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AlexFokin [52]
3 years ago
8

How would different ratios of fuel and oxygen affect the altitude the rocket reaches?

Chemistry
1 answer:
ElenaW [278]3 years ago
5 0

For the rocket engine a lean or excessively rich air-fuel ratio will lead to a compromise in the power output and ultimately the altitude the rocket will reach. Hence which ever the ratio the altitude is greatly affected, unless the ideal ratio is archived.

The air-fuel ratio is the ratio of the amount of air need to burn a part of fuel in a combustion cycle, the ratio is basically a compromise between fuel economy and  optimum power output.

For a gasoline engine for example, the ideal air-fuel mixture of 14.7 parts air to 1 part fuel.

Learn more:

brainly.com/question/4741454

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Non motile in an sentences
allsm [11]

Answer:

Motile bacteria have flagella, while nonmotile bacteria do not.

Explanation:

4 0
3 years ago
if i add 25 ml of water to 135 ml of a 0.25 M NaOH solution what will the molarity of the diluted solution be​
DENIUS [597]

Answer:

0.21 M. (2 sig. fig.)

Explanation:

The molarity of a solution is the number of moles of the solute in each liter of the solution. The unit for molarity is M. One M equals to one mole per liter.

How many moles of NaOH in the original solution?

n = c \cdot V,

where

  • n is the number of moles of the solute in the solution.
  • c is the concentration of the solution. c = 0.25 \;\text{M} = 0.25\;\text{mol}\cdot\textbf{L}^{-1} for the initial solution.
  • V is the volume of the solution. For the initial solution, V = 135\;\textbf{mL} = 0.135\;\textbf{L} for the initial solution.

n = c\cdot V = 0.25\;\text{mol}\cdot\textbf{L}^{-1} \times 0.135\;\textbf{L} = 0.03375\;\text{mol}.

What's the concentration of the diluted solution?

\displaystyle c = \frac{n}{V}.

  • n is the number of solute in the solution. Diluting the solution does not influence the value of n. n = 0.03375\;\text{mol} for the diluted solution.
  • Volume of the diluted solution: 25\;\text{mL} + 135\;\text{mL}  = 160\;\textbf{mL} = 0.160\;\textbf{L}.

Concentration of the diluted solution:

\displaystyle c = \frac{n}{V} = \frac{0.03375\;\text{mol}}{0.160\;\textbf{L}} = 0.021\;\text{mol}\cdot\textbf{L}^{-1} = 0.021\;\text{M}.

The least significant number in the question comes with 2 sig. fig. Keep more sig. fig. than that in calculations but round the final result to 2 sig. fig. Hence the result: 0.021 M.

8 0
3 years ago
Read 2 more answers
Calculate for the following electrochemical cell at 25°C, Pt H2(g) (1.0 atm) H (0.010 M || Ag (0.020 M) Ag if E (H) - +0.000 V a
viva [34]

Answer : The correct option is, (b) +0.799 V

Solution :

The values of standard reduction electrode potential of the cell are:

E^0_{[H^{+}/H_2]}=+0.00V

E^0_{[Ag^{+}/Ag]}=+0.799V

From the cell representation we conclude that, the hydrogen (H) undergoes oxidation by loss of electrons and thus act as anode. Silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.

The half reaction will be:

Reaction at anode (oxidation) : H_2\rightarrow 2H^{+}+2e^-    

Reaction at cathode (reduction) : Ag^{+}+1e^-\rightarrow Ag    

The balanced cell reaction will be,  

H_2+2Ag^{+}\rightarrow 2H^{+}+2Ag

Now we have to calculate the standard electrode potential of the cell.

E^o_{cell}=E^o_{cathode}-E^o_{anode}

E^o_{cell}=E^o_{[Ag^{+}/Ag]}-E^o_{[H^{+}/H_2]}

E^o_{cell}=(+0.799V)-(+0.00V)=+0.799V

Therefore, the standard cell potential will be +0.799 V

4 0
3 years ago
Which statement describes how the binary ionic compound cacl2 is named?
Oksi-84 [34.3K]
Calcium Chloride because it is a type 1 so the the anion ends with -ide
6 0
3 years ago
Rsy-feiu-xad j0in only girlś​
kirill [66]

Answer:

nahhhhhhhhhhhhhhhhhhhhhhh xD

have a good day :)

Explanation:

3 0
3 years ago
Read 2 more answers
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