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olya-2409 [2.1K]
3 years ago
5

2. _____ is the distance and direction of an object’s change in position from the starting   a.  Speed b.  Displacement   c.  Ve

locity d.  Rate
Chemistry
1 answer:
Llana [10]3 years ago
3 0
Displacement is the distance and direction of an object's change in position from the starting.
Hence option B is correct.

Hope this helps!
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The standard cell potential of the following galvanic cell is 1.562 V at 298 K. Zn(s) | Zn2+(aq) || Ag+(aq) | Ag(s) What is the
myrzilka [38]

Answer:

E = 1.602v

Explanation:

Use the Nernst Equation => E(non-std) = E⁰(std) – (0.0592/n)logQc …

             Zn⁰(s) => Zn⁺²(aq) + 2 eˉ

2Ag⁺(aq) + 2eˉ=> 2Ag⁰(s)          

_____________________________

Zn⁰(s) + 2Ag⁺(aq) => Zn⁺²(aq) + 2Ag(s)

Given E⁰ = 1.562v

Qc = [Zn⁺²(aq)]/[Ag⁺]² = (1 x 10ˉ³)/(0.150)² = 0.044

E = E⁰ -(0.0592/n)logQc = 1.562v – (0.0592/2)log(0.044) = 1.602v

4 0
3 years ago
Read 2 more answers
35.2 J of heat is
ruslelena [56]

The specific heat : c = 0.306 J/g K

<h3>Further explanation</h3>

Given

Heat = 35.2 J

Mass = 16 g

Temperature difference : 7.2 K =

Required

The specific heat

Solution

Heat can be calculated using the formula:  

Q = mc∆T  

Q = heat, J  

m = mass, g  

c = specific heat, joules / g ° C  

∆T = temperature difference, ° C / K  

Input the value :

c = Q / m.∆T  

c = 35.2 / 16 x 7.2

c = 0.306 J/g K

7 0
3 years ago
Hi i need help. Select all the statements that are true.
motikmotik

Answer:

A, C, E

Explanation:

hope this helps

7 0
3 years ago
Read 2 more answers
How many dL in 1,000 mL
Anna007 [38]

1,000 mL is the same as 10 dL.

6 0
3 years ago
Read 2 more answers
An insulated container contains 0.3 kg of water at 20 degrees C. An alloy with a mass of 0.090 kg and an initial temperature of
Lorico [155]

Answer:

The specific heat of the alloy is 2.324 J/g°C

Explanation:

<u>Step 1:</u> Data given

Mass of water = 0.3 kg = 300 grams

Temperature of water = 20°C

Mass of alloy = 0.090 kg

Initial temperature of alloy = 55 °C

The final temperature = 25°C

The specific heat of water = 4.184 J/g°C

<u>Step 2:</u> Calculate the specific heat of alloy

Qlost = -Qwater

Qmetal = -Qwater

Q = m*c*ΔT

m(alloy) * c(alloy) * ΔT(alloy) = -m(water)*c(water)*ΔT(water)

⇒ mass of alloy = 90 grams

⇒ c(alloy) = the specific heat of alloy = TO BE DETERMINED

⇒ ΔT(alloy) = The change of temperature = T2 - T1 = 25-55 = -30°C

⇒ mass of water = 300 grams

⇒ c(water) = the specific heat of water = 4.184 J/g°C

⇒ ΔT(water) = The change of temperature = T2 - T1 = 25 - 20 = 5 °C

90 * c(alloy) * -30°C = -300 * 4.184 J/g°C * 5°C

c(alloy) = 2.324 J/g°C

The specific heat of the  alloy is 2.324 J/g°C

3 0
3 years ago
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