Q: find Molar mass - K 2 SO 4 ⋅Al 2 (SO 4 ) 3 ⋅24H 2 O

The correct answer is: $948.76\text{ g/mol}$ Explanation We compute the molar mass by counting each element in the formula K2SO4·Al2(SO4)3·24H2O, multiplying by atomic masses, and summing. Steps: Count atoms in the full formula: $$\text{counts: }K:2,\ Al:2,\ S:4,\ O:40,\ H:48$$ (Explanation: K2SO4 contributes K2,S1,O4; Al2(SO4)3 contributes Al2,S3,O12; 24H2O contributes H48,O24. Summing S:1+3=4 and O:4+12+24=40.) Write the molar-mass sum using standard atomic masses used here: $M_K=39.0983$, $M_{Al}=26.9815$, $M_S=32.065$, $M_O=15.999$, $M_H=1.00794$ (units g/mol): $$M = 2M_K + 2M_{Al} + 4M_S + 40M_O + 48M_H$$ $$M = 2(39.0983)+2(26.9815)+4(32.065)+40(15.999)+48(1.00794)\ \text{g/mol}$$ Compute each contribution and sum: $$2\times39.0983=78.1966\ \text{g/mol}$$ $$2\times26.9815=53.9630\ \text{g/mol}$$ $$4\times32.065=128.2600\ \text{g/mol}$$ $$40\times15.999=639.9600\ \text{g/mol}$$ $$48\times1.00794=48.38112\ \text{g/mol}$$ $$M=78.1966+53.9630+128.2600+639.9600+48.38112=948.76072\ \text{g/mol}$$ Therefore, the molar mass of K2SO4·Al2(SO4)3·24H2O ≈ $948.76\text{ g/mol}$ (rounded to five significant figures). If you prefer using slightly different atomic weights (e.g., rounded atomic masses), the result will vary by a few tenths of a gram per mole — let me know if you want the value with different atomic mass table conventions.

Q: Why 1 coulomb is taken 1.6 into 10^-19

The correct statement is: the elementary charge (the magnitude of charge of one electron or proton) is $e = 1.602176634\times10^{-19}\text{ C}$. So 1 coulomb is the charge of about $6.241509074\times10^{18}$ electrons. Explanation First, a clarification: 1 coulomb is not equal to $1.6\times10^{-19}$. Rather, one elementary charge (the charge on a single electron) equals about $1.6\times10^{-19}\text{ C}$. A coulomb is a much larger unit of charge. The numerical value of $e$ was determined experimentally (famous early work: the Millikan oil‑drop experiment measured the electron charge). Later, more precise methods (Faraday constant divided by Avogadro’s number, measurements using the quantum Hall effect, shot noise, etc.) refined the value. You can see the relation using the Faraday constant $F$ (charge per mole of electrons) and Avogadro’s number $N_A$ (number of particles per mole): $$e=\frac{F}{N_A}$$Using $F\approx96485\text{ C/mol}$ and $N_A\approx6.02214076\times10^{23}\text{ mol}^{-1}$ gives $e\approx1.602\times10^{-19}\text{ C}$. Since the 2019 SI redefinition, the elementary charge was fixed exactly by definition to $$e=1.602176634\times10^{-19}\text{ C}$$so that value is now exact in the SI system. Example calculation: how many electrons make 1 coulomb?$$\frac{1\text{ C}}{e}\approx\frac{1}{1.602176634\times10^{-19}}\approx6.241509074\times10^{18}\text{ electrons.}$$ Bottom line: the value $1.602\times10^{-19}\text{ C}$ is the (measured and now defined) amount of electric charge carried by a single elementary charge; 1 coulomb contains roughly $6.24\times10^{18}$ such charges.

Q: The mass of the earth is 5,976,000,000,000,000,000,000,000 Kg and the radius of the earth is 6.37 x 106m. Moon is the natural satellite of earth which revolves around the sun due to strong gravitational force of the earth. The mass of the moon is 7.36 x 10 raised to 22Kg. The radius of the moon is 1.74 x 10 raised to 6 m. The distance between the earth and moon is 3.84 x 10 raised to 5 km.(AO2)a. Write the mass of earth in standard form?b. Express the distance between earth and moon in m and find its square. Write your answer in standard form?c. Express the double of difference of radius of earth and moon in standard form

