A balloon contains 0.950 mol of nitrogen gas and has a volume of 25.5 L. How many grams of N2 should be released from the balloo
1 answer:
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Answer:
82.9 mL
Explanation:
1. Volume of silver
2. Volume of gold
3. Total volume of silver and gold
V = 4.766 mL + 2.591 mL = 7.36 mL
4 New reading of water level
V = 75.5 mL + 7.36 mL = 82.9 mL
Your compound is 
.
Remember that the oxidation numbers in a neutral compound must add up to zero. Cl has an oxidation number of -1 because it is a halogen K has an oxidation number of +1 because it is an alkali metal, which exhibits an oxidation state of +1 in compounds.
Since you have 6 atoms of Cl, you have -1(6) = -6 for the Cl. Since you 2 atoms of K, you have +1(2) = +2 for the K. The oxidation number of Pt must make all the oxidation numbers add up to zero:
+2 + (-6) + oxidation number of Pt = 0
-4 + oxidation number of Pt = 0
Oxidation number of Pt = 4
Remember that the oxidation numbers in a neutral compound must add up to zero. Cl has an oxidation number of -1 because it is a halogen K has an oxidation number of +1 because it is an alkali metal, which exhibits an oxidation state of +1 in compounds.
Since you have 6 atoms of Cl, you have -1(6) = -6 for the Cl. Since you 2 atoms of K, you have +1(2) = +2 for the K. The oxidation number of Pt must make all the oxidation numbers add up to zero:
+2 + (-6) + oxidation number of Pt = 0
-4 + oxidation number of Pt = 0
Oxidation number of Pt = 4
Answer:
(a) 0.22 mol Cl₂ and 15.4g Cl₂
(b) 2.89.10⁻³ mol O₂ and 0.092g O₂
(c) 8 mol NaNO₃ and 680g NaNO₃
(d) 1,666 mol CO₂ and 73,333 g CO₂
(e) 18.87 CuCO₃ and 2,330g CuCO₃
Explanation:
In most stoichiometry problems there are a few steps that we always need to follow.
- Step 1: Write the balanced equation
- Step 2: Establish the theoretical relationship between the kind of information we have and the one we are looking for. Those relationships can be found in the balanced equation.
- Step 3: Apply conversion factor/s to the data provided in the task based on the relationships we found in the previous step.
(a)
Step 1:
2 Na + Cl₂ ⇄ 2 NaCl
Step 2:
In the balanced equation there are 2 moles of Na, thus 2 x 23g = 46g of Na. <u>46g of Na react with 1 mol of Cl₂</u>. Since the molar mass of Cl₂ is 71g/mol, then <u>46g of Na react with 71g of Cl₂</u>.
Step 3:
(b)
Step 1:
HgO ⇄ Hg + 0.5 O₂
Step 2:
<u>216.5g of HgO</u> form <u>0.5 moles of O₂</u>. <u>216.5g of HgO</u> form <u>16g of O₂</u>.
Step 3:
(c)
Step 1:
NaNO₃ ⇄ NaNO₂ + 0.5 O₂
Step 2:
<u>16g of O₂</u> come from <u>1 mol of NaNO₃</u>. <u>16g of O₂</u> come from <u>85g of NaNO₃</u>.
Step 3:
(d)
Step 1:
C + O₂ ⇄ CO₂
Step 2:
<u>12 g of C</u> form <u>1 mol of CO₂</u>. <u>12 g of C</u> form <u>44g of CO₂</u>.
Step 3:
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(e)
Step 1:
CuCO₃ ⇄ CuO + CO₂
Step 2:
<u>79.5g of CuO</u> come from <u>1 mol of CuCO₃</u>. <u>79.5g of CuO</u> come from <u>123.5g of CuCO₃</u>.
Step 3: