Which of your isolated compounds is more soluble in pure water?

What is the volume of 7.50 x 10^24 molecules of NH3 at STP?

Allushta [10]

27.9L

Explanation:

Given parameters:

Number of molecules = 7.5 x 10²⁴molecules

Condition = STP

Unknown:

Volume of gas = ?

Solution;

The volume of gas at STP is expressed below;

Volume of gas = number of moles x 22.4

Number of moles = $\frac{Number of molecules }{6.02 x 10^{23} }$

Now if we can obtain the value of the number of moles from the expression above, we can plug it back into the equation of the volume of gas at STP;

Number of moles of NH₃ = $\frac{[tex]7.5 x 10^{24}$ }{6.02 x 10^{23} }[/tex]

Number of moles of NH₃ = 12.5moles

Volume of gas = 12.5 x 22.4 = 27.9L

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3 0

3 years ago

How many grams of H2O would be made from 7.9 moles of H2? (Round to one decimal place)

Paraphin [41]

Answer:

2.01588 grams

Explanation:

4 0

3 years ago

Which model was proposed as a result of Rutherford's scattering experiment where positive particles did not pass straight throug

dimulka [17.4K]

Answer:

the nuclear model of the atom

6 0

3 years ago

15.0 mL of 0.050 M Ba(NO3)2 M and 100.0 mL of 0.10 M KIO3 are added together in a 250 mL erlenmeyer flask. In this problem, igno

timama [110]

Answer:

Yes, precipitation of barium iodate will occur.

Explanation:

Molarity of barium nitrate solution = 0.050 M

Volume of barium nitrate solution =15.0 mL = 0.0150 L

1 mL = 0.001 L

Moles of barium nitrate = n

$n=0.050 M\times 0.0150L=0.00075 mol$

$Ba(NO_3)_2(aq)\rightarrow Ba^{2+}(aq)+2NO_3^{-}(aq)$

Moles of barium ions: $1\times 0.00075 mol=0.00075 mol$

Molarity of potassium iodate solution = 0.10 M

Volume of potassium iodate solution =100.0 mL = 0.1000 L

1 mL = 0.001 L

Moles of potassium iodate = n'

$n'=0.10 M\times 0.1000 L=0.01 mol$

$KIO_3(aq)\rightarrow K^{+}(aq)+IO_3^{-}(aq)$

Moles of iodate ions = $1\times 0.01 mol=0.01 mol$

After mixing of both solution in 250 mL in erlenmeyer flask

Volume of the final solution = 250 mL = 0.250 L

Concentration of barium ions in 250 mL solution :

$[Ba^{2+}]=\frac{0.00075 mol}{0.250 L}=0.003 M$

Concentration of iodate ions:

$[IO_3^{-}]=\frac{0.01 mol}{0.250 L}=0.04 M$

Solubility product of barium iodate, $K_{sp}=4.01\times 10^{-9}$

Ionic product of the barium iodate in solution : $K_i$

$Ba(IO_3)_2\rightleftahrpoons Ba^{2+}+2IO_3^{-}$

$K_i=[Ba^{2+}][IO_2^{-}]^2$

$K_i=0.003 M\times (0.04 M)^2=4.8\times 10^{-6}$

$K_{sp}$ ( precipitation)

As we can see, the ionic product of the barium iodate is greater than the solubility product of the barium iodate precipitation of barium iodiate will occur in 250 mL of final solution.

5 0

3 years ago

Subtracting atomic number from atomic mass will give the number of

faltersainse [42]

Answer:

:)

Explanation:

If we subtract the atomic number from the atomic mass: atomic mass - atomic number = number of protons + number of neutrons - number of protons. Thus we get the number of neutrons present in an atom when we subtract the atomic number from the atomic mass.

8 0

3 years ago

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