<h3>
Answer:</h3>
0.111 J/g°C
<h3>
Explanation:</h3>
We are given;
- Mass of the unknown metal sample as 58.932 g
- Initial temperature of the metal sample as 101°C
- Final temperature of metal is 23.68 °C
- Volume of pure water = 45.2 mL
But, density of pure water = 1 g/mL
- Therefore; mass of pure water is 45.2 g
- Initial temperature of water = 21°C
- Final temperature of water is 23.68 °C
- Specific heat capacity of water = 4.184 J/g°C
We are required to determine the specific heat of the metal;
<h3>Step 1: Calculate the amount of heat gained by pure water</h3>
Q = m × c × ΔT
For water, ΔT = 23.68 °C - 21° C
= 2.68 °C
Thus;
Q = 45.2 g × 4.184 J/g°C × 2.68°C
= 506.833 Joules
<h3>Step 2: Heat released by the unknown metal sample</h3>
We know that, Q = m × c × ΔT
For the unknown metal, ΔT = 101° C - 23.68 °C
= 77.32°C
Assuming the specific heat capacity of the unknown metal is c
Then;
Q = 58.932 g × c × 77.32°C
= 4556.62c Joules
<h3>Step 3: Calculate the specific heat capacity of the unknown metal sample</h3>
- We know that, the heat released by the unknown metal sample is equal to the heat gained by the water.
4556.62c Joules = 506.833 Joules
c = 506.833 ÷4556.62
= 0.111 J/g°C
Thus, the specific heat capacity of the unknown metal is 0.111 J/g°C
Potassium oxide: K₂O.
There's no need for prefixes since K₂O is an ionic compound.
<h3>Explanation</h3>
Find the two elements on a periodic table:
- Potassium- K- on the left end of period four.
- Oxygen- O- near the right end of periodic two.
Elements on the bottom-left corner of the periodic table are metals. Those on the top-right corner are nonmetals.
- Potassium is a metal,
- Oxygen is a nonmetal.
A metal and a nonmetal combine to form an ionic compound. Potassium oxide is likely to be an ionic compound. It contains two types of ions:
- Potassium ions: Potassium is group 1 of the periodic table. It is an alkaline metal. Like other alkaline metals such as sodium Na, potassium K tends to lose one electron and form ions of charge +1 in compounds. The ion would be K⁺.
- Oxide ions from oxygen: Oxygen is the second most electronegative element on the periodic table. It tends to gain two electrons and form the oxide ion
when it combines with metals.
The two types of ions carry opposite charges. They shall pair up at a certain ratio such that they balance the charge on each other. The charge on each
ion is twice that on a
ion. Each
would pair up with two
. Hence the subscript in the formula:
.
There are two classes of compounds:
- Covalent compounds, which need prefixes, and
- Ionic compounds, which need no prefix.
Prefixes are needed only in covalent compounds. For instance in the covalent compound carbon dioxide
, the prefix di- indicates that there are two oxygen atoms in the formula
. However, there's no need for prefix in ionic compounds such as
.
Answer:
Scandium
Titanium
Vanadium
Chromium
Manganese
Iron
Cobalt
Nickel
Copper
Zinc
Yttrium
Zirconium
Niobium
Molybdenum
Technetium
Ruthenium
Rhodium
Palladium
Silver
Cadmium
Lanthanum
Hafnium
Tantalum
Tungsten
Rhenium
Osmium
Iridium
Platinum
Gold
Mercury
Actinium
Rutherfordium
Dubnium
Seaborgium
Bohrium
Hassium
Meitnerium
Darmstadtium
Roentgenium
Copernicium
Explanation:
all of those are transition metals lol
Answer:

<h3>
Explanation:</h3>
In Analytical Chemistry chromatography is widely used for the separation of samples.
- In thin layer chromatography, the mixture of components are separated on the basis of their polarity.
- The solvent solution(mobile phase) that we use are non polar & silica gel( TLC paper made of/stationary phase) are polar.
- Consider the mixture we have taken consist of two samples having large polar difference.
- Due to opposite nature of silica gel(polar) & solvent solution (non polar) the movement become easy & due to capillary action solvent solution rise to the top.
- The mixture of sample we have taken, the sample have less polarity have high peak or they travel more distance than that of more polar sample when they dipped into the solution.
In the given diagram, mixture of 8 samples are separated on the basis of their polarity, the distance travelled by solvent is 35 mm, distance travelled by sample 1 is 11 mm & similarly distance travelled by sample 2,3,4,5,6,7 are 15,31,4,22,25,33 in mm respectively.
Rf Value: Rf value is retention factor which tells about relative absorption of each sample & range of Rf value is 0-1.
Formula to calculate Rf value is

Now, solving for Rf value of sample 1
<em>Given:</em>
Distance moved by sample 1 = 11 mm
Distance movedby solvent = 35 mm
<em>To find:</em>
Rf value of sample 1 = ?
<em>Solution:</em>
Substituting the given data in above formula,


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