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Snezhnost [94]
3 years ago
14

Which sequence represents the correct ranking of the amount of nitrogen per molecule, from least to most nitrogen? Urea, uric ac

id, ammonia Uric acid, ammonia, urea Ammonia, urea, uric acid Ammonia, uric acid, urea
Chemistry
1 answer:
Reptile [31]3 years ago
5 0

Answer:

Ammonia, urea, uric acid

Explanation:

The given compounds are:-

Urea which has a molecular formula of CH_4N_2O and has 2 atoms of nitrogen per molecule.

Ammonia which has a molecular formula of NH_3 and has 1 atom of nitrogen per molecule.

Uric acid which has a molecular formula of C_5H_4N_4O_3 and has 4 atoms of nitrogen per molecule.

Thus,

The order from least to most nitrogen is:-

Ammonia, urea, uric acid

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W(OH)2 + 2 HCl → WCl2 + 2 H2O
Afina-wow [57]

The amount of W(OH)2 needed would be 448.126 g

<h3>Stoichiometric calculation</h3>

From the equation of the reaction:

W(OH)2 + 2 HCl → WCl2 + 2 H2O

The mole ratio of W(OH)2 to HCl is 1:2

Mole of 150g HCl = 150/36.461

                                 = 4.11 moles

Equivalent mole of W(OH)2 = 4.11/2

                                           = 2.06 moles

Mass of 2.06 moles W(OH)2 = 2.06 x 217.855

                                                = 448.188g

More on stoichiometric calculations can be found here: brainly.com/question/8062886

7 0
2 years ago
What happens to a glass of pure fruit juice when water is added to it?
Vesna [10]

Answer:

D is the answer i think

8 0
3 years ago
Explain how you would test the presence of oxygen and hydrogen gases. ​
Vedmedyk [2.9K]
Place a burning splint near the opening of a test tube. If a popping noise occurs, it's probably hydrogen. Place a glowing splint in the test tube, and if it reignites, it could be oxygen. Place a burning splint into a test tube, and if it goes out, it could be carbon dioxide.
5 0
3 years ago
how many ml of a 22.5% (v/v) ethanol solution would you need to measure out in order to have 12.5 ml of ethanol ?
Aleks [24]
Volume percent<span> or </span>volume/volume percent<span> (v/v%) is used when preparing solutions of liquids. It will have units of volume of the smaller composition substance over the volume of the solution. We calculate as follows:

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Hope this answers the question.</span>
8 0
3 years ago
In Part A, you found the amount of product (1.80 mol P2O5 ) formed from the given amount of phosphorus and excess oxygen. In Par
Finger [1]

First step is to balance the reaction equation. Hence we get P4 + 5 O2 => 2 P2O5

Second, we calculate the amounts we start with

P4: 112 g = 112 g/ 124 g/mol – 0.903 mol

O2: 112 g = 112 g / 32 g/mol = 3.5 mol

Lastly, we calculate the amount of P2O5 produced.

2.5 mol of O2 will react with 0.7 mol of P2O5 to produce 1.4 mol of P2O5.

This is 1.4 * (31*2 + 16*5) = 198.8 g

3 0
3 years ago
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