96.1 g

Molar mass of Ca(NO3)2 = 164 g/mol

# moles of ca(NO3)2 = 96.1/164 = 0.5859 moles

Therefore, # moles of Ca3(PO4)2 produced = 0.0589 * 1/3 = 0.0196 moles

Molar mass of Ca3(PO4)2 = 310 g/mol

Mass of Ca3(PO4)2 produced = 0.0196 * 310 = 6.076 g

Ans: Theoretical yield of Ca3(PO4)2 = 6.08 g

7 0

2 years ago

Read 2 more answers

What is the freezing point of a solution of 465 g of sucrose c12h22o11 dissolved in 575 ml of water?

Amanda [17]

Answer:

The freezing point of the solution is - 4.39 °C.

Explanation:

We can solve this problem using the relation:

ΔTf = (Kf)(m),

where, ΔTf is the depression in the freezing point.

Kf is the molal freezing point depression constant of water = -1.86 °C/m,

density of water = 1 g/mL.

So, the mass of 575 mL is 575 g = 0.575 kg.

m is the molality of the solution (m = moles of solute / kg of solvent = (465 g / 342.3 g/mol)/(0.575 kg) = 2.36 m.

∴ ΔTf = (Kf)(m) = (-1.86 °C/m)(2.36 m) = - 4.39 °C.

∵ The freezing point if water is 0.0 °C and it is depressed by - 4.39 °C.

∴ The freezing point of the solution is - 4.39 °C.

6 0

2 years ago