Answer:
0.095 moles of Calcium is there in 5.74 x 1022 atoms of calcium.
Explanation:
- As we know, 6.023*10^23 atoms of an element is equal to its atomic weight.
And, 6.023*10^23 atoms of an element is also equal to 1 mole of the element.
We have,
- 6.023*10^23 atoms of element calcium equals to 1 mole of Calcium
- 5.74*10^22 atoms of element calcium equals to
(1/(6.023*10^.23)) * 5.74*10^22 moles of calcium
Therefore,
- 5.74 x 1022 atoms of calcium= 0.095 moles of calcium.
Answer:
<h2>6.75 g</h2>
Explanation:
The mass of a substance when given the density and volume can be found by using the formula
mass = Density × volume
From the question we have
mass = 2.7 × 2.5
We have the final answer as
<h3>6.75 g</h3>
Hope this helps you
Answer:
ΔH⁰(11.4g NH₄NO₃) = -30.59Kj (4 sig. figs. ~mass of NH₄NO₃(s) given) (exothermic)
Explanation:
3NH₄NO₃(s) + C₁₀H₂₂(l) + 14O₂(g) => 3N₂(g) + 17H₂O(g) + 10CO₂(g)
ΔH⁰(f): 3(-365.6)Kj 1(-301)Kj 14(0)Kj 3(0)Kj 17(-241.8)Kj 10(-393.5)Kj
= -1096.8Kj = -301Kj = 0Kj = 0Kj = -4110.6Kj = -3930.5Kj
ΔHₙ°(rxn) = ∑
(ΔH˚(f)products) - ∑(ΔH˚(f)reactants)
= [3(0)Kj + 17(-241.8)Kj + (-393.5)Kj] - [(-(1096.8)Kj + (-301)Kj + (0)Kj]
= [-(8041.1) - (-1397.8)]Kj
= -6643.3Kj (for 3 moles NH₄NO₃ used in above equation)
∴ Standard Heat of Rxn = -6643.3Kj/3moles = -214.8Kj/mole NH₄NO₃(s)
ΔH°(rxn for 14.11g NH₄NO₃(s)) = (11.4g/80.04g·mol⁻¹)(-214.8Kj/mol) = 30.5937Kj ≅ 30.59Kj (4 sig. figs. ~mass of NH₄NO₃(s) given)
To determine the concentration of one solution which is specifically basic or acidic solution through taking advantage on its points of equivalence, titration analysis is done.
Let us determine the reaction for the titration below:
2NaOH +2H2SO4 = Na2SO4 +2H2O
So,
0.0665 mol NaOH (2 mol H2SO4/ 2mol NaOH) / .025 L solution
= 2.62 M H2SO4
The answer is the fourth option:
<span>2.62 M</span>