Q1: The correct answer is: $5.976\times10^{24}\text{ kg}$ Explanation Steps: Write the given mass: $$5,976,000,000,000,000,000,000,000\text{ kg}$$ Move the decimal after the first nonzero digit to get standard form: $$5,976,000,000,000,000,000,000,000=5.976\times10^{24}\text{ kg}$$ Q2: The correct answer is: distance $=3.84\times10^{8}\text{ m}$; square $=1.47456\times10^{17}\text{ m}^2$ Explanation Steps: Convert km to m: $$3.84\times10^{5}\text{ km}=3.84\times10^{5}\times10^{3}\text{ m}=3.84\times10^{8}\text{ m}$$ Square the distance: $$(3.84\times10^{8})^{2}=3.84^{2}\times10^{16}=14.7456\times10^{16}=1.47456\times10^{17}\text{ m}^2$$ Q3: The correct answer is: $9.26\times10^{6}\text{ m}$ Explanation Steps: Difference of radii: $$6.37\times10^{6}\text{ m}-1.74\times10^{6}\text{ m}=(6.37-1.74)\times10^{6}=4.63\times10^{6}\text{ m}$$ Double that difference: $$2\times4.63\times10^{6}=9.26\times10^{6}\text{ m}$$ If you want, I can also give these answers rounded to a different number of significant figures.

Q: What is the speed of sound in air? Choose from the options: 332 m/s, 996 m/s, 100 m/s, none of these.

The correct answer is: 332 m/s Explanation The speed of sound in air depends mainly on temperature. For an ideal diatomic gas (like dry air) the formula is$$v = \sqrt{\gamma R T}$$where $\gamma = 1.4$ (ratio of specific heats), $R = 287\text{ J/(kg·K)}$ (specific gas constant for air), and $T$ is the absolute temperature in kelvin. At $0^\circ\text{C}$ ($T=273.15\text{ K}$) this gives$$v = \sqrt{1.4 \times 287 \times 273.15}\approx 331.3\text{ m/s}.$$Values are commonly rounded to about $331\text{–}332\text{ m/s}$ at $0^\circ\text{C}$. A convenient approximation for ordinary temperatures in °C is$$v \approx 331 + 0.6\,T_{^\circ\text{C}},$$so at $20^\circ\text{C}$ $v\approx343\text{ m/s}$. Given the choices, 332 m/s is the physically correct and commonly quoted value (close to the 331.3 m/s theoretical value). 996 m/s and 100 m/s are not realistic for sound in air at ordinary temperatures, and “none of these” is therefore incorrect.

Q: How do you convert 4 light years to earth days?

The correct answer is: $1461\text{ days}$ Explanation A light-year is the distance light travels in one Julian year. So "4 light years" means light takes 4 years to travel that distance. Using the Julian year length (commonly used for the light‑year), one year = $365.25\text{ days}$. Steps: One light‑year corresponds to one year of light travel: $$1\text{ ly}=1\text{ year}=365.25\text{ days}$$ Multiply by 4: $$4\text{ ly}=4\times365.25\text{ days}=1461\text{ days}$$ Therefore, light takes $1461\text{ days}$ to travel $4$ light‑years. Note: If you use a calendar year of $365\text{ days}$ instead of the Julian year, you get $4\times365=1460\text{ days}$; the $365.25$ value accounts for the extra quarter day averaged in leap years.

Question

Why you put K_ 273 in the conventional

Accepted Answer

Because absolute zero is at −273.15°C, we add 273.15 (often rounded to 273) to convert Celsius to Kelvin: $T(	ext{K}) = T(°	ext{C}) + 273.15$.

Explanation

The Kelvin scale is an absolute temperature scale whose zero point (0 K) is absolute zero — the theoretical temperature where particles have minimal thermal motion.  
Absolute zero equals −273.15°C, so to move from the Celsius origin (0°C) to the Kelvin origin you must add 273.15 degrees. That gives the exact conversion formula:

$T(	ext{K}) = T(°	ext{C}) + 273.15$.    Conversely: $T(°	ext{C}) = T(	ext{K}) - 273.15$.

Why sometimes people use 273 instead of 273.15

273 is a convenient approximation for quick/rough calculations (error of 0.15 K, usually negligible for many problems).  
For precise science and engineering you should use 273.15 (or more digits if required). By definition of the Celsius/Kelvin scales, the offset 273.15 is exact.

If you meant a different notation (for example $K_{273}$ or something else), tell me and I’ll explain that specific usage.