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bazaltina [42]
3 years ago
10

Even though Dakota texts his girlfriend "I love you" every night, she only reciprocates sometimes.

Physics
1 answer:
Jet001 [13]3 years ago
7 0
Either she doesn’t love him or just isn’t the most romantic type
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Please help, I am stumped.
ehidna [41]

30 is the correct answer your looking for


6 0
3 years ago
Read 2 more answers
Joanna claims that a large block of ice will cool a substance more than a small block of ice will at the same temperature. To su
Levart [38]

She puts each block of ice in the same 3000 mL beaker, each with 2000 mL of water at room temperature, and measures the temperature before and after adding ice. Therefore, small blocks of ice will have the same temperature.

Joanna puts two blocks of ice (one larger than the other) into separate cups and fills each with water. She compares the final water temperature of the two cups after each block of ice melts.

Put each block of ice in the same 3000 mL beaker, each at room temperature, put 2000 mL of water in it, and measure the temperature before and after adding ice. This way you keep the water at the same temperature in the beginning, then the temperature changes after you add the ice, giving you a better idea of ​​the final temperature reading.

Learn more about Temperature here brainly.com/question/24746268

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3 0
1 year ago
Isabela is an astronomer studying stars A and B. Star A is farther away from Isabela and yet when Isabela observes the stars, th
kakasveta [241]

Answer: star A is more luminous

Explanation:

7 0
3 years ago
Consider the following reaction proceeding at 298.15 K: Cu(s)+2Ag+(aq,0.15 M)⟶Cu2+(aq, 1.14 M)+2Ag(s) If the standard reduction
lutik1710 [3]

Answer : The cell potential for this cell 0.434 V

Solution :

The balanced cell reaction will be,  

Cu(s)+2Ag^{+}(aq)\rightarrow Cu^{2+}(aq)+2Ag(s)

Here copper (Cu) undergoes oxidation by loss of electrons, thus act as anode. silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.

First we have to calculate the standard electrode potential of the cell.

E^o_{[Cu^{2+}/Cu]}=0.34V

E^o_{[Ag^{+}/Ag]}=0.80V

E^o=E^o_{[Ag^{+}/Ag]}-E^o_{[Cu^{2+}/Cu]}

E^o=0.80V-(0.34V)=0.46V

Now we have to calculate the concentration of cell potential for this cell.

Using Nernest equation :

E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Cu^{2+}][Ag]^2}{[Cu][Ag^+]^2}

where,

n = number of electrons in oxidation-reduction reaction = 2

E_{cell} = ?

Now put all the given values in the above equation, we get:

E_{cell}=0.46-\frac{0.0592}{2}\log \frac{(1.14)\times (1)^2}{(1)\times (0.15)}

E_{cell}=0.434V

Therefore, the cell potential for this cell 0.434 V

8 0
3 years ago
A motorcycle skids to a stop on a road.
Svetlanka [38]

Answer:

Friction of the road on the motorcycle in the opposite direction

Explanation:

Khanacademy

5 0
2 years ago
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