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katrin2010 [14]
3 years ago
7

Question 9 (1 point)

Chemistry
2 answers:
Bond [772]3 years ago
8 0

Answer:

.

Explanation:

Dependent and Independent variables are variables in mathematical modeling, statistical modeling and experimental sciences. Dependent variables receive this name because, in an experiment, their values are studied under the supposition or demand that they depend, by some law or rule, on the values of other variables.

Kaylis [27]3 years ago
6 0

▪▪▪▪▪▪▪▪▪▪▪▪▪  {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪

The Independent variable here is ~

  • time

And the dependent variable is ~

  • score on the final test

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How many grams of Al will be deposited from molten AlCl3 by a current of 15.0 amp flowing for 24.0 hr ( 1 Faraday = 96500 C)
zimovet [89]

Answer:

121 aprox.

Explanation:

mass/molar mass = Q/CF

where mass=?

molar mass of Al=27

Q=It = 15×24×60×60

C=3

F=96500

7 0
4 years ago
Aluminum:
Scorpion4ik [409]
Whiteish silver
shiny
ductile
conductor
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3 0
3 years ago
A 5.00g piece of metal is heated to 100.0°C, then placed in a beaker containing 20.0 of water at 10.0°C. The temperature of the
Ahat [919]
Answer:

This metal has a specific heat of 0.9845J/ g °C

Explanation:
Step 1: Given data
q = m*ΔT *Cp
⇒with m = mass of the substance
⇒with ΔT = change in temp = final temperature T2 - initial temperature T1
⇒with Cp = specific heat (Cpwater = 4.184J/g °C) (Cpmetam = TO BE DETERMINED)

Step 2: Calculate specific heat
For this situation : we get for q = m*ΔT *Cp
q(lost, metal) = q(gained, water)

- mass of metal(ΔT)(Cpmetal) = mass of water (ΔT) (Cpwater)
-5 * (15-100)(Cpmetal) = 20* (15-10) * (4.184J/g °C =
-5 * (-85)(Cpmetal) = 418.4

Cpmetal = 418.4 / (-5*-85) = 0.9845 J/g °C

This metal has a specific heat of 0.9845J/ g °C
7 0
3 years ago
You have been given an orange liquid and told to describe its properties.
chubhunter [2.5K]

Answer:

The correct answer would be - observing with the help of five senses.

Explanation:

To find and describe the physical properties of the given substance or the solution or liquid students can observe using their five senses. By looking at the liquid one can find its state and color, by smelling students can find the odor of the sample, by touching it one can observe and describe the texture.

Fluidity can also be measure by the touch if the solution is viscous or free-floating. By using a thermometer and using a graduated cylinder one can find the temperature at room temperature and the weight of substance respectively.

5 0
4 years ago
A quantity of 0.0250 mol of a gas initially at 0.050 L and 19.0°C undergoes a constant-temperature expansion against a constant
KiRa [710]

Answer:

V_2=2.995L\\\\W=248.5J

Explanation:

Hello,

In this case, for us to compute the final volume we apply the Boyle's law that analyzes the pressure-volume temperature as an inversely proportional relationship:

P_1V_1=P_2V_2

So we solve for V_2 by firstly computing the initial pressure:

P_1=\frac{nRT}{V_1}=\frac{0.025mol*0.082\frac{atm*L}{mol*K}*(19+273.15)K}{0.050L}  =11.98atm

V_2=\frac{P_1V_1}{P_2}=\frac{11.98atm*0.050L}{0.200atm}\\ \\V_2=2.995L

Finally, we can compute the work by using the following formula:

W=nRTln(\frac{V_2}{V_1} )=0.025mol*8.314\frac{J}{mol*K}*(19.0+273.15)K*ln(\frac{2.995L}{0.050L}) \\\\W=248.5J

Best regards.

4 0
4 years ago
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