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DiKsa [7]
3 years ago
9

Which statement describes a controlled experiment? A.) It has one group in which the controlled variable is tested, and another

in which the manipulated variable is tested. B.) It has one group in which the responding variable is tested, and another in which the responding variable is not tested. C.) It has one group in which the manipulated variable is tested, and another in which the responding variable is tested. D.) It has one group in which the manipulated variable is tested, and another in which the manipulated variable is not tested.
Chemistry
1 answer:
scoundrel [369]3 years ago
6 0

Answer:C.) It has one group in which the manipulated variable is tested, and another in which the responding variable is tested

Explanation: hope this helps

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D) force,cause, displacement

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What is the mass of one mole of Cl2
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70.906 g/mol

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Which of the following is not a physical property of a liquid? indefinite volume indefinite shape definite mass
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The heat of fusion for water is 80. cal/g. How many calories of heat are needed to melt a 35 g ice cube that has a temperature o
HACTEHA [7]

Answer: 2800 calories

Explanation:

Latent heat of fusion is the amount of heat required to convert 1 mole of solid to liquid at atmospheric pressure.

Amount of heat required to fuse 1 gram of water = 80 cal

Mass of ice given = 35 gram

Heat required to fuse 1 g of ice at 0^0C = 80 cal

Thus Heat required to fuse 35 g of ice =\frac{80}{1}\times 35=2800cal

Thus 2800 calories of energy is required to melt 35 g ice cube

6 0
3 years ago
Consider a voltaic cell where the anode half-reaction is Zn(s) → Zn2+(aq) + 2 e− and the cathode half-reaction is Sn2+(aq) + 2 e
notsponge [240]

<u>Answer:</u> The concentration of Sn^{2+} in the cell is 9.0\times 10^{-3}M

<u>Explanation:</u>

We are given:

<u>Oxidation half reaction:</u>  Zn(s)\rightarrow Zn^{2+}(aq.)+2e^-   E^o_{Zn^{2+}/Zn}=-0.76V

<u>Reduction half reaction:</u>  Sn^{2+}(aq.)+2e^-\rightarrow Sn(s)   E^o_{Sn^{2+}/Sn}=-0.136V

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

Here, tin will undergo reduction reaction and will get reduced.

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}=-0.136-(-0.76)=0.624V

To calculate the EMF of the cell, we use the Nernst equation, which is:

E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Mn^{2+}]}{[Cu^{2+}]}

where,

E_{cell} = electrode potential of the cell = 0.660 V

E^o_{cell} = standard electrode potential of the cell = +0.624 V

n = number of electrons exchanged = 2

[Zn^{2+}]=2.5\times 10^{-3}M

[Sn^{2+}] = ?

Putting values in above equation, we get:

0.660=0.624-\frac{0.059}{2}\times \log(\frac{2.5\times 10^{-3}}{[Sn^{2+}})

[Sn^{2+}]=9.0\times 10^{-3}M

Hence, the concentration of Sn^{2+} ions is 9.0\times 10^{-3}M

3 0
3 years ago
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