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3 years ago

Which of the following elements has properties different than the rest?

1 answer:

hjlf3 years ago

7 0

The correct answer is: [D]: "S" (sulfur).
___________________________________________________
Note: The other answer choices:

[A]: "K" (potassium) ;
[B]: "Na" (sodium) ;
[C] "Mg" (magnesium) ;
___________________________________________________
are all metals.
___________________________________________________
Answer choice: [D]: "S" (sulfur) ; is NOT a metal; and may exist in gaseous form.
___________________________________________________

You might be interested in

Which type of molecule is shown below?

zubka84 [21]

Answer:

$\boxed{\mathrm{Alkene}}$

Explanation:

The hydrocarbon shown has a double bond. Hydrocarbons with double bonds are known as alkenes.

Cyclic alkanes have cyclic structure.

Alkanes only have single bonds.

Alkynes have triple bonds.

8 0

3 years ago

Read 2 more answers

A chemistry student weighs out 0.027 kg of an unknown solid compound X and adds it to 550. mL of distilled water at 30.° C. Afte

QveST [7]

Answer:

1. Yes

2.The solubility of X is 34.55g/L

Explanation:

Solubility of solute refers to how readily a solute will dissolve in a solvent at a particular temperature. Its the amount of moles or grams required to saturate 1dm $^{3}$ or 1 Litre of water.

From the problem, when the liquid was drained off and amount of X which didn't dissolve was measured, it weighed 0.008kg, this means out of 0.027kg, 0.027-0.008 actually dissolved

= 0.019kg*1000 = 19g.

if 19g is required to saturate 550mL at 30°C,

then $\frac{19*1000}{550}$ will saturate 1L

= 34.545g will saturate 1Litre

The solubility thus is 34.55g/L

5 0

3 years ago

Why do you think that tsunamis occur most commonly in the Pacific Ocean?

zhenek [66]

Tsunamis are most common in the Pacific Ocean Basin. The entire Pacific Ocean is ringed by areas known as subduction zones, where the tectonic plates of the earth are moving relative to one another. Subduction zones are where two or more plates are colliding and one is going underneath another.

5 0

3 years ago

A plot of binding energy per nucleon (Eb/ A) versus the mass number (A) shows that nuclei with a small mass number have a small

juin [17]

Answer:

a) 1.12 MeV / nucleon

b) 5.62 MeV / nucleon

c) 8.80 MeV / nucleon

d) 8.56 MeV / nucleon

we can conclude that the binding energy has a maximum value for nuclei with a mass around 60

Explanation:

Binding energy = ( Δm * 931.5 ) MeV

Binding energy per nucleon = Binding energy in / Number of nucleon

a) ²H = 1 neutron , 1 proton = 2 nucleons

Given that the theoretical mass = 2.0141 u

Actual mass = 1.0078 u + 1.0087 u = 2.0165 u

Δm = 2.0165 u - 2.0141 u = 2.4 * 10^-3 u

∴ Binding energy per nucleon = ( 2.4 * 10^-3 * 931.5 ) MeV / 2 nucleons

= 1.12 MeV / nucleon

b) ⁷Li = 3 protons , 4 neutrons = 7 nucleons

theoretical mass = 7.0160 u

Actual mass = ( 3 * 1.0078 ) + ( 4 * 1.0087 ) = 7.0582 u

Δm = ( 7.0582 u - 7.0160 u ) = 0.0422 u

∴ Binding energy per nucleon = ( 0.0422 * 931.5 ) / 7

= 5.62 MeV / nucleon

C) ⁶²Ni = 28 protons , 34 neutrons = 62 nucleons

Theoretical mass = 61.9283 u

Actual mass = ( 28 * 1.0078 ) u + ( 34 * 1.0087 ) u

= 62.5142 u

Δm = 0.5859 u

∴ Binding energy per nucleon = ( 0.5859 * 931.5 ) / 62

= 8.80 MeV / nucleon

D) ¹¹⁰Cd = 48 protons , 62 neutrons = 110 nucleons

Theoretical mass = 109.9030 u

Actual mass = ( 48 * 1.0078 ) + ( 62 * 1.0087 )

= 110.9138 u

Δm = ( 110.9138 - 109.9030 ) = 1.0108 u

∴ Binding energy per nucleon = ( 1.0108 * 931.5 ) / 110

= 8.56 MeV / nucleon

hence we can conclude that the binding energy has a maximum value for nuclei with a mass around 60

3 0

3 years ago

Calculate the ph of a dilute solution that contains a molar ratio of potassium acetate to acetic acid (pka ???? 4.76) of (a) 2:1

givi [52]

According to Hasselbach-Henderson equation:

$pH=pK_{a}+log\frac{[A^{-}]}{[HA]}$

Here, $[A^{-}]$ is concentration of conjugate base and [HA] is concentration of acid.

In the given problem, conjugate base is $CH_{3}COOK$ and acid is $CH_{3}COOH$ thus, Hasselbach-Henderson equation will be as follows:

$pH=pK_{a}+log\frac{[CH_{3}COOK]}{[CH_{3}COOH]}$ ...... (1)

(a) Ratio of concentration of potassium acetate and acetic acid is 2:1 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=2$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{2}{1}=5.06$

Therefore, pH of solution is 5.06.

(b) Ratio of concentration of potassium acetate and acetic acid is 1:3 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{1}{3}$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{1}{3}=4.28$

Therefore, pH of solution is 4.28.

(c)Ratio of concentration of potassium acetate and acetic acid is 5:1 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{5}{1}$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{5}{1}=5.45$

Therefore, pH of solution is 5.45.

(d) Ratio of concentration of potassium acetate and acetic acid is 1:1 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=1$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{1}{1}=4.76$

Therefore, pH of solution is 4.76.

(e) Ratio of concentration of potassium acetate and acetic acid is 1:10 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{1}{10}$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{1}{10}=3.76$

Therefore, pH of solution is 3.76.

4 0

3 years ago