Empirical Formulae for;
Compound 1- K5 Mn5 O16
Compound 2- Na2 Cr2 O7
Compound 3- C3 H4 O4
Compound 4- C3 H3 O1
Explanation:
Step 1; as all the element percentages are given in percent assume the total mass of the compound is 100g and take each percentage as grams i.e., 27.5% of K as 27.5 g of K and so on.
Step 2; convert the mass of each element into their mole values by dividing available mass by molar masses.
Molar masses of required elements are as follows; K=39, Mn=55, O=16 C=12, H=1, Na=23, Cr=52.
Step 3; Divide all the values by the smallest mole value. I.e. for compound 1 after dividing the masses by molar masses we get 0.705, 0.681, and 2.187 for elements K, Mn, O respectively. Divide all three values with the least value which is 0.681 and write these values down.
Step 4; Convert all the numbers available into whole numbers by multiplying with suitable values. i.e. 3 if values are 0.33, 2 if values are 0.5 etc.
Step 5; Assign these values to corresponding elements and you will get the above empirical formula.
Answer:
Matter is the Stuff Around You
Matter is everything around you. Atoms and compounds are all made of very small parts of matter. ... Matter is defined as anything that has mass and takes up space (it has volume).
Explanation:
hope it was helpful
Taking into account the stoichiometry of the reaction, 34.12 grams of LiOH are produced from 9.89 g of Li.
In first place, the balanced reaction is:
2 Li + 2 H₂O ⇒ 2 LiOH + H₂
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
- Li= 2 moles
- H₂O= 2 moles
- LiOH= 2 moles
- H₂= 1 mole
The molar mass of each compound is:
- Li= 6.94 g/mole
- H₂O= 18 g/mole
- LiOH= 23.94 g/mole
- H₂= 2 g/mole
By reaction stoichiometry, the following amounts of mass of each compound participate in the reaction:
- Li= 2 moles× 6.94 g/mole= 13.88 g
- H₂O= 2 moles× 18 g/mole= 36 g
- LiOH= 2 moles× 23.94 g/mole= 47.88 g
- H₂= 1 mole× 2 g/mole= 2 g
Then you can apply the following rule of three: if by reaction stoichiometry 13.88 g of Li produce 47.88 g of LiOH, 9.89 g of Li produce how much mass of LiOH?
Solving:
<u><em>mass of LiOH= 34.12 grams</em></u>
Finalli, 34.12 grams of LiOH are produced from 9.89 g of Li.
Learn more:
Answer:
the heat rate required to cool down the gas from 535°C until 215°C is -2.5 kW.
Explanation:
assuming ideal gas behaviour:
PV=nRT
therefore
P= 109 Kpa= 1.07575 atm
V= 67 m3/hr = 18.6111 L/s
T= 215 °C = 488 K
R = 0.082 atm L /mol K
n = PV/RT = 109 Kpa = 1.07575 atm * 18.611 L/s /(0.082 atm L/mol K * 488 K)
n= 0.5 mol/s
since the changes in kinetic and potencial energy are negligible, the heat required is equal to the enthalpy change of the gas:
Q= n* Δh = 0.5 mol/s * (- 5 kJ/mol) =2.5 kW
Tip for next time, when you are looking at these things, rather than just posting it, take a look at your options. A solid to a gas, this means that melting cant be an option because that is solid to a liquid, evaporation cant be an option because that is liquid to a gas and condensation cant be because that is gas to a l. Therefore the answer must be sublimatio.
