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Masteriza [31]
1 year ago
13

If two unidentified solids of the same texture and color have different solubilities in 100 grams of water at 20°C, you could co

nclude that
a.
they are the same substance.
b.
they are different substances.
c.
they have different melting points.
d.
their solubilities will be the same if the water temperature is increased.
Chemistry
2 answers:
AlekseyPX1 year ago
3 0
I'm certain it's "D"

...because it can't be "A" or "B" because solubility IS a property but to actually determine whether these two substances are the same or different we would need at least two-three properties (like boiling point or specific heat).

and it can't be "C" because the melting point is just simply irrelevant when comparing the solubility of two substances.
Licemer1 [7]1 year ago
3 0

they have different melting points.

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As he began to teach inorganic chemistry, Mendeleev could not find a textbook that met his needs. Since he had already published a textbook on organic chemistry in 1861 that had been awarded the prestigious Demidov Prize, he set out to write another one. The result was Osnovy khimii (1868–71; The Principles of Chemistry), which became a classic, running through many editions and many translations. When Mendeleev began to compose the chapter on the halogen elements (chlorine and its analogs) at the end of the first volume, he compared the properties of this group of elements to those of the group of alkali metals such as sodium. Within these two groups of dissimilar elements, he discovered similarities in the progression of atomic weights, and he wondered if other groups of elements exhibited similar properties. After studying the alkaline earths, Mendeleev established that the order of atomic weights could be used not only to arrange the elements within each group but also to arrange the groups themselves. Thus, in his effort to make sense of the extensive knowledge that already existed of the chemical and physical properties of the chemical elements and their compounds, Mendeleev discovered the periodic law.

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2 years ago
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Give the number of significant figures in this number: 40.00
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If you have a case with zeros in front, you take to the first non zero digits.

For example, 0.071004 you wold express as 0.071 and those 7, and 1 are the significant ones.

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3 0
1 year ago
Acetone has a boiling point of 56.5 celcius. How many grams of the acetone vapor would occupy the 250 mL Erlenmeyer flask at 57
vodomira [7]

Answer:

0.515 g

Explanation:

<em>Acetone (C₃H₆O) has a boiling point of 56.5 °C. How many grams of the acetone vapor would occupy the 250 mL Erlenmeyer flask at 57 °C and 730 mmHg?</em>

<em />

Step 1: Given data

Temperature (T): 57°C

Pressure (P): 730 mmHg

Volume (V): 250 mL

Step 2: Convert "T" to Kelvin

We will use the following expression.

K = °C + 273.15 = 57°C + 273.15 = 330 K

Step 3: Convert "P" to atm

We will use the conversion factor 1 atm = 760 mmHg.

730 mmHg × (1 atm/760 mmHg) = 0.961 atm

Step 4: Convert "V" to L

We will use the conversion factor 1 L = 1,000 mL.

250 mL × (1 L/1,000 mL) = 0.250 L

Step 5: Calculate the moles (n) of acetone

We will use the ideal gas equation.

P × V = n × R × T

n = P × V/R × T

n = 0.961 atm × 0.250 L/(0.0821 atm.L/mol.K) × 330 K

n = 8.87 × 10⁻³ mol

Step 6: Calculate the mass corresponding to 8.87 × 10⁻³ moles of acetone

The molar mass of acetone is 58.08 g/mol.

8.87 × 10⁻³ mol × 58.08 g/mol = 0.515 g

8 0
1 year ago
Draw all of the constitutional isomers of the molecule with formula C3H5Br. Ignore geometric and stereoisomers.
Kay [80]

Answer:

-) 3-bromoprop-1-ene

-) 2-bromoprop-1-ene

-) 1-bromoprop-1-ene

-) bromocyclopropane

Explanation:

In this question, we can start with the <u>I.D.H</u> (<em>hydrogen deficiency index</em>):

I.D.H~=~\frac{(2C)+2+(N)-(H)-(X)}{2}

In the formula we have 3 carbons, 5 hydrogens, and 1 Br, so:

I.D.H~=~\frac{(2*3)+2+(0)-(5)-(1)}{2}~=~1

We have an I.D.H value of one. This indicates that we can have a cyclic structure or a double bond.

We can start with a linear structure with 3 carbon with a double bond in the first carbon and the Br atom also in the first carbon (<u>1-bromoprop-1-ene</u>). In the second structure, we can move the Br atom to the second carbon (<u>2-bromoprop-1-ene</u>), in the third structure we can move the Br to carbon 3 (<u>3-bromoprop-1-ene</u>). Finally, we can have a cyclic structure with a Br atom (<u>bromocyclopropane</u>).

See figure 1

I hope it helps!

7 0
1 year ago
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