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Elenna [48]
1 year ago
11

Could use some help, thanks. Complete each row of the table below by filling in the missing prefix or missing exponent.

Chemistry
1 answer:
Minchanka [31]1 year ago
3 0

Answer:

if eel u gngg ferr

Explanation:

asdads

You might be interested in
Name one element with similar chemical properties to lithium
marysya [2.9K]

Answer:

Sodium

Explanation:

Its neighbor to the bottom (sodium) has similar properties to lithium. Group trends are very strong for the Group 1 (alkali) and Group 2 (alkaline earths) metals. Group 1 elements behave very similarly to each other.

4 0
3 years ago
The compound tyrosine contains 59.66% C, 6.12% H, 7.73% N, and 26.49% O by mass. What is the empirical formula of tyrosine?
Bezzdna [24]

Answer: C9H11O3N

Explanation:

The percentage by mass of each element is divided by its relative atomic mass. The lowest ratio is now observed. Each ratio is now divided by this lowest ratio and approximated to whole numbers. All these are shown in detail in the image attached.

3 0
3 years ago
A set of headphones produces sounds that cancel out sound waves from the
Basile [38]

Answer:

C. Destructive interference

Explanation:

<em>Destructive interference</em> is when there are two waves <u>with the same frequency</u>, and the peaks of one wave (the highest points) line up with the valleys (the lowest points) of the second one.

Constructive interference would cause the sound to be louder.

Absorption and reflection are interactions that would not take into account the fact that the headphones produce sounds.

4 0
3 years ago
O, li, c which electrons experience the greatest effective nuclear charge? express your answer as a chemical formula.
Nataly_w [17]
<span>The effective nuclear charge of an atom = total electrons - inner electrons For O, ENC = 8 - 2 = 6 For Li, ENC = 3 - 2 = 1 For C, ENC = 6 - 2 = 4 The electrons in O experience the greatest effective nuclear charge and that is why O is smaller than C (which is smaller than Li).</span>
5 0
4 years ago
A solution contains 0.25 M Ni(NO3)2 and 0.25 M Cu(NO3)2. Can the metal ions be separated by slowly adding Na2CO3? Assume that fo
amid [387]

Explanation:

Ksp of NiCO3 = 1.4 x 10^-7

Ksp of CuCO3 = 2.5 x 10^-10

Ionic equations:

NiCO3 --> Ni2+ + CO3^2-

CuCO3 --> Cu2+ + CO3^2-

[Cu2+][CO3^2-]/[Ni2+][CO3^2-]

= (2.5* 10^-10)/(1.4* 10^-7)

= 0.00179.

[Cu2+]/[Ni2+]

= 0.00179

= 0.00179*[Ni2+]

If all of Cu2+ is precipitated before Na2CO3 is added.

= 0.00179 * (0.25)

The amount of Cu2+ not precipitated = 0.000448 M

The percent of Cu2+ precipitated before the NiCO3 precipitates = concentration of Cu2+ unprecipitated/initial concentration of Cu2+ * 100

= 0.000448/0.25 * 100

= 0.18%

Therefore, percentage precipitated = 100 - 0.18

= 99.8%

The two metal ions can be separated by slowly adding Na2CO3. Thus that is the unpptd Cu2+.

8 0
3 years ago
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