Answer:
3120.75J
Explanation:
So, we have the formula 
. For this example, q is the heat energy in Joules, m is the mass in grams, c is the specific heat capacity in 
, and 
t is the change in temperature. In this case, m = 47.5g, c = 0.9 , and t = 94-21 = 73°C. Plugging in the values, we get the joules of heat required to raise 47.5g of Al from 21°C to 94°C which is stated above. You can double check my answer but that should be it. An important thing to be aware of are the units. Sometimes, the heat capacity may not be . I may be in Kelvin or something. Anyways, hope this helps.
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Sodium want to give up an electron and "will always want to give up an electron* because less energy is needed to remove the one valence electron of sodium in order to have an octet configuration. This is the reason why sodium and other alkali metals form positive ions so easily.
<h2>Case study</h2>
For instance: In forming an ionic bond, the sodium atom, which is electropositive, loses its valence electron to chlorine. The resulting sodium ion has the same electron configuration as neon (1s2 2s22p6). It has a +1 charge, because there are 11 protons in the nucleus, but only 10 electrons around the nucleus of the ion
Learn more about sodium:
brainly.com/question/25832904
The balanced equation for the above neutralisation reaction is as follows;
2KOH + H₂SO₄ --> K₂SO₄ + 2H₂O
stoichiometry of KOH to H₂SO₄ is 2:1
neutralisation is the reaction between H⁺ ions and OH⁻ ions to form water which is neutral
number of KOH moles - 1.56 mol
2 mol of KOH require 1 mol of H₂SO₄ for neutralisation
therefore 1.56 mol of KOH require - 1/2 x 1.56 mol = 0.78 mol
0.78 mol of H₂SO₄ are required for neutralisation
Answer:
Al³⁺, S²⁻; Al₂S₃
Explanation:
Al is a metal, so it loses its three valence electrons in a reaction to form Al³⁺ ions.
S is a nonmetal. It needs two more electrons to complete its octet, so it tends to form S²⁻ ions.
We can use the criss-cross method to work out the formula of the compound.
The steps are
- Write the symbols of the anion and cation.
- Criss-cross the numbers of the charges to become the subscripts of the other ion.
- Write the formula with the new subscripts.
- Divide the subscripts by their highest common factor.
- Omit all subscripts that are 1.
When we use this method with Al³⁺ and S²⁻, the formula for the compound formed becomes Al₂S₃ .
