-<u><em>Oxygen</em></u>
According to Google these are the percentages of the <em>Earths Atmosphere</em>
<em>1</em> 78% - Nitrogen
<u>2</u> 21% - Oxygen
<em>3</em> 0.9% - Argon
<em>4 </em>0.3 - Carbon Dioxide with very small percentage of other elements.
Answer: Gold.
Explanation: Water, CO2, and table salt are compounds. They are composed of two or more separate elements; a mixture, where as gold is neither.
Answer:
a. 5.9 × 10⁻³ M/s
b. 0.012 M/s
Explanation:
Let's consider the following reaction.
2 N₂O(g) → 2 N₂(g) + O₂(g)
a.
Time (t): 12.0 s
Δn(O₂): 1.7 × 10⁻² mol
Volume (V): 0.240 L
We can find the average rate of the reaction over this time interval using the following expression.
r = Δn(O₂) / V × t
r = 1.7 × 10⁻² mol / 0.240 L × 12.0 s
r = 5.9 × 10⁻³ M/s
b. The molar ratio of N₂O to O₂ is 2:1. The rate of change of N₂O is:
5.9 × 10⁻³ mol O₂/L.s × (2 mol N₂O/1 mol O₂) = 0.012 M/s
I know what you're asking but I don't think the question is stated properly. Technically, an atom will not join with an "oxide" ion; i.e., the oxide ion is an atom of oxygen to which two electrons have been added. An oxide ion will add to 2 K ions or 1 Ca ion. The K ion has lost just one electron so it takes two of them to equal the 2- charge on the oxide ion whereas the Ca ion has lost two electrons and it takes only one of them to equal the charge on the oxide ion.
V(NaOH)=15 mL =0.015 L
C(NaOH)=0.1 mol/L
C(H₂SO₄)=0.05 mol/L
2NaOH + H₂SO₄ = Na₂SO₄ + 2H₂O
n(NaOH)=V(NaOH)C(NaOH)=2n(H₂SO₄)
n(H₂SO₄)=V(H₂SO₄)C(H₂SO₄)
V(NaOH)C(NaOH)=2V(H₂SO₄)C(H₂SO₄)
V(H₂SO₄)=V(NaOH)C(NaOH)/{2C(H₂SO₄)}
V(H₂SO₄)=0.015*0.1/{2*0.05}=0.015 L = 15 mL