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3 years ago

Determine what elements are denoted by the following electrons: [Kr] 5s2 4d10 5p3

Chemistry

1 answer:

RoseWind [281]3 years ago

4 0

Answer:

Antimony Sb

Explanation:

ntimony atoms have 51 electrons and the shell structure is 2.8.18.18.5.

The ground state electron configuration of ground state gaseous neutral antimony is [Kr].4d10.5s2.5p3

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Answer:

Sounds travels in transverse waves requires a medium to travel through

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3 years ago

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Excess Ca(OH)2 is shaken with water to produce a saturated solution. The solution is filtered, and a 50.00 mL sample titrated wi

julia-pushkina [17]

<u>Answer:</u> The $K_{sp}$ for calcium hydroxide is $5.324\times 10^{-6}$

<u>Explanation:</u>

To calculate the concentration of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $HCl$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is $Ca(OH)_2$

We are given:

$n_1=1\\M_1=0.0983M\\V_1=11.22mL\\n_2=2\\M_2=?M\\V_2=50mL$

Putting values in above equation, we get:

$1\times 0.0983\times 11.22=2\times M_2\times 50\\\\M_2=0.011M$

The concentration of $Ca(OH)_2$ comes out to be 0.011 M.

The balanced equilibrium reaction for the ionization of calcium hydroxide follows:

$Ca(OH)_2\rightleftharpoons Ca^{2+}+2OH^-$

The expression for solubility constant for this reaction follows:

$K_{sp}=[Ca^{2+}][OH^-]^2$

Putting the values in above equation, we get:

$K_{sp}=(0.011)\times (2\times 0.11)^2$

$K_{sp}=5.324\times 10^{-6}$

Hence, the $K_{sp}$ for calcium hydroxide is $5.324\times 10^{-6}$

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3 years ago

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The melting and boiling points of a substance are independent of

lubasha [3.4K]

Answer is D.

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4 years ago

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Ou have a 5 ml sample of a protein in 0.5 m nacl. you place the protein/salt sample inside dialysis tubing (see fig. 2-14) and p

Oduvanchick [21]

Answer:

Procedure (2)

Explanation:

Assume the dialyses come to equilibrium in the allotted times.

Procedure (1)

If you are dialyzing 5 mL of sample against 4 L of water, the concentration of NaCl will be decreased by a factor of

$\dfrac{5}{4000} = \dfrac{1}{800}$

Procedure (2)

For the first dialysis, the factor is

$\dfrac{5}{1000} = \dfrac{1}{200}$

After a second dialysis, the original concentration of NaCl will be reduced by a factor of

$\dfrac{1}{200} \times \dfrac{1}{200} = \dfrac{1}{40000}$

Procedure (2) is more efficient by a factor of

$\dfrac{40000}{800} = \mathbf{50}$

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3 years ago

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rewona [7]

Answer:

True

Explanation:

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3 years ago