answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
My name is Ann [436]
2 years ago
14

A 14.3-cm3 sample of tin has a mass of 0.104 kg.

Chemistry
1 answer:
VladimirAG [237]2 years ago
7 0
This is a true statement if it is density you are looking for... Density problem.....

Density is the ratio of the mass of an object to its volume.
D = m / V
D = 104g / 14.3 cm³ = 7.27 g/cm³ .............. to three significant digits

The conventions for the units of density is that grams per cubic centimeter (g/cm³) are usually used for solids, but will work for anything. Grams per milliliter (g/mL) are usually used for liquids and grams per liter (g/L) are for gases. Therefore, by convention, the units for tin (a solid) should be in grams per cubic centimeter.

Since 1 mL is equivalent to 1 cm³, then the density could be expressed as 7.27 g/mL.

The accepted value for the density of tin is 7.31 g/cm³
You might be interested in
Calculate the mass in grams of 0.800 mole of H2CO3. g
PilotLPTM [1.2K]
MH₂CO₃: (1g×2) + 12g + (16g×3) = 62 g/mol

1 mol --- 62g
0,8 mol -- X
X = 0,8×62
X = 49,6g
6 0
2 years ago
Read 2 more answers
Rank the following amine derivatives from highest acidity (lowest pKa value) to lowest acidity (highest pKa value).
stira [4]

Answer:

anilinium ion > ammonium ion > amide > aniline > secondary amine

Explanation:

Acidity of amine derivatives can derived from their pKa values.

The rule of thumb for acidity with relation to pKa values is that:

As the pKa decreases the acid strength increases and the conjugate base decreases. Similarly, as the pKa increases, the acid strength decreases and the conjugate base increase.

Hence the stronger the acid , the  lower pKa value  and the weaker the acid , the stronger the pKa value.

So the pKa value for anilinium ion = 4.6

ammonium ion = 9.4

Amide = 15

Similarly, for aniline and secondary amine, in order to determine the derivative with the higher acidity, we will consider the electron withdrawing substituent group.

The more difficult the electron are being withdraw from the electron withdrawing substituent , the more acidic the compound.

In aniline , the stabilized benzene ring attached to NH₂ makes it a less electron withdrawing group compared to the straight chains structure found in secondary amine where electron are easily withdraw by nucleophilic substitution reactions.

Thus, from highest acidity (lowest pKa value) to lowest acidity (highest pKa value).

the amine derivatives ranking is as follows:

anilinium ion > ammonium ion > amide > aniline > secondary amine

8 0
2 years ago
A molecule contains 23.24 g iodine (I),
gavmur [86]
Convert each amount of grams into moles:

I: 23.24g x 1 mol / 126.90g = 0.1831 mol I

C: 2.198 x 1 mol / 12.01g = 0.1830 mol C

N: 2.562 x 1 mol / 14.01g = 0.1829 mol N

Each element has roughly the same amount of moles, which means the whole number ratio between the elements is 1:1:1

Therefore the empirical formula is ICN
7 0
1 year ago
Unit Conversion Help Thank you
AlekseyPX

Answer : 1721.72 g/qt are in 18.2 g/cL

Explanation :

As we are given: 18.2 g/cL

Now we have to convert 18.2 g/cL to g/qt.

Conversions used are:

(1) 1 L = 100 cL

(2) 1 L = 1000 mL

(3) 1 qt = 946 qt

The conversion expression will be:

\frac{18.2g}{1cL}\times \frac{100cL}{1L}\times \frac{1L}{1000mL}\times \frac{946mL}{1qt}

=1721.72\text{ g/qt}

Therefore, 1721.72 g/qt are in 18.2 g/cL

5 0
2 years ago
Use the following half-reactions to design a voltaic cell: Sn4+(aq) + 2 e− → Sn2+(aq) Eo = 0.15 V Ag+(aq) + e−→ Ag(s) Eo = 0.80
AVprozaik [17]

Answer:

E° = 0.65 V

Explanation:

Let's consider the following reductions and their respective standard reduction potentials.

Sn⁴⁺(aq) + 2 e⁻ → Sn²⁺(aq) E°red = 0.15 V

Ag⁺(aq) + e⁻ → Ag(s) E°red = 0.80 V

The reaction with the highest reduction potential will occur as a reduction while the other will occur as an oxidation. The corresponding half-reactions are:

Reduction (cathode): Ag⁺(aq) + e⁻ → Ag(s) E°red = 0.80 V

Oxidation (anode): Sn²⁺(aq) → Sn⁴⁺(aq) + 2 e⁻ E°red = 0.15 V

The overall cell potential (E°) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.

E° = E°red, cat - E°red, an = 0.80 V - 0.15 V = 0.65 V

4 0
2 years ago
Other questions:
  • How can the existence of spectra help to prove that energy levels in atoms exist?
    9·1 answer
  • A gas sample is at 25°C and 1.0 atmosphere. Which changes in temperature and pressure will cause this sample to behave more like
    9·1 answer
  • Sulfurous acid, h2so3, breaks down into water (h2o) and sulfur dioxide (so2). if only one molecule of sulfurous acid was involve
    15·2 answers
  • A student places a small amount of water in a glass dish and then carefully places a small paperclip on the surface of the water
    8·2 answers
  • A 1.000-g sample of lead shot reacted with oxygen to give 1.077 g of product. Calculate the empirical formula of the lead oxide
    6·2 answers
  • Acetonitrile (ch3c≡n) is deprotonated by very strong bases. part a draw resonance forms to show the stabilization of the carbani
    8·1 answer
  • Automobile bodies contain significant amounts of iron. The iron is protected by the addition of zinc. This is called galvanizati
    9·1 answer
  • Phthalic acid is a diprotic acid having the formula HO2CC6H4CO2H that can be converted to a salt by reaction with base. Which of
    8·1 answer
  • HCl can be produced by the exothermic reaction H2(g)+2ICl(g)→2HCl(g)+I2(g) . A proposed mechanism for the reaction has two eleme
    12·1 answer
  • A 3.81-gram sample of NaHCO3 was completely decomposed in an experiment. 2NaHCO3 → Na2CO3 + H2CO3 In this experiment, carbon dio
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!