<span>The molar heat of solution of NaOH is -445,100 J/mol. To compute much heat (in J) will be released if 40.00 g of NaOH are dissolved in water, we first convert the given grams of NaOH to moles of NaOH, and use the given molar heat of solution to compute for the energy. (Using dimensional analysis):
40 g NaOH x (1 mol NaOH/ 40 g NaOH) x (-445100 J / 1 mol NaOH) = -445100 J of energy.</span>
Answer:
element
Explanation:
we know that helium is a pure substance although helium atoms are sometimes mixed with their isotopes it is still the same element. since there is no other element combined with helium this makes it an element.
250-128 = 122J is reflected
<h2>The required "option is b) potential energy".</h2>
Explanation:
- The energy which is associated with the motion of the object is called kinetic energy.
- The energy which is associated with the amount of heat released is known as thermal energy.
- The energy which is released during a chemical reaction is known as chemical energy.
Therefore, option (a), (c), and (d) are incorrect.
- The energy which is possessed by a stationary object or associated with the position of an object is called as potential energy.
- Hence, the correct option is potential energy.
Here's how to do it:
<span>Balanced equation first: </span>
<span>Mg + HCl = H2 + MgCl2 unbalanced </span>
<span>Mg + 2 HCl = H2 = MgCl balanced </span>
<span>Therefore 1 mole Mg reacts with 2 moles Hcl. </span>
<span>50g Mg = ? moles (a bit over 2; you work it out) </span>
<span>75 g HCl = ? moles (also a bit over 2; you work it out) </span>
<span>BUT, you need twice the moles HCl; therefore it is the Mg that is in excess. (you can now work out how many moles are in excess, and therefore how much mg is left over). </span>
<span>So, 2 moles HCl produce 1 mole H2(g) </span>
<span>therefore, the amount of H2 produced is half the number of moles of HCl </span>
<span>At STP, there are X litres per mole of gas (look it up - about 22 from memory) </span>
<span>Therefore, knowing the moles of H2, you can calculate the volume</span>
