The number of calories that are required to change the temperature of 2.18 g of water from 15.3 c to 69.5 c is <u>118.16 cal</u>
<u><em> calculation</em></u>
- Heat in calories = MCΔ T where,
- M(mass)= 2.18 g
- C(specific heat capacity)= 1.00 cal/g/c
- ΔT( change in temperature)= 69.5- 15.3 =54.2 c
heat is therefore= 2.18 g x 1.00 cal/g/c x 54.2 c=118.16 cal
Answer:
0.098 moles
Explanation:
Let y represent the number of moles present
1 mole of Ba(OH)₂ contains 2 moles of OH- ions.
Hence, 0.049 moles of Ba(OH)2 contains y moles of OH- ions.
To get the y moles, we then do cross multiplication
1 mole * y mole = 2 moles * 0.049 mole
y mole = 2 * 0.049 / 1
y mole = 0.098 moles of OH- ions.
1 mole of OH- can neutralize 1 mole of H+
Therefore, 0.098 moles of HNO₃ are present.
<span>H2SO4 gives 2 moles oh H+ per mole of acid
[H2SO4] = 2M so [H+] = 4M
pH = -log(4) = -.6
Therefore, the pH </span><span>of a 2.0 M H2SO4 solution is -0.6
</span>
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