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swat32
3 years ago
10

3 NO2(g) + H2O(ℓ) −→

Chemistry
1 answer:
gulaghasi [49]3 years ago
7 0
The balanced equation for the above reaction is as follows;
3NO₂ + H₂O --> 2HNO₃ + NO
stoichiometry of NO₂ to NO is 3:1
molar volume is where 1 mol of any gas occupies a volume of 22.4 L
volume of gas is directly proportional to number of moles of gas.
therefore stoichiometry can be applied for volume as well.
volume ratio of NO₂ to NO is 3:1
volume of NO₂ reacted - 854 L
therefore volume of NO formed - 854 L /3 = 285 L
volume of NO formed - 285 L
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The flask contains 10.0 mL of HCl and a few drops of phenolphthalein indicator. The buret contains 0.160 M NaOH. It requires 18.
olchik [2.2K]

Answer:

Approximately 0.291\; \rm M (rounded to two significant figures.)

Explanation:

The unit of concentration \rm M is the same as \rm mol \cdot L^{-1} (moles per liter.) On the other hand, the volume of both the \rm NaOH solution and the original \rm HCl solution here are in milliliters. Convert these two volumes to liters:

  • V(\mathrm{NaOH}) = 18.2\; \rm mL = 18.2 \times 10^{-3}\; \rm L = 0.0182\; \rm L.
  • V(\text{$\mathrm{HCl}$, original}) = 10.0\; \rm mL = 10.0\times 10^{-3}\; \rm L = 0.0100\; \rm L.

Calculate the number of moles of \rm NaOH in that 0.0182\; \rm L of 0.160\; \rm M solution:

\begin{aligned} n(\mathrm{NaOH}) &= c(\mathrm{NaOH})\cdot V(\mathrm{NaOH})\\ &= 0.160\; \rm mol \cdot L^{-1} \times 0.0182\; \rm L \approx 0.00291\; \rm mol\end{aligned}.

\rm HCl reacts with \rm NaOH at a one-to-one ratio:

\rm HCl\; (aq) + NaOH\; (aq) \to NaCl\; (aq) + H_2O\; (l).

Coefficient ratio:

\displaystyle \frac{n(\mathrm{HCl})}{n(\mathrm{NaOH})} = 1.

In other words, one mole of \rm NaOH would neutralize exactly one mole of \rm HCl. In this titration, 0.291\; \rm mol of \rm NaOH\! was required. Therefore, the same amount of \rm HC should be present in the original solution:

\begin{aligned}&n(\text{$\mathrm{HCl}$, original})\\ &= n(\mathrm{NaOH})\cdot \frac{n(\mathrm{HCl})}{n(\mathrm{NaOH})} \\ &\approx 0.00291\; \rm mol \times 1 = 0.00291\; \rm mol\end{aligned}.

Calculate the concentration of the original \rm HCl solution:

\displaystyle c(\text{$\mathrm{HCl}$, original}) = \frac{n(\text{$\mathrm{HCl}$, original})}{V(\text{$\mathrm{HCl}$, original})} \approx \frac{0.00291\; \rm mol}{0.0100\; \rm L} \approx 0.291\; \rm M.

5 0
3 years ago
Please help me, thank you
Bezzdna [24]

Answer:

it's C

Explanation:

Yes

5 0
3 years ago
Read 2 more answers
What is the partial pressure of 0.50 mol Ne gas combined with 1.20 mol Kr gas at a final pressure of 730 torr?
WARRIOR [948]
  The partial pressure  of 0.50  Ne  gas    is   214.71  torr


      calculation

 the partial   pressure  of Ne  = moles  of Ne/total moles  x  final  pressure

 
find  the total  moles  of the air mixture  

that is moles of   Ne  +  moles of K=  0.50 + 1.20  =  1.70 moles
  

The partial  pressure  is therefore =   0.50 /1.70  x  730  =  214.71  torr
   
8 0
3 years ago
Read 2 more answers
1. A 225-L barrel of white wine has an initial free SO2 concentration of 22 ppm and a pH of 3.70. How much SO2 (in grams) should
Alexandra [31]

Answer:

The appropriate answer is "9.225 g".

Explanation:

Given:

Required level,

= 63 ppm

Initial concentration,

= 22 ppm

Now,

The amount of free SO₂ will be:

= Required \ level -Initial \ concentration

= 63-22

= 41 \ ppm

The amount of free SO₂ to be added will be:

= 41\times 225

= 9225 \ mg

∵ 1000 mg = 1 g

So,

= 9225\times \frac{1}{1000}

= 9.225

Thus,

"9.225 g" should be added.

3 0
3 years ago
Please help!!
prisoha [69]
A word equation is a written description of a chemical reaction.
All word equations start with the reactants.
Then, what comes next is the word "react to form."
Finally, the products of the equation are mentioned.
An example is,
Zinc and Hypochondriac acid react to form hydrogen gas and zinc chloride.
3 0
3 years ago
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