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KiRa [710]
3 years ago
6

How many moles of oxygen are required to react with methanol in order to produce 13 moles of formaldehyde?

Chemistry
1 answer:
ddd [48]3 years ago
3 0

Answer:

6.5 moles of Oxygen are required

Explanation:

Based on the reaction:

CH3OH + 1/2 O2 → CH2O + H2O

1 mole of methanol reacts with 1/2 moles O2 to produce 1 mole of formaldehyde and 1 mole of water.

Thus, to produe 13 moles of formaldehyde, CH2O, are needed:

13 moles CH2O * (1/2mol O2 / 1mol CH2O) =

<h3>6.5 moles of Oxygen are required</h3>
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Find the ph of a 0.250 m solution of nac2h3o2. (the ka value of hc2h3o2 is 1.80×10−5). express your answer numerically to four s
Slav-nsk [51]
Answer is: pH value of solution of NaC₂H₃O₂ is 9.07.
Chemical reaction: C₂H₃O₂⁻ + H₂O ⇄ HC₂H₃O₂ + OH⁻.
Ka(HC₂H₃O₂) = 1,8·10⁻⁵.<span>
Ka · Kb = Kw.
</span>1,8·10⁻⁵ mol/dm³ · Kb = 1·10⁻¹⁴ mol²/dm⁶; the ionic product of water at 25°C.<span>
Kb(</span>C₂H₃O₂⁻) = 1·10⁻¹⁴ mol²/dm⁶ ÷ 1,8·10⁻⁵ mol/dm³.<span>
Kb(</span>C₂H₃O₂⁻) = 5,56·10⁻¹⁰ mol/dm³.
c(C₂H₃O₂⁻) = 0,25 M.
[OH⁻] = [HC₂H₃O₂] = x.
[C₂H₃O₂⁻] = 0,25 M - x.
Kb = [OH⁻] · [HC₂H₃O₂] / [C₂H₃O₂⁻].
5,56·10⁻¹⁰ = x² / (0,25 M -x).
Solve quadratic equation: x = [OH⁻] = 0,0000118 M.
pOH = -log[OH⁻] = -log(0,0000118M) = 4,93.
pH + pOH = 14.
pH = 14 - 4,93 = 9,07.
6 0
3 years ago
Must show units and how they cancelli 1.) 175 km to um 3.) 385 nm to dm 5.) 492 um tom 7.) 52 x 103 dm to mm 9.) 321x 1035 mm to
morpeh [17]

Explanation:

1.) 175 km to μm

1 km=10^9 \mu m

175 km=175\times 10^9\mu m=1.75\times 10^{11} \mu m

3.) 385 nm to dm

1 nm=10^{-8} dm

385 nm=385\times 10^{-8} dm=3.85\times 10^{-6} dm

5.) 492 μm  to m

1 μm =  10^{-6} m

492 \μm=492\times 10^{-6} m=4.92\times 10^{-4} m

7.) 52\times 10^3 dm to mm

1 dm = 100 mm

52\times 10^3 dm=52\times 10^3\times 100 mm=5.2\times 10^{6}dm

9.) 321\times 10^{35} mm to km

1 mm = 10^{-6} km

321\times 10^{35} mm=321\times 10^{35}\times 10^{-6} km=3.21\times 10^{31} km

11.) 456\times 10^3 m to km

m = 0.001 km

456\times 10^3m =456\times 10^3 m\times 0.001 km=456 km

13.) 422\times 10^3 m to nm

1 m = 10^{9} nm

422\times 10^3 m=422\times 10^3\times 10^{9} nm=4.22\times 10^{14} nm

15.) 4.87\times 10^{30} m to pm

1 m = 10^{12} pm

4.87\times 10^{30} m=4.87\times 10^{30}\times 10^{12} pm=4.82\times 10^{42} pm

17.) 5.26\times 10^3 m to um

1 m =  10^{6} \mu m

5.26\times 10^3 m=5.26\times 10^3\times 10^6 \mu m=5.26\times 10^{9} \mu m

19.) 1.25\times 10^{35}m to Mm

1 m =  10^{-6} Mm

1.25\times 10^{35} m=1.25\times 10^{35}\times 10^{-6} Mm=1.25\times 10^{-29} Mm

21.) 4.22\times 10^3 Tm to nm

1 Tm = 10^{21} nm

4.22\times 10^3 Tm=4.22\times 10^3\times 10^{21} nm=4.22\times 10^{24} nm

6 0
3 years ago
KHP is a monoprotic acid which provides one H+ ion. How would your results be affected if a diprotic acid (such as sulfuric acid
kari74 [83]

Explanation:

A.

In a diprotic acid, 2 moles of H+ ions is released. Therefore, number of moles of H+ in a diprotic acid = 2 × number of moles of H+ of monoprotic acid.

B.

Equation of the reaction

2NaOH + H2SO4 --> Na2SO4 + 2H2O

Number of moles of H2SO4 = molar concentration × volume

= 0.75 × 0.0105

= 0.007875 moles.

By stoichiometry, since 1 mole of H2SO4 reacts with 2 moles of NaOH. Therefore, number of moles of NaOH = 2 × 0.007875

= 0.01575 moles.

Molar concentration of NaOH = number of moles ÷ volume

= 0.01575 ÷ 0.0175

= 0.9 M of NaOH.

7 0
3 years ago
Which of the following acids is in household vinegar? A nitric acid hydrochloric acid C fluoric acid D acetic acid
Free_Kalibri [48]

Answer: Out of the given options acetic acid is in household vinegar.

Explanation:

Household vinegar is the one that is commonly used in our home while cooking a number of dishes.

The common name of vinegar is acetic acid and its chemical formula is CH_{3}COOH.

For example, vinegar (acetic acid) is used while making noodles.

Thus, we can conclude that out of the given options acetic acid is in household vinegar.

7 0
3 years ago
Plz help ill give brainlist to you
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3 0
3 years ago
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