Answer:
CasH52(1) + 38 O2(g) → 25 CO2(g) + 26 H2O(g)
Explanation:
Answer: 4) Concentrate it
Explanation:
<span>Fischer projection for D-2-ketotetrose is in Word document below.
</span>D-2-ketotetrose is monosaccharide, having both a ketone (a ketose) and four carbons (a tetrose). There are two ketotetroses (the enantiomers) L and D-erythrulose, this is D-erytrhrulose (1,3,4-trihydroxy-2-butanone).
The <span>Fischer projection is </span>two-dimensional<span> representation of a </span>three dimensional organic molecule.
Answer:
The answer to your question is Molarity = 0.0708
Explanation:
Data
NaOH 0.05 M Volume 1 = 3.87 ml Volume 2 = 25.11 ml
HCl 15 ml
Process
1.- Find the volume used of NaOH
25.11 - 3.87 = 21.24 ml = 0.02124 l
2.- Write the balanced equation of the reaction
NaOH + HCl ⇒ NaCl + H₂O
3.- Calculate the moles of NaOH in the solution
Molarity = 
moles = Molarity x volume
moles = 0.05 x 0.02124
moles = 0.001062
4.- From the reaction we know that NaOH and HCl react in a proportion 1:1.
1 mol of NaOH ------------- 1 mol of HCl
0.001062 moles of NaOH ------------ x
x = (0.001062 x 1) / 1
x = 0.001062 moles of HCl
5.- Find the molarity of HCl
Molarity = 
Molarity = 0.0708
Answer:
Explanation:
<u>1) Balanced chemical equation:</u>
<u>2) Mole ratio:</u>
- 2 mol S : 3 mol O₂ : 2 mol SO₃
<u>3) Limiting reactant:</u>
n = 6.0 g / 32.0 g/mol = 0.1875 mol O₂
n = 7.0 g / 32.065 g/mol = 0.2183 mol S
Actual ratio: 0.1875 mol O₂ / 0.2183 mol S =0.859
Theoretical ratio: 3 mol O₂ / 2 mol S = 1.5
Since there is a smaller proportion of O₂ (0.859) than the theoretical ratio (1.5), O₂ will be used before all S be consumed, and O₂ is the limiting reactant.
<u>4) Calcuate theoretical yield (using the limiting reactant):</u>
- 0.1875 mol O₂ / x = 3 mol O₂ / 2 mol SO₃
- x = 0.1875 × 2 / 3 mol SO₃ = 0.125 mol SO₃
<u>5) Yield in grams:</u>
- mass = number of moles × molar mass = 0.125 mol × 80.06 g/mol = 10.0 g
<u>6) </u><em><u>Percent yield:</u></em>
- Percent yield, % = (actual yield / theoretical yield) × 100
- % = (7.9 g / 10.0 g) × 100 = 79%
