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Sonja [21]
3 years ago
12

An experiment was conducted to see which metal pot would be better for cooking food quickly. The table below shows how quickly s

mall pieces of wax melted during the experiment. Type of metal Time for the wax to melt (seconds) Aluminum 27 Bronze 45 Copper 15 Steel 80 The following statement was made after the data were examined: “Copper pots are the best pots for cooking food quickly.” Which term best describes this statement?
Chemistry
1 answer:
LenaWriter [7]3 years ago
4 0

Answer:

The term conclusion best illustrates the given statement.

Explanation:

Conclusion refers to a decision or judgment that can be acquired by reasoning. From the observation mentioned in the given table:  

1. The steel, aluminum, and bronze took 27, 45 and 80 seconds to dissolve small section of wax, this signifies that more time will be needed by them to transfer heat to the food while cooking. Thus, food will take more time to cook in them.  

2. Copper consumed less time, that is, 15 seconds to melt the section of wax than the other metal pots, which signifies that it is a good conductor of heat than the others. Thus, the copper pot would be good for cooking food briskly.  

After evaluating the results and observations in an experiment, the conclusion was made that pots made of copper are best for cooking foods.  

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Which of the following best defines Avogadro's number?
Furkat [3]

Answer:

b

Explanation:

3 0
3 years ago
Read 2 more answers
How many moles are in 128.9 grams of Cr2(SO3)2?
Dominik [7]

Answer: 0.5 moles

Explanation:

Cr2(SO3)2 is the chemical formula for chromium sulphate.

Given that,

Amount of moles of Cr2(SO3)2 (n) = ?

Mass of Cr2(SO3)2 in grams = 128.9g

For molar mass of Cr2(SO3)2, use the atomic masses:

Chromium, Cr = 52g;

Sulphur, S = 32g;

Oxygen, O = 16g

Cr2(SO3)2 =

(52g x 2) + [(32g + 16g x 3) x 2]

= 104g + [(32g + 48g) x 2]

= 104g + [80g x 2]

= 104g + 160g

= 264g/mol

Since, n = mass in grams / molar mass

n = 128.9g / 264g/mol

n = 0.488 mole [Round the value of n to the nearest tenth which is 0.5

Thus, there are 0.5 moles in 128.9 grams of Cr2(SO3)2

7 0
3 years ago
Which is not a characteristic of minerals
dem82 [27]

I think d is wrong because  Everything else is correct for sure. also, minerals vary in the position they hold and have different chemical compounds.

6 0
3 years ago
How many carbon atoms are there in a diamond (pure carbon) with a mass of "48" mg?
blagie [28]

Answer:

              3.47 × 10²³ C atoms

Solution:

Data Given:

                 Mass of Diamond  =  48 mg  =  0.048 g

                 M.Mass of Diamond  =  12.01 g.mol⁻¹ (as it is purely Carbon)

Step 1: Calculate Moles of Diamond as,

                 Moles  =  Mass ÷ M.Mass

Putting values,

                 Moles  =  0.048 g ÷ 12.01 g.mol⁻¹

                 Moles  =  0.576 mol

Step 2: Calculate number of Carbon atoms,

As 1 mole of any substance contains 6.022 × 10²³ particles (Avogadro's Number) then the relation for Moles and Number of Carbon atoms can be written as,

                 Moles  =  Number of C Atoms ÷ 6.022 × 10²³ Atoms.mol⁻¹

Solving for Number of C atoms,

                 Number of C atoms  =  Moles × 6.022 × 10²³ Atoms.mol⁻¹

Putting value of moles,

                 Number of C atoms  =  0.576 mol × 6.022 × 10²³ Atoms.mol⁻¹

                 Number of C atoms =  3.47 × 10²³ C atoms

7 0
3 years ago
Mercury’s atomic emission spectrum is shown below. Estimate the wavelength of the orange line. What is its frequency? What is th
lions [1.4K]

The wavelength of the orange line is 610 nm, the frequency of this emission is 4.92 x 10¹⁴ Hz and the energy of the emitted photon corresponding to this <em>orange line</em> is 3.26 x 10⁻¹⁹ J.

<em>"Your question is not complete, it seems to be missing the diagram of the emission spectrum"</em>

the diagram of the emission spectrum has been added.

<em>From the given</em><em> chart;</em>

The wavelength of the atomic emission corresponding to the orange line is 610 nm = 610 x 10⁻⁹ m

The frequency of this emission is calculated as follows;

c = fλ

where;

  • <em>c is the speed of light = 3 x 10⁸ m/s</em>
  • <em>f is the frequency of the wave</em>
  • <em>λ is the wavelength</em>

f = \frac{c}{\lambda } \\\\f = \frac{3\times 10^8}{610 \times 10^{-9}} \\\\f = 4.92 \times 10^{14} \ Hz

The energy of the emitted photon corresponding to the orange line is calculated as follows;

E = hf

where;

  • <em>h is Planck's constant = 6.626 x 10⁻³⁴ Js</em>

<em />

E = (6.626 x 10⁻³⁴) x (4.92 x 10¹⁴)

E = 3.26 x 10⁻¹⁹ J.

Thus, the wavelength of the orange line is 610 nm, the frequency of this emission is 4.92 x 10¹⁴ Hz and the energy of the emitted photon corresponding to this <em>orange line</em> is 3.26 x 10⁻¹⁹ J.

Learn more here:brainly.com/question/15962928

6 0
2 years ago
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