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patriot [66]
2 years ago
6

How many grams of sodium nitrate (NaNO2) will dissolve in 100g of water at 20°C?​

Chemistry
1 answer:
inn [45]2 years ago
8 0
88g ( so sorry if this isn’t correct )
You might be interested in
Compound that contains a terminal carbonyl?
malfutka [58]

Many compunds have a terminal carbonyl

Aldehyde, Ketone, Carboxylic acid, Amide, Imide, Acid anhydride are the first that come to my mind.

7 0
3 years ago
Explain why, when the imidazole ring of histidine is protonated, the double-bonded nitrogen is the nitrogen that accepts the pro
eduard
The important thing to note is the reason why electron react is due to the instability of the electrons. All elements wants to aim the electron configuration of the noble gases. This is the most stable form in which each of the orbitals are sufficiently filled. When it comes to bonding, the order of reactivity is: alkynes > alkenes > alkanes. Alkynes are compounds with triple bonds, alkenes with double bonds and alkanes with single bonds. The single bonds are called saturated hydrocarbons. This is because they have reached stability, so it is quite difficult to react this with reducing or oxidizing agents. Alkynes and alkenes are unsaturated hydrocarbons. They readily react with reducing and oxidizing agents so as to become saturated, as well. The underlying principle for this is that single bonds contain sigma bonds which is the head-on overlapping of electrons. These is the strongest type of covalent bond. Double and triple bonds contain pi bonds which is the side overlapping of electrons orbitals. Hence, these electrons would be easily separated making it more reactive especially during protonation.
4 0
3 years ago
Ammonia can be produced in the laboratory by heating ammonium chloride
AnnyKZ [126]

Answer:

Mass = 2.89 g

Explanation:

Given data:

Mass of NH₄Cl = 8.939 g

Mass of Ca(OH)₂ = 7.48 g

Mass of ammonia produced = ?

Solution:

2NH₄Cl   +  Ca(OH)₂     →    CaCl₂ + 2NH₃ + 2H₂O

Number of moles of NH₄Cl:

Number of moles = mass/molar mass

Number of moles = 8.939 g / 53.5 g/mol

Number of moles = 0.17 mol

Number of moles of Ca(OH)₂ :

Number of moles = mass/molar mass

Number of moles = 7.48 g / 74.1 g/mol

Number of moles = 0.10 mol

Now we will compare the moles of ammonia with both reactant.

                      NH₄Cl          :          NH₃

                          2              :           2

                         0.17          :          0.17

                   Ca(OH)₂         :          NH₃

                        1                :           2

                    0.10              :          2/1×0.10 = 0.2 mol

Less number of moles of ammonia are produced by ammonium chloride it will act as limiting reactant.

Mass of ammonia:

Mass = number of moles × molar mass

Mass = 0.17 mol × 17 g/mol

Mass = 2.89 g

6 0
2 years ago
(30 Points) You need to produce a buffer solution that has a pH of 5.02. You already have a solution that contains 10. mmol (mil
chubhunter [2.5K]
<span>By definition:

pH = pKa + log [acetate]/ [acetic acid]

so

5.02 = 4.74 + log [acetate] / 10 mmole

10mmole = 10/1000 = 0.01 mole

5.02 = 4.74 + log [acetate] / 0.01

5.02 - 4.74 = 0.28 = log [acetate] /0.01


10^0.28 = </span><span>1.90546</span> = [acetate] / 0.01 <span>
[acetate] = 0.019 mole 

= 19 millimoles

</span>
8 0
3 years ago
Read 2 more answers
What is the concentration of a solution containing 1.11 g sugar (sucrose, C12H22O11, MW = 342.3 g/mol, d = 1.587 g/cm3) in 432 m
djyliett [7]

Answer:

0.0075 M

0.0060 m

Explanation:

Our strategy here is to use the definition of molarity and molality to solve this question.

The molarity is the number of moles of solute, sucrose in this case, per liter of solution.

The molality is the number of moles of solute per kilogram of solvent.

So the molarity of the  solution is

M = moles of solute/ V solution

As we see we need the volume of solution since we are only given the volume of solvent, but this will be easy to compute since we have the density of  sucrose.

So determine the moles of sucrose , and the volume of solution:

Moles sucrose = 1.11 g/342.3 g/mol = 3.24 x 10⁻³ M

Volume of solution = Vol Sucrose + Vol glycerine

d = m/V ⇒ Vsucrose = m / d = 1.11 g/ 1.587 g/cm³ = 0.70 cm³

Vol solution = 432 mL + 0.70 mL = 432.7 mL  (1cm³  = 1 mL)

Vol solution = 432.7 mL x 1 L / 1000 mL = 0.4327 L

⇒ M = 3.2 x 10⁻³  mol / 0.4327 L = 0.0075  M

For the molarity what we need is to first calculate the kilograms of glycerine from the given density:

d = m/v ⇒ m = d x v = 1.261 g/cm³ x  432 cm³ = 544.75 g

Converting to Kg:

544.75 g x 1 Kg/ 1000 g = 0.544 kg

Now the molality is

m = mol sucrose/ kg solvent = 3.24 x 10⁻³ mol / 0.544 Kg = 0.0060 m

Note: In the calculation for  volume of solution we could have approximated it to that of just glycerine, but since the density of sucrose was given we calculated the total volume of solution to be more rigorous.

8 0
3 years ago
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