Answer:
1.25mole of Mg(OH)₂
Explanation:
The reaction is between Mg(OH)₂ and HCl, this is a neutralization reaction between an acid and a base.
Mg(OH)₂ + 2HCl → MgCl₂ + 2H₂O
The balanced reaction equation is given above.
2 mole of HCl reacts with 1 mole of Mg(OH)₂
So, 2.5mole of HCl will react with 
= 1.25mole of Mg(OH)₂
The number of moles of Mg(OH)₂ is given as 1.25mol
<span>C2H5
First, you need to figure out the relative ratios of moles of carbon and hydrogen. You do this by first looking up the atomic weight of carbon, hydrogen, and oxygen. Then you use those atomic weights to calculate the molar masses of H2O and CO2.
Carbon = 12.0107
Hydrogen = 1.00794
Oxygen = 15.999
Molar mass of H2O = 2 * 1.00794 + 15.999 = 18.01488
Molar mass of CO2 = 12.0107 + 2 * 15.999 = 44.0087
Now using the calculated molar masses, determine how many moles of each product was generated. You do this by dividing the given mass by the molar mass.
moles H2O = 11.5 g / 18.01488 g/mole = 0.638361 moles
moles CO2 = 22.4 g / 44.0087 g/mole = 0.50899 moles
The number of moles of carbon is the same as the number of moles of CO2 since there's just 1 carbon atom per CO2 molecule.
Since there's 2 hydrogen atoms per molecule of H2O, you need to multiply the number of moles of H2O by 2 to get the number of moles of hydrogen.
moles C = 0.50899
moles H = 0.638361 * 2 = 1.276722
We can double check our math by multiplying the calculated number of moles of carbon and hydrogen by their respective atomic weights and see if we get the original mass of the hydrocarbon.
total mass = 0.50899 * 12.0107 + 1.276722 * 1.00794 = 7.400185
7.400185 is more than close enough to 7.40 given rounding errors, so the double check worked.
Now to find the empirical formula we need to find a ratio of small integers that comes close to the ratio of moles of carbon and hydrogen.
0.50899 / 1.276722 = 0.398669
0.398669 is extremely close to 4/10, so let's reduce that ratio by dividing both top and bottom by 2 giving 2/5.
Since the number of moles of carbon was on top, that ratio implies that the empirical formula for this unknown hydrocarbon is
C2H5</span>
The elements in Groups 1A(1) and 7A(17) are all quite reactive.
<h3>Major difference between Groups 1A(1) and 7A(17) : </h3>
Group 7's halogens, which are non-metal elements, become less reactive as you move down the group. In contrast to the alkali metals in Group 1 of the periodic table, this trend is the opposite. The most reactive element in Group 7 is fluorine.
Alkali metals are soft and reactive metals. They react vigorously with water and become more reactive. And other hand halogens are reactive non metals.
- Elements of group 1A are known as alkali metals. Elements of this group are lithium, sodium, potassium, rubidium, cesium.
- Reactivity increase down group 1 but decrease up group 7 this is because group 7 elements react by gaining an electron. As one move down the group, the amount of electron shielding increases, meaning that the electron is less attracted to the nucleus.
To know more about Groups 1A(1) and 7A(17) please click here :
brainly.com/question/13063502
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Hey buddy I am here to help!
1. C
2. A
3. A & B
4. C
5. C
6. A
7. A
8. A & C
Hope it helps!
Plz mark brainlist!
I believe that would be oil and vinegar.
