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miskamm [114]
3 years ago
10

When Na2O is mixed with Mn(NO3)4 it forms NaNO3 and MnO2. We start with 25.0 g

Chemistry
1 answer:
coldgirl [10]3 years ago
5 0
I don’t know what have you heard that with na20
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Buck's Turf Care Company mowed 233 lawns over a 11 week period. What is the average weekly rate of mowing lawns?
masya89 [10]

Answer: 21 lawns per week

Explanation:

The average weekly rate refers to how many lawns were mowed per week given that 233 were done in 11 weeks.

Rate will be given by;

= Lawns mowed / Weeks taken

= 233 / 11

= 21 lawns per week

6 0
3 years ago
The Olympian swam 200 meters in 20 seconds. <br> 1) velocity<br> 2) acceleration<br> 3) speed
irakobra [83]

Answer: velocity

Explanation:

3 0
3 years ago
Read 2 more answers
We get 1700 Tonnes of Ammonia every day. How many tonnes of 63% Nitric acid we can get?
jeyben [28]
There are 3 equations involved in manufacturing Nitric Acid from Ammonia. 

First the ammonia is oxidized:
4NH3 + 5O2 = 4NO + 6H2O

Then for the absorption of the nitrogen oxides.
2NO + O2 = N2O4

Lastly, the N2O4 is further oxidized into Nitric acid.
3N2O4 + 2H2O = 4HNO3 + 2NO

Then run stoichiometry through these equations.
The first equation produces roughly 271,722,938 grams of NO
The second equation produces roughly 416,606,944 grams of N2O4
The last equation produces roughly 380,412,294 grams of HNO3 (nitric acid)

Convert the exact number back into tons, and your answer is: <span>419.332775 tons.
</span>
Rounded, I'm going to say that's 419.33 tons.
Hope this helps! :)

Also, it seems that commercially, Nitric Acid is commonly made by bubbling NO2 into water, rather than using ammonia.
 
8 0
3 years ago
When 70.4 g of benzamide (C7H7NO) are dissolved in 850. g of a certain mystery liquid X, the freezing point of the solution is 2
Arlecino [84]

Answer:

1.62

Explanation:

From the given information:

number of moles of benzamide  =\dfrac{70.4 \ g}{121.14 \ g/mol}

= 0.58 mole

The molality = \dfrac{mass \ of \ solute (i.e. \ benzamide )}{mass \ of \ solvent  }

= \dfrac{0.58 }{0.85 }

= 0.6837

Using the formula:

\mathbf {dT  = l   \times  k_f  \times m}

where;

dT = freezing point = 27

l = Van't Hoff factor = 1

kf = freezing constant of the solvent

∴

2.7 °C = 1 × kf ×  0.6837 m

kf = 2.7 °C/ 0.6837m

kf = 3.949 °C/m

number of moles of NH4Cl = \dfrac{70.4 \ g}{53.491 \  g /mol}

= 1.316 mol

The molality = \dfrac{1.316 \ mol}{0.85 \ kg}

= 1.5484

Thus;

the above kf value is used in determining the  Van't Hoff factor for  NH4Cl

i.e.

9.9 = l × 3.949 × 1.5484 m

l = \dfrac{9.9}{3.949 \times 1.5484 \ m}

l = 1.62

5 0
3 years ago
Structural formula for 4-nonene and fluorine gas
goldenfox [79]
I THINK it's <span>1,1-Difluorononane, or </span>C_9H_{18}F_2.<span>
</span>

6 0
3 years ago
Read 2 more answers
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