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Oxana [17]
3 years ago
7

The law of segregation increases genetic variation true or false

Chemistry
1 answer:
Karo-lina-s [1.5K]3 years ago
4 0
True is dose incress


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Which Mixture is Homogeneous?
lana66690 [7]
The correct answer is:  [D]:  " milk " .
__________________________________________________________
Choice  [A]:  "soil" is incorrect;  since "soil" is "heterogeneous" {composed of many different "ingredients" .].
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Same with "Choice [B]: "granola" [composed of many different ingredients—clumps of sweetened oats, raisins, coconuts, etc.].
__________________________________________________________
Same with "Choice [C]:  "salad dressing".  {Notice how we usually have to "shake the bottle" ?  Composed of multiple ingredients, (e.g. oil, vinegar, and spices, or oil and other spices, and more ingredientes). 
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Choice: [D]: "milk", as a liquid, is a single, well-mixed, uniform, mixture;  as such, it is "homogeneous".   Note:  "homo-" means "same".
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4 0
2 years ago
Whats the molar mass of tetraphosphorus decoxide
Savatey [412]
<span>283.89 g/mol is the molar mass of tetraphosphorus decoxide</span>
4 0
3 years ago
Why are sodium ions good for making batteries?
Wewaii [24]

The largest advantage of sodium-ion batteries is the high natural abundance of sodium. This could make commercial production of sodium-ion batteries less expensive than lithium-ion batteries. As of 2020, sodium ion batteries have very little share of the battery market.

3 0
2 years ago
What is the mass of 1.71 ✕ 1023 molecules of h2so4?
Anuta_ua [19.1K]
The equation for calculating a mass is as follows:

m=n×M

Molar mass (M) we can determine from Ar that can read in a periodical table, and a number of moles we can calculate from the available date for N:

n(H2SO4)=N/NA

n(H2SO4)= 1.7×10²³ / 6 × 10²³

n(H2SO4)= 0.3 mole

Now we can calculate a mass of H2SO4:

m(H2SO4) = n×M = 0.3 × 98 = 27.8 g



7 0
3 years ago
Use the following half-reactions to construct a voltaic cell:
velikii [3]

<u>Answer:</u> The correct answer is 3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.);E^o_{cell}=+1.53V

<u>Explanation:</u>

We are given:

Cr^{3+}(aq.)+3e^-\rightarrow Cr(s);E^o=-0.73V\\\\Ag^+(aq.)+e^-\rightarrow Ag(s);E^o=+0.80V

The substance having highest positive E^o potential will always get reduced and will undergo reduction reaction. Here, silver will always undergo reduction reaction will get reduced.

Chromium will undergo oxidation reaction and will get oxidized.

The half reactions for the above cell is:

Oxidation half reaction: Cr(s)\rightarrow Cr^{3+}+3e^-;E^o_{Cr^{3+}/Cr}=-0.73V

Reduction half reaction: Ag^{+}+e^-\rightarrow Ag(s);E^o_{Ag^{+}/Ag}=0.80V       ( × 3)

Net equation:  3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.)

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

Putting values in above equation, we get:

E^o_{cell}=0.80-(-0.73)=1.53V

Hence, the correct answer is 3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.);E^o_{cell}=+1.53V

4 0
3 years ago
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