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qwelly [4]
4 years ago
14

Consider the ionization constants cyanic acid (HOCN) : Ka= 3.5×10−4; ammonia (NH3): Kb= 1.8×10−5. A solution

Chemistry
1 answer:
elixir [45]4 years ago
6 0
The correct answer for this question is this one:

<span>2. acidic, because the anion hydrolyzes to a greater extent than the cation.
3. basic, because the cation hydrolyzes to a greater extent than the anion.
5. neutral, because the cation hydrolyzes to a greater extent than the anion.
7. neutral, because NH4OCN is a weak base/weak acid salt. 8. basic, because the cation and the anion hydrolyze to the same extent.
10. neutral, because the cation and the anion hydrolyze to the same extent.</span>
Hope this helps answer your question and have a nice day ahead.
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What would be the mass in grams of 8.94 x 10 22 formula units of copper(II) fluoride, CuF 2
Nadya [2.5K]

The mass in grams of 8.94×10²² formula units of CuF₂ is 15.07 g

<h3>Avogadro's hypothesis </h3>

6.02×10²³ formula units = 1 mole of CuF₂

But

1 mole of CuF₂ = 101.5 g

Thus,

6.02×10²³ formula units = 101.5 g of CuF₂

<h3>How to determine the mass of 8.94×10²² formula units of CuF₂</h3>

6.02×10²³ formula units = 101.5 g of CuF₂

Therefore,

8.94×10²² formula units = (8.94×10²² × 101.5 ) / 6.02×10²³

8.94×10²² formula units = 15.07 g of CuF₂

Thus, 8.94×10²² formula units are present in 15.07 g of CuF₂

Learn more about Avogadro's number:

brainly.com/question/26141731

#SPJ1

5 0
2 years ago
Hurry please im being timed. i will give brainliest!!!!
Monica [59]
Answer is C, continents were once all joined together
3 0
3 years ago
Read 2 more answers
Calcium Oxide will react with ammonium chloride to produce ammonia gas, water vapor, and calcium chloride. If only 16.3 g of amm
Vaselesa [24]

Answer:

                     %age Yield  =  22.72 %

Explanation:

                    The balance chemical equation is as follow;

                           2 NH₄Cl + CaO → 2 NH₃ + CaCl₂ + H₂O

Step 1: <u>Calculate Moles of CaO, NH₄Cl and NH₃;</u>

CaO:

        Moles  =  Mass / M.Mass

        Moles  =  112 g / 56.07 g/mol

        Moles  =  1.99 moles of CaO

NH₄Cl:

        Moles  =  Mass / M.Mass

        Moles  =  224 g / 53.49 g/mol

        Moles  =  4.18 moles of NH₄Cl

NH₃:

        Moles  =  Mass / M.Mass

        Moles  =  16.3 g / 17.03 g/mol

        Moles  =  0.95 moles of NH₃

Step 2: <u>Calculate Limiting reagent as:</u>

According to equation.

                   2 moles of NH₄Cl reacts with  =  1 mole of CaO

So,

              4.18 moles of NH₄Cl will react with  =  X moles of CaO

Solving for X,

                     X =  4.18 mol × 1 mol / 2 mol

                     X  =  2.09 mol of CaO

This means that none of the given reagent is limiting reagent. They both are almost equal in number of moles.

Step 3: <u>Calculate Theoretical Yield of NH₃ as;</u>

According to equation.

                   2 moles of NH₄Cl produced  =  2 moles of NH₃

So,

              4.18 moles of NH₄Cl will produce  =  X moles of NH₃

Solving for X,

                     X =  4.18 mol × 2 mol / 2 mol

                     X  =  4.18 mol of NH₃

Step 4: <u>Calculate Percentage Yield as;</u>

             %age Yield  =  Actual Yield / Theoretical Yield × 100

             %age Yield  =  0.95 mol / 4.18 × 100

             %age Yield  =  22.72 %

5 0
3 years ago
Suppose 13L of gas is known to contain .965 Mol. If the amount of gas is Increased to 3.20 mol, what new volume will result? (Av
Ostrovityanka [42]

Answer &Explanation:

From Avogadro's lawa equal volume of gas contain equal number of moles

V=N

Hence

13L=965

XL=3.2mol

Hint:as the question state is increase to iteans final mol was 3.2 but if it could state by it would mean initial moles plus adde moles ie 3.2mol)

Cross multiplication

The new volume will be

=(3.2mol×13L÷965mol)

=0.043L

8 0
3 years ago
A wooden block has the following measured dimensions: height 1.25 cm; width 2.5 cm and length of 15.956 cm. Calculate its volume
never [62]

Answer:

V=50mL

Explanation:

Hello,

In this case, by knowing that the volume of an object is computed by considering its dimensions, width, length and height, for the given measurements, we obtain:

V=W*H*L=1.25cm*2.5cm*15.956cm\\\\V=49.86cm^3

Moreover, since one cubic centimetre equals one millilitre, the required volume is:

V=49.86cm^3*\frac{1mL}{1cm^3}\\ \\V=49.86mL

Finally, since 2.5 cm has the fewest significant figures (2), the proper result is:

V=50mL

Regards.

3 0
4 years ago
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