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sergeinik [125]
3 years ago
8

Topic 2 - Evidence Ticket

Chemistry
1 answer:
choli [55]3 years ago
5 0
Fatmagul neushe mode
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You are given an unknown type of clothing dye. how could you use the procedures in this lab to see if this dye is a mixture?
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One separation technique to be used is the paper chromatography. This works by separating the components of the mixture through the difference of their concentrations. There is a stationary phase and the mobile phase, which flows through the stationary phase. The components travel at different rates and is usually signified by the colors. If more than one color would appear, that means that the dye is a mixture.
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List four physical properties of toothpaste
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Abrasiveness is the most common property found in almost every toothpaste. The abrasiveness of a toothpaste is responsible for whitening action and removal of debris stuck on tooth enamel. The same abrasiveness can worsen the teeth sensitivity by damaging the tooth enamel. Relative Dentin Abrasivity (RDA) is used to measure a toothpaste’s abrasiveness.

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4 years ago
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7 0
3 years ago
A volume of 60.0 mL of a 0.120 M HNO3 solution is titrated with 0.840 M KOH. Calculate the volume of KOH required to reach the e
german

Answer: 8.57 ml of KOH is required to reach the equivalence point.

Explanation:

To calculate the volume of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is HNO_3

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is KOH.

We are given:

n_1=1\\M_1=0.120M\\V_1=60.0mL\\n_2=1\\M_2=0.840M\\V_2=?

Putting values in above equation, we get:

1\times 0.120\times 60.0=1\times 0.840\times V_2\\\\V_2=8.57mL

Thus 8.57 ml of KOH is required to reach the equivalence point.

8 0
4 years ago
A 100 W light bulb is placed in a cylinder equipped with a moveable piston. The light bulb is turned on for 2.0×10−2 hour, and t
drek231 [11]

Answer:

(a) ΔU = 7.2x10²

(b) W = -5.1x10²

(c) q = 5.2x10²

Explanation:

From the definition of power (p), we have:

p = \frac {\Delta W}{\Delta t} = \frac {\Delta U}{\Delta t} (1)

<em>where, p: is power (J/s = W (watt)) W: is work = ΔU (J) and t: is time (s) </em>  

(a) We can calculate the energy (ΔU) using equation (1):

\Delta U = p \cdot \Delta t = 100 \frac{J}{s} \cdot 2.0\cdot 10^{-2} h \cdot \frac{3600s}{1h} = 7.2 \cdot 10^{2} J  

(b) The work is related to pressure and volume by:

\Delta W = -p \Delta V

<em>where p: pressure and ΔV: change in volume = V final - V initial      </em>

\Delta W = - p \cdot (V_{fin} - V_{ini}) = - 1.0 atm (5.88L - 0.85L) = - 5.03 L \cdot atm \cdot \frac{101.33J}{1 L\cdot atm} = -5.1 \cdot 10^{2} J

(c) By the definition of Energy, we can calculate q:

\Delta U = \Delta W + \Delta q

<em>where Δq: is the heat transfer </em>

\Delta q = \Delta U - \Delta W = 7.2 J - (-5.1 \cdot 10^{2} J) = 5.2 \cdot 10^{2} J    

I hope it helps you!  

6 0
4 years ago
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